NCERT solutions for class 10 maths Chapter 4 Quadratic equations 2 Solution by factorization | SureSolv

NCERT solutions for class 10 maths Chapter 4 Quadratic equations 2 Solution by factorization

Solving a quadratic equation by factorization

NCERT solutions class 10 maths ch 4 pt 2

This is the second of the four part series on Quadratic equations. It complements Chapter 4 of NCERT Class 10 Maths. This session will cover,

  1. Roots of a quadratic equation as solution,
  2. How to solve a quadratic equation by factorization.

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 2nd exercise of NCERT Chapter 4.

Solution and Roots of a quadratic equation

Let's assume you have to solve the quadratic equation,

$x^2-10x+21=0$.

Let's also assume that you know how to factorize the quadratic expression, or to be more formal, the quadratic polynomial of degree 2, $x^2-10x+21$ in the LHS of the equation.

Factorizing the LHS expression and equating to 0, you'll get,

$x^2-10x+21=(x-3)(x-7)=0$.$\qquad....................(1)$

If $x=3$, the LHS evaluates to,

$3^2-10\times{3}+21=9-30+21=0$,

Or, $LHS=RHS$.

That is, the value $x=3$ satisfies the quadratic equation and so is a solution of the equation.

Similarly, if $x=7$, the LHS evaluates to,

$7^2-10\times{7}+21=49-70+21=0$,

Or, $LHS=RHS$.

Again, the value $x=7$ satisfies the quadratic equation and so is the second solution of the equation.

More formally, these two values of $x$ are called the ROOTS of the quadratic equation.

Whenever you are solving a quadratic equation, you are actually trying to find its roots.


Note:

As at $x=3$ and $x=7$ the quadratic polynomial in the LHS evaluates to 0, these two values are called the ZEROES of the quadratic polynomial.


Maximum number of roots of a quadratic equation

As the highest power of $x$ in a quadratic equation is 2, it can have AT MOST 2 ROOTS.

Can you think of any quadratic equation with less than 2 roots?

Well, consider the quadratic equation,

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

What are its factors? The LHS is simply $(x-2)^2$. So the equation is transformed to,

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

In this special case, both the roots have equal value of 2—effectively, number of roots is 1.

This is true for any quadratic equation of the form,

$(x\pm a)^2=0$, where $a$ is a real number.

Let's now see briefly why the factorization gives us the roots of a quadratic equation.

Reason behind factorization yielding roots of a quadratic equation

Let us take up again the quadratic equation,

$x^2-10x+21=0$.

It is factorized to,

$x^2-10x+21=(x-3)(x-7)=0$.

For satisfying the equation, that is, for the LHS to be equal to 0, it is a necessary condition that,

Either the factor $(x-3)=0$ or the second factor $(x-7)=0$.

[Note: both cannot be 0 simultaneously]

If $x-3=0$, $x=3$, the first root.

And if $x-7=0$, $x=7$, the second root.

That's why, either $x=3$ or $x=7$ satisfies the quadratic equation and are the roots of the quadratic equation.

We have learnt some terminologies, but not how to factorize the quadratic polynomial in the LHS of a quadratic equation.

How to factorize a quadratic polynomial

The generally accepted method to factorize a quadratic polynomial such as, $x^2-10x+21$ is by splitting the middle term of the three term expression in $x$.

The middle term—to be more specific—the coefficient of the middle term is split into two parts so that the sum of the two parts forms the coefficient of the middle term and the product forms the constant third term (in a special case when coefficient of $x^2$ is 1).

This not really is a systematic method and is largely dependent on guessing and trial and error.

For simple problems it may work, but for more complex forms of quadratic polynomials, you may waste considerable amount of time guessing and on trial and error.

As a rule, we would follow a much more systematic method that requires minimum amount of guessing.

Instead of splitting the coefficient of the middle term, we would split the product of the first term coefficient and the third constant term.

If you desire, you may skip the following section on mathematical background of this powerful systematic method and directly go over to using the method on actual problems.


Mathematical background of factorization of a quadratic polynomial by splitting the product of the coefficient of the first term and the third term

The general form of a quadratic equation is,

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

Assuming that the two roots to be $x=-\displaystyle\frac{p}{m}$ and $x=-\displaystyle\frac{q}{n}$, the quadratic equation is expressed as,

$ax^2+bx+c=(mx+p)(nx+q)=0$,

where $m$, $n$, $p$ and $q$ are non-zero real numbers in the two linear factors of the quadratic polynomial in the LHS.

