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NCERT solutions for class 10 maths Chapter 4 Quadratic equations 3 Solution by completing a square

Solving a quadratic equation by Completing a square

NCERT solutions class 10 maths ch 4 pt 3

This is the third of the four part series on Quadratic equations. It complements the section with Exercise 3 of Chapter 4 of NCERT Class 10 Maths, in short, Ex 4.3 class 10. In this session, we will cover,

  1. How to solve a quadratic equation by completing a square, and,
  2. The quadratic equation formula by Sreedhar Acharya.

Discussions are divided into three sections:

  • First: concepts and methods explained in an easy to understand way,
  • Second: example problems used for establishing the concepts and methods clearly and,
  • Third: the solutions to the problems in the 3rd exercise of NCERT Chapter 4.

To Skip the Concept Explanations and Go straight to the Solutions, click here.

How the method of solving a quadratic equation by completing a square works algebraically

This is a more systematic method with no guessing involved, but is more cumbersome with more calculations involved.

Let us first explain how the method works algebraically using a problem example.

Problem example 1

Find the roots of the following quadratic equation by completing a square.

$x^2-2x-35=0$.$\qquad....................(1)$

Our goal is,

To convert this quadratic equation as a square of linear polynomial in $x$ on the LHS equated to a real constant on the RHS.

This is the crux of this method, and forms the primary objective.

In a general form, the goal expression is,

$x^2-2x-35=0$

Or, $(x+p)^2=q^2$,

where $p$ and $q$ are real constants that may take negative or positive values.

In such a case then we would easily be able to take the square root of both sides of the equation getting,

$x+p=\pm q$,

Or, $x=-p\pm q$.

In essence this is the method algebraically. The only question remains is—how to convert the quadratic equation to the desired form!

We have to convert our following quadratic polynomial first to a whole square of a linear expression in $x$,

$x^2-2x-35$.

Step 1: Consume the first two terms of the quadratic polynomial to form the first two terms $x^2 +2xp$ of the expanded three term form of a square,

$(x+p)^2=x^2+2px+p^2$.

Thus, the first two terms in this problem become,

$x^2+2(-1)(x)$, where $p=-1$.

Step 2: Add the third term of $p^2=(-1)^2=1$ to the first two terms to complete the three term expansion of $(x+p)^2$. The changed three term polynomial becomes,

$x^2+2(-1)(x)+1$.

Step 3: To compensate this addition of the new term, subtract the same $p^2$ from the polynomial and keep the original equation unchanged,

$x^2+2(-1)(x)+1-1-35=0$.

Step 4. Simplify and express the first three terms as a square of a linear polynomial in $x$ equal to $q^2=6^2$,

$(x-1)^2=36=6^2$

So you have converted the quadratic equation in the desired form.

Now take the square root of both sides of the equation,

$x-1=\pm 6$

The roots will be given by,

$x-1=+6$ or, $x=1+6=7$, and,

$x-1=-6$ or, $x=1-6=-5$.

We'll present now the method of finding roots of a quadratic equation by completion of the square with the help of geometric visualization using a second problem example.

Step-wise systematic representation of the method of finding roots of a quadratic equation by completion of the square—Geometric visualization

Problem example 2.

Find the roots of the following quadratic equation by the method of completion of the square.

$x^2+4x-5=0$.

The following shows the geometric visualization of conversion of the first two terms in $x$ as a three term expansion of a square of linear expression in $x$.

ncert-class-10-ch-4-pt-3-completing-square-geometry

The steps are:

Step 1: The middle term is expanded as $2.(p).x=2.2.x$ where $p=2$ to convert it to the form of the middle term $2px$ of the expansion $x^2+2px+p^2$.

Step 2: Add a new term $p^2=2^2$ to the first two terms to complete the three term expansion of $(x+p)^2$.

Step 3: To compensate this addition of the new term, subtract the same $p^2$ from the polynomial and keep the original equation unchanged.

Step 4: Express the square of the linear expression in $x$ in the LHS and take the rest to the RHS.

The result of these steps is,

$x^2+4x-5=x^2+2.2.x+2^2 - 2^2-5=0$,

Or, $(x+2)^2=9$.

The square is completed and it is time to solve for the roots.

Take square root of the two sides of the equation. That was the main objective for completing the square. Result is,

$x+2=\pm 3$.

The roots are given by, $x=-2+3=1$, and,

by $x=-2-3=-5$.

Let's solve another problem example, but this time with the coefficient of $x$ not 1 or a perfect square.

Problem example 3.

Find the roots of the following quadratic equation by the method of completing the square.

$5x^2-6x-2=0$.

To make the coefficient of $x$ a perfect square, just multiply the equation by 5. Result is,

$25x^2-30x-10=0$.

You could have divided the equation by 5 also, but that would have forced you to deal with awkward fraction computation.

Change the transformed equation to introduce the third term of a three term expansion of the square of a linear polynomial and compensate this introduction of the new term. Result is,

$(5x)^2-2(5x)(3)+3^2-3^2-10=0$,

Or, $(5x-3)^2=19$.

Take the square root of both sides of the equation,

$5x-3=\pm \sqrt{19}$.

The two roots are,

$x=\displaystyle\frac{3+\sqrt{19}}{5}$, and,

$x=\displaystyle\frac{3-\sqrt{19}}{5}$.

Though the roots involve surd, still the problems are not difficult to solve. Now we'll solve a few quadratic equations of different types to show you how this new method works even on such forms of quadratic equations.

Finding the roots of a quadratic equation that has surd coefficients by completion of the square method

Problem example 4.

Find the roots of the following quadratic equation by the method of completing the square.

$\sqrt{2}x^2+7x+5\sqrt{2}=0$

Knowing that surd cannot be avoided in the roots, we decide to multiply the equation by $\sqrt{2}$ to convert the coefficient of the first term as $(\sqrt{2})^2$,

$(\sqrt{2}x)^2+2(\sqrt{2}x)\left(\displaystyle\frac{7}{2}\right)+\left(\displaystyle\frac{7}{2}\right)^2-\left(\displaystyle\frac{7}{2}\right)^2+10=0$

Or, $\left(\sqrt{2}x+\displaystyle\frac{7}{2}\right)^2=\displaystyle\frac{9}{4}$.

Take square root on both sides of the equation,

$\sqrt{2}x+\displaystyle\frac{7}{2}=\pm \displaystyle\frac{3}{2}$.

The roots of the equation are given by,

$\sqrt{2}x=-\displaystyle\frac{7}{2}+\displaystyle\frac{3}{2}=-2$, or, $x=-\sqrt{2}$, and,

$\sqrt{2}x=-\displaystyle\frac{7}{2}-\displaystyle\frac{3}{2}=-5$, or, $x=-\displaystyle\frac{5}{\sqrt{2}}$.

Finding the roots of a more difficult quadratic equation by completing the square method

Problem example 5.

Find the roots of the following quadratic equation by completing the square method.

$14x^2+23x-30=0$.

To convert the coefficient of the first term in $x^2$, multiply the equation by 14. Result is,

$14^2x^2+14(23x)-14(30)=0$,

Or $(14x)^2+2.(14x).\displaystyle\frac{23}{2}+\left(\displaystyle\frac{23}{2}\right)^2-\left(\displaystyle\frac{23}{2}\right)^2-420=0$,

Or, $\left(14x+\displaystyle\frac{23}{2}\right)^2=\displaystyle\frac{529+1680}{4}=\frac{2209}{4}$.

Take the square root of both sides of the equation. Result is,

$14x+\displaystyle\frac{23}{2}=\pm \displaystyle\frac{47}{2}$.

The roots of the equation are given by,

$14x=-\displaystyle\frac{23}{2}+\displaystyle\frac{47}{2}=12$, Or, $x=\displaystyle\frac{6}{7}$, and,

$14x=-\displaystyle\frac{23}{2}-\displaystyle\frac{47}{2}=-35$, Or, $x=-\displaystyle\frac{5}{2}$.

You can appreciate how much calculation had to be done in this method.

But the importance of the method lies in the fact that it leads us to the standard formula for finding roots of a quadratic equation ascribed to Sreedhar Acharya.

The method of completing the square leading to the formula for finding square roots of a general quadratic equation

General Problem.

Find the roots of the general form of quadratic equation,

$ax^2+bx+c=0$.

To make the coefficient of $x^2$ a perfect square multiply the equation by $a$. Result is,

$a^2x^2+2.(ax).\left(\displaystyle\frac{b}{2}\right)+\left(\displaystyle\frac{b}{2}\right)^2-\left(\displaystyle\frac{b}{2}\right)^2+ac=0$,

Or, $\left(ax+\displaystyle\frac{b}{2}\right)^2=-ac+\displaystyle\frac{b^2}{4}=\displaystyle\frac{b^2-4ac}{4}$.