Expanding the product you get,

$ax^2+bx+c=mnx^2+(mq+pn)x+pq$.

Comparing the coefficients of like terms,

$a=mn$, $c=pq$ so that,

$ac=mnpq$.$\qquad \qquad \qquad \qquad...........................(2)$

And, $b=mq+pn$.$\qquad \qquad \qquad.........................(3)$

Equation (2) is intuitive—the product of the coefficient of first $x^2$ term and the constant third term must equal the product of all four coefficients of the two linear polynomials of the two factors.

Basically, the set of four coefficients of the two linear factors are to be split up into two subsets one of which equals $a$ and the other $c$.

But again, if the product of these same four coefficients of the two linear polynomials $(mx+p)$ and $(nx+q)$ is split into two suitable numbers, you get their sum as the middle term coefficient by equation (3).

So in this more efficient method,

The product of the coefficient of the first term and the third term is first expressed as a product of prime factors. Then by examining the prime factors two suitable subsets are identified each generating a number, sum which gives the coefficient of the middle term.

This is multiplicative splitting.

If you start from equation (3) instead, you would be doing summative splitting of the same set of four coefficients into two parts, sum of which gives the coefficient of the middle terms $b$.

It is a fact by nature of numbers that,

Splitting a number as a sum of two numbers that fulfills a certain goal generates many more combinations and involves much more guessing than splitting a set of prime factors into two sets each generating a number, sum of which is a known number.

That's why for factorization of a quadratic polynomial, as a rule we split the product of first term coefficient and the third term into two sets of prime factors of the product so that the sum of the two resulting numbers is the coefficient of the middle term in $x$.

We would demonstrate the method using a few problem examples.


Problem example 1.

Find the roots of the following quadratic equation by factorization.

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

Solution example 1.

The product of the coefficient of the first term in $x^2$ and the constant third term is,

$-14\times{30}=-2\times{2}\times{3}\times{5}\times{7}$.

These 5 prime factors are to be split into two groups each forming a number so that sum of the two numbers becomes 23, the coefficient of the middle term.

This is the first step.

Dealing with the sign of the two split numbers:

Though you have to guess, you would follow a systematic path.

So next you must take care of the fact that one of the two split numbers would be positive and the other negative (as their product is $-420$).

As the coefficient 23 of middle term is positive,

The larger split number would be positive and the smaller negative.$ \qquad...(C1)$

This is the second step.

With this knowledge, it should take a few seconds for you to identify,

$2\times{2}\times{3}=12$ as the negative split part, and the rest of the prime factors,

$5\times{7}=35$ as the positive split part, so that,

$35-12=23$, the coefficient of the middle term.

This is the third step.

Now we'll factorize 14 into two parts as the coefficients of $x$ of the two linear factors of the quadratic polynomial.

There is only one possible choice, $14=2\times{7}$.

To aid forming the final linear factors, at this stage we express the partially formed information as,

$14x^2+23x-35=(7x+ p)(2x+ q)$, where one of $p$ and $q$ would be negative.

This is the fourth step.

From condition (C1), the multiplicative factor of 7 which is the coefficient of $x$ in first linear factor, must be $q=5$, the constant term of the second linear factor, and its sign must be positive. The product of the two would have to be 35.

That leaves constant term of the first linear factor as,

$p=-\displaystyle\frac{30}{5}=-6$, as $pq=-30$.

These two values of $p$ and $q$ would also satisfy the condition,

$7q+2p=35-12=23$, the coefficient of the middle term.

This is the fifth and final step.

In factorized form the quadratic equation is then expressed as,

$14x^2+23x-30=(7x-6)(2x+5)=0$.

The two roots are given by $(7x-6)=0$, or, $x=\displaystyle\frac{6}{7}$, and,

by $(2x+5)=0$ or, $x=-\displaystyle\frac{5}{2}$.

Verification:

Substituting the value of $x=\displaystyle\frac{6}{7}$ you get,

$14x^2+23x-30$

$=14\times{\left(\displaystyle\frac{6}{7}\right)^2}+23\times{\displaystyle\frac{6}{7}}-30$

$=14\times{36}+23\times{6}\times{7}-30\times{49}$

$=6(84+161-245)$

$=6(245-245)=0$.