Taking square root of both sides of the equation,

$ax+\displaystyle\frac{b}{2}=\pm \displaystyle\frac{\sqrt{b^2-4ac}}{2}$,

Or, $x=\displaystyle\frac{-b \pm \sqrt{b^2-4ac}}{2a}$.

These are the two roots of the general quadratic equation,

$ax^2+bx+c=0$.

This unique formula for roots of a general quadratic equation is ascribed to Sreedhar Acharya and gives you the third method to find roots of a quadratic equation directly from a formula.

We'll now examine the results of applying this formula for finding the roots for the problem examples 1, 2, 3 and 4.

Problem example 1 had the quadratic equation,

$x^2-2x-35=0$.

In this case, $a=1$, $b=-2$ and $c=-35$.

Substituting the values you get the roots as,

$x=\displaystyle\frac{2 \pm\sqrt{4+140}}{2}=\frac{2 \pm \sqrt{144}}{2}=\displaystyle\frac{2 \pm 12}{2}$

The two roots are, $x=7$, or, $x=-5$.

Problem example 2 had the quadratic equation,

$x^2+4x-5=0$.

In this case, $a=1$, $b=4$ and $c=-5$,

he roots are given by the formula as,

$x=\displaystyle\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$=\displaystyle\frac{-4 \pm \sqrt{16+20}}{2}$

$=\displaystyle\frac{-4 \pm \sqrt{36}}{2}$

$=\displaystyle\frac{-4 \pm 6}{2}$

So, $x=1$ or $x=-5$.

Problem example 3 had the quadratic equation,

$5x^2-6x-2=0$.

In this case,

$a=5$, $b=-6$ and $c=-2$.

The roots are given by the formula as,

$x=\displaystyle\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$=\displaystyle\frac{6 \pm \sqrt{36+40}}{10}$

$=\displaystyle\frac{6 \pm \sqrt{76}}{10}$

$=\displaystyle\frac{6 \pm 2\sqrt{19}}{10}$

$=\displaystyle\frac{3 \pm \sqrt{19}}{5}$.

Problem example 4 had the quadratic equation,

$\sqrt{2}x^2+7x+5\sqrt{2}=0$.

In this case, $x=\sqrt{2}$, $b=7$ and $c=5\sqrt{2}$.

Substituting in the formula you get the roots as,

$x=\displaystyle\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$=\displaystyle\frac{-7 \pm \sqrt{49-40}}{2\sqrt{2}}$

$=\displaystyle\frac{-7 \pm 3}{2\sqrt{2}}$.

The roots are,

$x=-\sqrt{2}$, and,

$x=-\displaystyle\frac{5}{\sqrt{2}}$.

Checking whether the roots of a quadratic equation are real or not

This is an important question that can be answered easily and quickly by using the quadratic equation roots formula.

Problem example 6.

Check whether the roots of the following quadratic equation are real or not.

$4x^2+3x+5=0$.

In this case, $a=4$, $b=3$ and $c=5$.

We need to just examine whether $(b^2-4ac)$ is negative or positive. If it is negative, in the formula for the roots,

$x=\displaystyle\frac{-b \pm \sqrt{b^2-4ac}}{2a}$,

square root of a negative number is invalid, and both roots become imaginary. We would say in this case, the equation does not have any real roots.

Let us see this result in our problem,

$b^2-4ac=3^2-4.4.5=9-80=-71$.

The quadratic equation does not have any real roots.

It is time to solve the exercise problems.

Solved problems: NCERT Class 10 maths chapter 4 exercise 3: Solution by completing the square

Problem 1.i.

Find the roots of the following quadratic equation, the they exist, by the method of completing the square.

$2x^2-7x+3=0$.

Solution to problem 1.i.

Multiplying the equation by 2 makes the coefficient of $x^2$ a perfect square. The equation becomes,

$4x^2-14x+6=0$,

Or, $(2x)^2-2.(2x)\left(\displaystyle\frac{7}{2}\right)+\left(\displaystyle\frac{7}{2}\right)^2-\left(\displaystyle\frac{7}{2}\right)^2+6=0$,

Or, $\left(2x-\displaystyle\frac{7}{2}\right)^2-\displaystyle\frac{49}{4}+6=0$,

Or, $\left(2x-\displaystyle\frac{7}{2}\right)^2=\displaystyle\frac{25}{4}=\left(\displaystyle\frac{5}{2}\right)^2$.

Take the square root of both sides of the equation,

$2x-\displaystyle\frac{7}{2}=\pm \displaystyle\frac{5}{2}$.

The two roots are then given by,

$x=\displaystyle\frac{7+5}{4}=3$, and

$x=\displaystyle\frac{7-5}{4}=\displaystyle\frac{1}{2}$.

Answer: $3$, $\displaystyle\frac{1}{2}$.

Problem 1.ii.

Find the roots of the following quadratic equation, the they exist, by the method of completing the square.

$2x^2+x-4=0$.

Solution to problem 1.ii.

Multiplying the equation by 2 makes the coefficient of $x^2$ a perfect square. The equation becomes,

$4x^2+2x-8=0$,

Or, $(2x)^2+2.(2x)\left(\displaystyle\frac{1}{2}\right)+\left(\displaystyle\frac{1}{2}\right)^2-\left(\displaystyle\frac{1}{2}\right)^2-8=0$,

Or, $\left(2x+\displaystyle\frac{1}{2}\right)^2=\displaystyle\frac{33}{4}$.

Taking the square root of both sides of the equation,

$2x+\displaystyle\frac{1}{2}=\pm \displaystyle\frac{\sqrt{33}}{2}$.

The roots are given by,

$x=\displaystyle\frac{-1+\sqrt{33}}{4}$, and,

$x=\displaystyle\frac{-1-\sqrt{33}}{4}$.

Answer: $\displaystyle\frac{-1+\sqrt{33}}{4}$, and, $\displaystyle\frac{-1-\sqrt{33}}{4}$.

Problem 1.iii.

Find the roots of the following quadratic equation, the they exist, by the method of completing the square.

$4x^2+4\sqrt{3}x+3=0$.

Solution to problem 1.iii.

The given equation is expressed in the form of,

$(2x)^2+2.(2x).(\sqrt{3})+(\sqrt{3})^2=0$,

Or, $(2x+\sqrt{3})^2=0$.

Taking square root,

$2x+\sqrt{3}=0$,

Or, $x=-\displaystyle\frac{\sqrt{3}}{2}$.

Answer: Both roots are of equal value, $-\displaystyle\frac{\sqrt{3}}{2}$.

Problem 1.iv.

Find the roots of the following quadratic equation, the they exist, by the method of completing the square.

$2x^2+x+4=0$.

Solution to problem 1.iv.

Multiplying the equation by 2 makes the coefficient of $x^2$ a perfect square. The equation becomes,

$4x^2+2x+8=0$,

Or, $(2x)^2 +2.(2x)\left(\displaystyle\frac{1}{2}\right)+\left(\displaystyle\frac{1}{2}\right)^2-\left(\displaystyle\frac{1}{2}\right)^2+8=0$,

Or, $\left(2x+\displaystyle\frac{1}{2}\right)^2=-\displaystyle\frac{31}{4}$.

Taking square root of both sides of the equation,

$2x+\displaystyle\frac{1}{2}=\displaystyle\frac{\sqrt{-31}}{2}$.

In the RHS the square root of a negative number makes both the roots imaginary.

The equation does not have any real roots.

Answer: No real roots.

Problem 2.i.

Find the roots of the following quadratic equation by applying the quadratic root formula.

$2x^2-7x+3=0$.

Solution to problem 2.i.

The quadratic equation root formula is,

$x=\displaystyle\frac{-b \pm \sqrt{b^2-4ac}}{2a}$,

where, $a=2$, $b=-7$ and $c=3$ in this problem.

The roots are then given by,

$x=\displaystyle\frac{7 \pm \sqrt{49-24}}{4}$

$=\displaystyle\frac{7 \pm \sqrt{25}}{4}$

$=\displaystyle\frac{7 \pm 5}{4}$,

Or, $x=3$, and,

$x=\displaystyle\frac{1}{2}$.

Answer: $3$, $\displaystyle\frac{1}{2}$.

Problem 2.ii.

Find the roots of the following quadratic equation by applying the quadratic root formula.

$2x^2+x-4=0$.

Solution to problem 2.ii.

The quadratic equation root formula is,

$x=\displaystyle\frac{-b \pm \sqrt{b^2-4ac}}{2a}$,

where, $a=2$, $b=1$ and $c=-4$ in this problem.

So the roots are then given by,

$x=\displaystyle\frac{-1 \pm \sqrt{1+32}}{4}$.