Or, $LHS=RHS$.

Verified.

Substituting the value of second root, $x=-\displaystyle\frac{5}{2}$, you get,

$14x^2+23x-30$

$=14\times{\left(\displaystyle\frac{5}{2}\right)^2}-23\times{\displaystyle\frac{5}{2}}-30$

$=350-230-120$

$=350-350=0$,

Or, $LHS=RHS$.

Verified. 

You should try to solve the problem by middle term splitting.

Problem example 2.

Find the roots of the following quadratic equation by factorization.

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

Solution example 2.

The product of the coefficient of the first term in $x^2$ and the constant third term is,

$2\times{3}=2\times{\sqrt{3}}\times{\sqrt{3}}$.

The reason why the integer $3$ is split into two factors each $\sqrt{3}$ is the middle term value of $-3\sqrt{3}$.

Though the product of first term coefficient and the third term is an integer 6, it must be split into two surd terms each having $\sqrt{3}$ as a factor so that their sum is the surd middle term coefficient $-3\sqrt{3}$.

This falls under what we call the Surd term factoring technique, a valuable technique for dealing with surd expressions.

When the quadratic polynomial involves surds, in addition to prime factors, surd techniques are to be used.

Obvious choices of product 6 split into two parts are, $-\sqrt{3}$ and $-2\sqrt{3}$ so that their sum is the middle term coefficient,

$-\sqrt{3}-2\sqrt{3}=-3\sqrt{3}$.

As the coefficient of the first term $x^2$ is 2, the intermediate stage of the two linear polynomial factors can be formed as,

$2x^2-3\sqrt{3}+3=(2x+p)(x+q)$, where both $p$ and $q$ must negative.

The multiplicative factor of 2, which is the coefficient of $x$ of the first linear factor, must be $q=-\sqrt{3}$, the constant term of the second linear factor and so, the constant term of the first linear factor must be,

$p=\displaystyle\frac{3}{-\sqrt{3}}=-\sqrt{3}$, as $pq=3$.

These values would also satisfy the condition,

$2q+p=-2\sqrt{3}-\sqrt{3}=-3\sqrt{3}$, the coefficient of the middle term.

That decides the final product of two linear factors as,

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

The two roots are given by $(2x-\sqrt{3})=0$, or, $x=\displaystyle\frac{\sqrt{3}}{2}$, and

by $(x-\sqrt{3})=0$, $x=\sqrt{3}$.

Verification:

Substituting the first root, $x=\displaystyle\frac{\sqrt{3}}{2}$, you get,

$2x^2-3\sqrt{3}x+3$

$=2\times{\displaystyle\frac{3}{4}}-3\sqrt{3}\times{\displaystyle\frac{\sqrt{3}}{2}}+3$

$=\displaystyle\frac{3}{2}-\displaystyle\frac{9}{2}+3$

$=\displaystyle\frac{9}{2}-\displaystyle\frac{9}{2}=0$.

Or, $LHS=RHS$.

Verified.

Substituting the second root value of $x=\sqrt{3}$, you get,

$2x^2-3\sqrt{3}x+3$

$=2\times{3}-3\sqrt{3}\times{\sqrt{3}}+3$

$=6-9+3=0$.

Or, $LHS=RHS$.

Verified.

Problem example 3.

Find the dimensions of a prayer hall of area 300 sq m, where the length is 1 metre more than twice the breadth.

Solution example 3.

Let $\text{breadth}=x$ m.

So, $\text{length}=2x+1$ m, and,

$\text{Area}=\text{length}\times{\text{breadth}}=x(2x+1)=300$,

Or, $2x^2+x-300=0$.

First you would find the two roots of this quadratic equation by factorization method.

The product of the coefficient of the first $x^2$ term and the constant third term is,

$-2\times{300}=-2\times{2}\times{2}\times{3}\times{5}\times{5}$

These six prime factors are to be split into two groups resulting in one positive integer and the other negative integer (as constant term is negative) so that the difference of the two is 1, the coefficient of middle term in $x$.

$-2\times{2}\times{2}\times{3}=-24$ and $5\times{5}=25$ can easily be identified to be the two numbers.