That is, the two roots are,

$x=\displaystyle\frac{-1 + \sqrt{33}}{4}$, and,

$x=\displaystyle\frac{-1 - \sqrt{33}}{4}$.

Answer: $\displaystyle\frac{-1 + \sqrt{33}}{4}$, $\displaystyle\frac{-1 - \sqrt{33}}{4}$.

Problem 2.iii.

Find the roots of the following quadratic equation by applying the quadratic root formula.

$4x^2+4\sqrt{3}x+3=0$.

Solution to problem 2.iii.

The quadratic equation root formula is,

$x=\displaystyle\frac{-b \pm \sqrt{b^2-4ac}}{2a}$,

where, $a=4$, $b=4\sqrt{3}$ and $c=3$ in this problem.

So the roots are then given by,

$x=\displaystyle\frac{-4\sqrt{3} \pm \sqrt{48-48}}{8}$

Or, $x=\displaystyle\frac{-4\sqrt{3}}{8}=-\frac{\sqrt{3}}{2}$.

Both the roots are of equal value.

Answer: The two roots are of same value, $-\displaystyle\frac{\sqrt{3}}{2}$.

Problem 2.iv.

Find the roots of the following quadratic equation by applying the quadratic root formula.

$2x^2+x+4=0$.

Solution to problem 2.iv.

The quadratic equation root formula is,

$x=\displaystyle\frac{-b \pm \sqrt{b^2-4ac}}{2a}$,

where, $a=2$, $b=1$ and $c=4$ in this problem.

So the roots are then given by,

$x=\displaystyle\frac{-1 \pm \sqrt{1-32}}{4}$,

Or, $x=\displaystyle\frac{-1 \pm \sqrt{-31}}{4}$.

With square root of a negative number involved, the equation does not have any real roots.

Answer: No real roots.

Problem 3.i.

Find the roots of the following equation,

$x-\displaystyle\frac{1}{x}=3$, $x\neq 0$.

Solution to problem 3.i.

Simplifying the equation to a quadratic equation form we get,

$x^2-3x-1=0$, multiplying the equation by $x$ and simplifying.

We'll apply completing the square method to find its roots.

The equation is transformed in the expanded square of linear polynomial form as,

$x^2-2.(x)\left(\displaystyle\frac{3}{2}\right)+\left(\displaystyle\frac{3}{2}\right)^2-\left(\displaystyle\frac{3}{2}\right)^2-1=0$,

Or, $\left(x-\displaystyle\frac{3}{2}\right)^2=\displaystyle\frac{13}{4}$.

Taking the square root on both sides of the equation, the roots are formed as,

$x=\displaystyle\frac{3}{2} \pm \displaystyle\frac{\sqrt{13}}{2}$.

Answer: $\displaystyle\frac{3+\sqrt{13}}{2}$, $\displaystyle\frac{3-\sqrt{13}}{2}$.

Problem 3.ii.

Find the roots of the following equation,

$\displaystyle\frac{1}{x+4}-\displaystyle\frac{1}{x-7}=\displaystyle\frac{11}{30}$, $x\neq 4\text{, }7$.

Solution to problem 3.ii.

Simplifying the equation to a quadratic equation form we get,

$30(x-7-x-4)=11(x+4)(x-7)$, cross-multiplying,

Or, $-30=x^2-3x-28$, canceling factor $11$,

Or, $x^2-3x+2=0$.

The LHS of the equation is straightway factorized to,

$x^2-3x+2=(x-1)(x-2)=0$.

The roots are, $1$ and $2$.

Answer: $1$, $2$.

Problem 4.

The sum of reciprocals of Rehman's ages (in years) 3 years ago and 5 years from now is $\displaystyle\frac{1}{3}$. Find his present age.

Solution to problem 4.

Let Rehman's present age in years be $x$.

3 years ago his age was, $(x-3)$ and 5 years from now his age will be $(x+5)$.

Sum of reciprocals of these two ages is,

$\displaystyle\frac{1}{x-3}+\displaystyle\frac{1}{x+5}=\displaystyle\frac{1}{3}$.

Combining the fraction terms in the LHS and cross-multiplying,

$3(x+5+x-3)=(x-3)(x+5)$,

Or, $6x+6=x^2+2x-15$,

Or, $x^2-4x-21=0$.

The LHS of the equation is easily factorized by observation as,

$(x-7)(x+3)=0$.

The two roots are, $x=7$ and $x=-3$, an invalid value.

Answer: Rehman's present age is 7 years.

Verify yourself.

Problem 5.

In a class test the sum of Shefali's marks in Mathematics and English is 30. Had she got 2 more marks in Mathematics and 3 marks less in English, the product of the two marks would have been 210. Find her marks in the two subjects.

Solution to problem 5.

Let Shefali's marks in Mathematics be $x$. So her marks in English is,

$(30-x)$.

With 2 marks more her marks in Mathematics would have been, $(x+2)$, and with 3 marks less her marks in English would have been,

$(30-x-3)=(27-x)$.

Product of these two marks would be,

$(x+2)(27-x)=210$,

Or, $-x^2+25x+54=210$,

Or, $x^2-25x+156=0$.

The LHS of the equation is easily factorized into,

$(x-12)(x-13)=0$.

The two roots are $x=12$ and $x=13$.

Both are valid values of Shefali's marks in Mathematics.

As sum of these two values is 30,

If her marks in Mathematics is 12, marks in English will be 18, and,

If her marks in Mathematics is 13, marks in English would be 17.

Answer: Mathematics 12, English 18 or, Mathematics 13, English 17.

Verify yourself.

Problem 6.

The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the length of the sides of the field.

Solution to problem 6.

Let length of the shorter side of the field be $x$ metres.

So, Length of the diagonal is, $(x+60)$ metres, and,

length of the longer side is, $(x+30)$ metres.

The following diagram shows the rectangle for visual understanding.

ncert-class-10-ch-4-pt-3-q6

As a diagonal divides a rectangle into two right-angled congruent triangles with corresponding sides equal, applying Pythagoras theorem on the three sides of any of the right-angled triangles,

$x^2+(x+30)^2=(x+60)^2$,

Or, $x^2+x^2+60x+900=x^2+120x+3600$,

Or, $x^2-60x-2700=0$.

Being a simple problem, we'll apply completing the square method,

$x^2-2.(x).(30)+30^2-30^2-2700=0$,

Or, $(x-30)^2=3600=60^2$.

Take the square root of both the sides of the equation.

$x-30=\pm 60$.

The two roots are given by,

$x=30+60=90$ metres, and,

$x=30-60=-30$ metres, an invalid value. 

Shorter side is 90 metres long and longer side is, $90+30=120$ metres long.

Answer: 90 metres, 120 metres.

Verify yourself.

Problem 7.

The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the numbers.

Solution to problem 7.

Let the larger number is $x$ and the smaller number is $y$.

By first condition,

$x^2-y^2=180$.

And by second condition,

$y^2=8x$.

Substituting for $y^2$ in the first equation, you get,

$x^2-8x=180$,

Or, $x^2-8x-180=0$.

It is straightforward to factorize the quadratic polynomial in the LHS as,

$(x-18)(x+10)=0$.

The two roots are, $x=18$ and $=-10$, an invalid value, as with this value $y^2=-80$, resulting in an invalid square root of a negative number.

With $x=18$, value of $y$ is given by,

$y^2=8x=8\times{18}=4\times{36}=2^2\times{6^2}=12^2$.

Or, $y=\pm 12$.

Answer: $18$, $12$ or $18$, $-12$.

Verify yourself.

Problem 8.

A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less to travel the same distance. Find the speed of the train.

Solution to problem 8.

Let the speed of the train be $x$ km/hr and the time taken to cover 360 km at this speed be $T$ hr.

The speed time distance relationship in this first case is,

$T=\displaystyle\frac{360}{x}$$\qquad \qquad \qquad ......(1)$

When the speed increases by 5 km/hr, it becomes, $(x+5)$ km/hr and at this speed the train takes 1 hour less to cover the same distance, that is, it takes $(T-1)$ hr.

The speed time distance relationship in this second case is then,

$T-1=\displaystyle\frac{360}{x+5}$$\qquad \qquad ......(2)$.

Subtract equation (2) from equation (1). Result is,

$1=\displaystyle\frac{360}{x}-\displaystyle\frac{360}{x+5}$,

Or, $x^2+5x=360(5)=1800$,

Or, $x^2+5x-1800=0$.

Though the calculations would be a bit heavy we will show you the solution using completing the square method.