The intermediate stage product of two linear factors can then be expressed as,

$(2x+ p)(x + q)$ where one of $p$ and $q$ must take one of the two values $25$ and $\displaystyle\frac{300}{25}=12$ with one of the values negative.

So the multiplier of 2, the coefficient of the first linear factor must be $-12$, the constant term of the second linear factor $q$, and the constant term of the first linear factor would be $25$.

These values would satisfy the condition,

$2q+p=-24+25=1$, the coefficient of the middle term.

The factorized quadratic expression would be,

$2x^2+x-300=(2x+25)(x-12)=0$.

The two roots would be given by $(2x+25)=0$, or $x=-\displaystyle\frac{25}{2}$, an invalid value as breadth cannot be negative, and

by $(x-12)=0$, or, $x=12$ m

Length would be $2x+1=25$ m.

Verification:

Substituting $x=-\displaystyle\frac{25}{2}$, you get,

$2x^2+x-300$

$=\displaystyle\frac{625}{2}-\displaystyle\frac{25}{2}-300$

$=\displaystyle\frac{600}{2}-300=0$

Or, $LHS=RHS$.

Verified.

Substituting $x=12$ you get,

$2x^2+x-300$

$=288+12-300=0$.

Or, $LHS=RHS$.

Verified.

We'll now take up solving the exercise problems.

Problems to solve: Exercise 2 Chapter 4 NCERT solutions for Class 10 maths: Solving Quadratic equations by Factorization

Problem 1.i.

Find the roots of the following quadratic equation by factorization.

$x^2-3x-10=0$.

Solution to Problem 1.i

The first coefficient being 1, the two linear factors can be expressed as,

$x^2-3x-10=(x+p)(x+q)=0$, where $pq=-10$, and $(p+q)=-3$.

The constant third term is factorized to,

$-10=-2\times{5}$.

As the sum of $p$ and $q$ is negative, the larger of the two will be $-5$ and the smaller $+2$ so that,

$p+q=-3$.

The factorized form of the quadratic equation would then be,

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

The roots are given by, $x-5=0$, or, $x=5$, and

by $x+2=0$, or, $x=-2$.

Answer: $-2$, $5$.

Verify yourself.

Problem 1.ii.

Find the roots of the following quadratic equation by factorization.

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

Solution to problem 1.ii.

Product of the coefficient of first $x^2$ term and constant third term is,

$2\times{(-6)}=-2\times{2}\times{3}$.

The middle term coefficient can easily be formed as two parts of these three prime factors as,

$2\times{2}-3=1$.

So the linear factors would be,

$2x^2+x-6=(2x-3)(x+2)=0$.

The roots are given by, $(2x-3)=0$, or, $x=\displaystyle\frac{3}{2}$, and,

$(x+2)=0$ or, $x=-2$.

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

Verify yourself.

Problem 1.iii.

Find the roots of the following quadratic equation by factorization.

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

Solution to problem 1.iii.

Product of the coefficient of first $x^2$ term and constant third term is,

$\sqrt{2}\times{5}\times{\sqrt{2}}=2\times{5}$.

So the two parts of the product will be $5+2=7$, the coefficient of the middle term.

The factorized quadratic equation would be,

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

The roots are given by,

$(\sqrt{2}x+5)=0$, or, $x=-\displaystyle\frac{5}{\sqrt{2}}$, and,

$(x+\sqrt{2})=0$, or, $x=-\sqrt{2}$.

Verify yourself.

Answer: $-\displaystyle\frac{5}{\sqrt{2}}$, and $-\sqrt{2}$.

Problem 1.iv.

Find the roots of the following quadratic equation by factorization.

$2x^2-x+\displaystyle\frac{1}{8}=0$.

Solution to problem 1.iv.

The product of the coefficient of the first $x^2$ term and the third constant term is,

$2\times{\displaystyle\frac{1}{8}}=\displaystyle\frac{1}{4}=\left(-\displaystyle\frac{1}{2}\right)\left(-\displaystyle\frac{1}{2}\right)$

The sum of these two factors would be, $-\displaystyle\frac{1}{2}-\displaystyle\frac{1}{2}=-1$, the coefficient of the middle term.