Expressing the above quadratic equation in expanded form of square of a linear polynomial in $x$, you get,

$x^2+2.(x)\left(\displaystyle\frac{5}{2}\right)+\left(\displaystyle\frac{5}{2}\right)^2-\left(\displaystyle\frac{5}{2}\right)^2-1800=0$,

Or, $\left(x+\displaystyle\frac{5}{2}\right)^2=1800+\displaystyle\frac{25}{4}$

$=\displaystyle\frac{7225}{4}=\displaystyle\frac{85^2}{2^2}$

Taking the square root,

$x+\displaystyle\frac{5}{2}=\pm \displaystyle\frac{85}{2}$,

Or, $x=-\displaystyle\frac{5}{2} \pm \displaystyle\frac{85}{2}$.

So the two roots of $x$ are given by,

$x=-\displaystyle\frac{5}{2}+ \displaystyle\frac{85}{2}=40$ km/hr, and,

$x=-\displaystyle\frac{5}{2}- \displaystyle\frac{85}{2}=-45$ km/hr, an invalid value.

Answer: The present speed of the train is  40 km/hr.

Verify yourself.

Problem 9.

Two water taps together can fill a tank in $9\frac{3}{8}$ hrs. The tap of larger diameter takes 10 hours less to fill the tank separately. Find the time in which each tap can separately fill the tank.

Solution to problem 9.

Let time taken by the smaller diameter tap to fill the tank separately be $x$ hr.

So the time taken by the larger diameter tap to fill the tank separately is $(x-10)$ hr.

When working together, in 1 hour they will then fill a portion of the tank given by,

$\displaystyle\frac{1}{x}+\displaystyle\frac{1}{x-10}$.

As the time taken by the taps together to fill the tank is $9\frac{3}{8}=\displaystyle\frac{75}{8}$ hr, you get,

$\displaystyle\frac{75}{8}\left(\displaystyle\frac{1}{x}+\displaystyle\frac{1}{x-10}\right)=1$, the full portion of the tank,

Or, $75(x-10+x)=8(x^2-10x)$,

Or, $8x^2-80x-150x+750=4x^2-115x+375=0$.

To avoid heavy calculations, we'll avoid the completing the square method. Instead we'll use the quickest method of splitting the product of coefficient of 1st term and the third term.

The product is,

$4\times{375}=3\times{4}\times{5}\times{5}\times{5}=15\times{100}$.

The factors are divided into two groups resulting in two numbers 15 and 100, sum of which is 115, the coefficient of the middle term.

As one of the two parts of the middle term is 15, an odd number, the coefficient of $x^2$ which is 4 cannot be split into two equal parts of value 2 each. That would make the two parts of the middle term coefficient both even. 

Thus, the factorized equation is formed as,

$(4x-15)(x-25)=0$.

The two roots are given by,

$4x-15=0$ or, $x=\displaystyle\frac{15}{4}$ hr, and,

$x-25=0$ or, $x=25$ hr.

But, $x=\displaystyle\frac{15}{4}$ hr is invalid as it is less than $9\frac{3}{8}$ hrs, the time to fill the tank by both taps together, which is not possible.

Answer: Smaller tap takes 25 hours and larger tap 15 hours to fill the tank separately.

Verify yourself.

Problem 10.

An express train takes 1 hr less than a passenger train to travel 132 km between Mysore and Bengaluru (without taking into consideration the time they stop at intermediate stations). If average speed of the express train is 11 km/hr more than that of the passenger train, find the average speed of the two trains.

Solution to problem 10.

Let the average speed of the passenger train be $x$ km/hr and it takes $T$ hrs to cover the distance of 132 km.

For this train, the speed time distance relationship is,

$T=\displaystyle\frac{132}{x}$ hr.$\qquad \qquad \qquad ......(1)$.

As the average speed of the express train is 11 km/hr more than that of the passenger train, it is, $x+11$ km/hr.

And at this speed it takes 1 hr less than the passenger train, that is, $T-1$ hr to cover the distance of 132 km.

For the express train the speed time distance relationship is then,

$T-1=\displaystyle\frac{132}{x+11}$ hr.$\qquad \qquad ......(2)$.

Subtract equation (2) from equation (1) to eliminate $T$. Result is,

$1=\displaystyle\frac{132}{x}-\displaystyle\frac{132}{x+11}$,

Or, $x(x+11)=11\times{132}$,

Or, $x^2+11x-11\times{132}=0$.

Factorizing the product of the coefficient of the 1st term and the third term, you get,

$-11\times{132}=-2\times{2}\times{3}\times{11}\times{11}=-33\times{44}$.

The larger split of 44 must be positive and smaller split 33 negative so their sum is 11, the coefficient of the middle term.