The intermediate factorized form of the quadratic equation can be formed as,

$2x^2-x+\displaystyle\frac{1}{8}=(2x+p)(x+q)$,

Where both $p$ and $q$ would be negative,

$pq=\displaystyle\frac{1}{8}$, and,

$p+2q=-1$.

The multiplicative factor of 2 which is the coefficient of the first linear factor must be $q=-\displaystyle\frac{1}{4}$ so that product of the two is $-\displaystyle\frac{1}{2}$.

This results in $p=-\displaystyle\frac{1}{2}$ so that,

$p+2q=-\displaystyle\frac{1}{2}-2\times{\displaystyle\frac{1}{4}}=-\displaystyle\frac{1}{2}-\displaystyle\frac{1}{2}=-1$, the coefficient of the middle term.

The constant term also is duly formed as,

$pq=\left(-\displaystyle\frac{1}{2}\right)\times{\left(-\displaystyle\frac{1}{4}\right)}=\displaystyle\frac{1}{8}$.

The factorized quadratic equation is,

$\left(2x-\displaystyle\frac{1}{2}\right)\left(x-\displaystyle\frac{1}{4}\right)=0$.

The roots are given by,

$2x-\displaystyle\frac{1}{2}=0$, or, $x=\displaystyle\frac{1}{4}$, and,

$x-\displaystyle\frac{1}{4}=0$, or, $x=\displaystyle\frac{1}{4}$.

The two roots of the quadratic equation will have equal value of $\displaystyle\frac{1}{4}$.

Answer: Two roots of same value, $\displaystyle\frac{1}{4}$.

The problem can be solved quickly by converting the given quadratic equation to a perfect square,

$2x^2-x+\displaystyle\frac{1}{8}=0$,

Or, $x^2-2\times{\displaystyle\frac{1}{4}}x+\left(\displaystyle\frac{1}{4}\right)^2=0$, dividing the equation by 2,

Or, $\left(x-\displaystyle\frac{1}{4}\right)^2=0$.

Two roots of same value, $\displaystyle\frac{1}{4}$.

Problem 1.v.

Find the roots of the following quadratic equation by factorization.

$100x^2-20x+1=0$.

Solution to problem 1.v.

The quadratic polynomial in LHS can be expressed as a perfect square,

$100x^2-20x+1=(10x-1)^2=0$.

The equation would have two roots of equal value given by,

$(10x-1)=0$,

Or, $x=\displaystyle\frac{1}{10}$.

Answer: Two roots of equal value, $\displaystyle\frac{1}{10}$.

Problem 2.i.

John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the marbles they now have is 124. Find how many marbles they had to start with.

Solution to Problem 2.i.

Let the number of marbles John had be $x$.

So the number of marbles Jivanti had to start with was, $(45-x)$.

After both of the lost 5 marbles each number of marbles with John reduced to, $x-5$,

And number of marbles of Jivanti reduced to $(45-x-5)=(40-x)$.

Product of the two became 124. So,

$(x-5)(40-x)=124$,

Or, $-x^2+45x-200=124$,

Or, $x^2-45x+324=0$.

The product of the coefficient of the first $x^2$ term and the constant third term is,

$324=2\times{2}\times{3}\times{3}\times{3}\times{3}$.

The six prime factors can be divided into two groups of, $3\times{3}=9$, and, $6\times{6}=36$ so that their sum is the absolute value pf middle term 45.

The quadratic polynomial in the LHS can then be expressed as,

$x^2-45x+324=(x-9)(x-36)=0$

The roots are given by, $(x-9)=0$, or, $x=9$, and, $(x-36)=0$ or, $x=36$.

As none of these two values of $x$ are invalid, John originally had 9 marbles and Jivanti 36, or John had 36 marbles and Jivanti 9,

Answer 9 and 36.

Problem 2.ii.

A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in Rs.) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs.750. What was the the number of of toys produced that day?

Solution to problem 2.ii.

Let the number of toys produced in a day be $x$ and so the cost of production of each toy in Rs. is,

$(55-x)$.

That is, total cost of production of the toys in a day is,

$x(55-x)$.

As on a particular day total cost of production is Rs.750, you have,

$750=x(55-x)$,

Or, $x^2-55x+750=0$.

The product of the coefficient of first term and the constant third term is,

$750=2\times{3}\times{5}\times{5}\times{5}$.