The LHS of the quadratic equation in factorized form is,

$(x-33)(x+44)=0$.

The two roots are given by,

$x-33=0$, or, $x=33$ km/hr, and,

$x+44=0$, or, $x=-44$ km/hr, an invalid value.

Answer: Average speed of passenger train is 33 km/hr and that of express train is 44 km/hr.

Verify yourself.

Problem 11.

Sum of the areas of two squares is 468 sq metre. If the difference of their perimeters is 24 metre, find the sides of the two squares.

Solution to problem 11.

Let $x$ metre be the side length of the larger square and $y$ be the side length of the smaller square.

By the first condition then,

$x^2+y^2=468$.

By the second condition,

$4(x-y)=24$ metre,

Or, $y=x-6$, common factor 4 is canceled out.

Substitute in the first equation. Result is,

$x^2+(x-6)^2=468$,

Or, $x^2+x^2-12x+36-468=0$,

Or, $2x^2-12x-432=0$,

Or, $x^2-6x-216=0$.

Factorize the third term,

$-216=2\times{2}\times{2}\times{3}\times{3}\times{3}=-12\times{18}$.

Sum of $-18$ and $12$ is $-6$, the coefficient of the middle term.

The LHS of the quadratic equation can then be factorized as,

$(x-18)(x+12)=0$.

The roots are, $x=18$ metre, and,

$x=-12$ metre, an invalid value.

The smaller square side value is, $y=x-6=18-6=12$ metre.

Answer: 12 metre, 18 metre.

Verify yourself.

End note

Finding the roots of a quadratic equation by the method of completing the square most times involves larger amount of calculations and hence takes more time compared to factorization method.

Nevertheless, the importance of this approach is in,

  1. leading to the quadratic equation root formula by Sreedhar Acharya, and
  2. determining the nature of the two roots of the equation quickly.

We'll deal with this second important use of this method in the next session.


NCERT Solutions for Class 10 Maths

Chapter 1: Real Numbers

NCERT Solutions for Class 10 Maths on Real numbers part 3, HCF and LCM by factorization and problem solutions

NCERT Solutions for Class 10 Maths on Real numbers part 2, Euclid’s division algorithms, HCF and problem solutions

NCERT Solutions for Class 10 Maths on Real numbers part 1, Euclid’s division lemma puzzle solutions

Chapter 2: Polynomials

NCERT Solutions for Class 10 Maths Chaper 2 Polynomials 1 Geometrical Meaning of Zeroes of Polynomials

Chapter 3: Linear Equations

NCERT solutions for class 10 maths Chapter 3 Linear equations 7 Problem Collection

NCERT solutions for class 10 maths Chapter 3 Linear equations 6 Reducing non-linear to linear form

NCERT solutions for class 10 maths Chapter 3 Linear equations 5 Algebraic solution by Cross Multiplication

NCERT solutions for class 10 maths Chapter 3 Linear Equations 4 Algebraic solution by Elimination

NCERT solutions for class 10 maths Chapter 3 Linear Equations 3 Algebraic solution by Substitution

NCERT solutions for class 10 maths Chapter 3 Linear Equations 2 Graphical solutions

NCERT solutions for class 10 maths Chapter 3 Linear Equations 1 Graphical representation.

Chapter 4: Quadratic equations

NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 1 What are quadratic equations

NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 2 Solving by factorization

NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 3 Solution by Completing the square

NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 4 Nature of roots of a quadratic equation

Chapter 6: Triangles

NCERT solutions for class 10 maths chapter 6 Triangles 1 Similarity of Triangles and Polygons

Solutions to Exercise 2 Chapter 6 NCERT X Maths, Characteristics of Similar triangles

Chapter 8: Introduction to Trigonometry, Concepts and solutions to exercise problems

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 1 Trigonometric Ratios

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 2 Ratio values for selected angles

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 3 Complementary angle relations

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 4 Trigonometric identities


Chapter 8: Introduction to Trigonometry, only solutions to selected problems

NCERT Solutions for Class 10 Maths on Trigonometry, solution set 6

NCERT Solutions for Class 10 Maths on Trigonometry, solution set 5

NCERT Solutions for Class 10 Maths on Trigonometry, solution set 4

NCERT Solutions for Class 10 Maths on Trigonometry, solution set 3

NCERT Solutions for Class 10 Maths on Trigonometry, solution set 2

NCERT Solutions for Class 10 Maths on Trigonometry, solution set 1


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