The five factors are split easily into two groups of,

$2\times{3}\times{5}=30$, and,

$5\times{5}=25$.

The sum of 25 and 30 is 55 and product is 750. 

So the quadratic polynomial is factorized as,

$x^2-55x+750=(x-25)(x-30)=0$.

The two roots are given by,

$(x-25)=0$, or, $x=25$, and,

$(x-30)=0$, or, $x=30$.

Answer: Number of toys produced on the day would be either 25 or 30.

Problem 3.

Find two numbers sum of which is 27 and product 182.

Solution to Problem 3.

Let one of the two numbers be $x$. So the second number is, $(27-x)$.

Their product is,

$x(27-x)=182$,

Or, $27x-x^2=182$,

Or, $x^2-27x+182=0$.

The roots of this quadratic equation is to be found out by factorization.

Product of the coefficient of the first term and the constant term is,

$182=2\times{7}\times{13}$.

So, these three prime factors can be divided into two groups of value $-13$ and $-14$ sum of which would be $-27$, the coefficient of the middle term.

As the coefficient of$x^2$ is 1, the two linear factors of the quadratic polynomial in the LHS can be expressed as,

$x^2-27x+182=(x-13)(x-14)=0$.

The two roots, and the two numbers would be given by,

$(x-13)=0$, or, $x=13$, and

$(x-14)=0$, or, $x=14$.

These are the two valid values of one of the numbers.

So if this number is 13, the other is $27-13=14$ and vice versa.

Answer: 13 and 14.

Problem 4.

Find two consecutive positive integers sum of squares of which is 365.

Solution to problem 4.

Let the smaller positive integer be $x$, so that the larger is $(x+1)$.

Sum of the squares of the two is,

$x^2+(x+1)^2=365$,

Or, $x^2+x^2+2x+1=365$,

Or, $2x^2+2x-364=0$,

Or, $x^2+x-182=0$.

The product of the coefficient of the first term and the constant term is,

$-182=-2\times{7}\times{13}$.

The three prime factors can be divided into two groups of, $2\times{7}=14$ and $-13$ so that their sum is,

$14-13=1$, the coefficient of the middle term.

The quadratic polynomial in the LHS can then be factorized as,

$x^2+x-182=(x-13)(x+14)=0$.

The two roots are given by, $(x-13)=0$ or, $x=13$, and,

$(x+14)=0$, or, $x=-14$, an invalid value as it is negative.

So the two positive integers are, $13$ and $13+1=14$.

Answer: Positive integers are 13 and 14.

Problem 5.

The altitude of a right-angled triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the length of the other two sides.

Solution to problem 5.

Let the base of the triangle be $x$ cm, so that altitude is, $x-7$ cm.

By Pythagoras theorem, the square of hypotenuse of length 13 cm equals the sum of squares of its altitude and base. So you have,

$x^2+(x-7)^2=13^2$,

Or, $2x^2-14x+49=169$,

Or, $x^2-7x-60=0$.

The product of the coefficient of the first term and the constant third term is,

$-60=-2\times{2}\times{3}\times{5}$.

The four prime factors can be split into two groups of $-2\times{2}\times{3}=-12$, and $5$, sum of which will be,

$-12+5=-7$, the coefficient of the middle term.

The quadratic polynomial can thus be factorized as,

$x^2-7x-60=(x-12)(x+5)=0$.

The two roots are given by, $(x-12)=0$, or, $x=12$, and,

$(x+5)=0$, or, $x=-5$.

As length cannot be negative, the second root of $-5$ is invalid and the base is 12 cm and the altitude 5 cm.

Answer: Altitude 5 cm and base 12 cm.

Problem 6.

A cottage industry produces certain number of pottery articles in a day. It was observed on a particular day that the cost of each article (in Rs.) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs.90, find the number of articles produced and cost of each article.

Solution to problem 6.

Let $x$ be the number of articles produced on the day so that cost of each article on the day (in Rs.) was, $2x+3$ and the total cost,

$x(2x+3)=90$,

Or, $2x^2+3x-90=0$.

The product of the coefficient of the first term and the constant third term is,

$-2\times{90}=-2\times{2}\times{3}\times{3}\times{5}$.

The five prime factors can be split into two groups of $3\times{5}=15$, and $-2\times{2}\times{3}=-12$ so that,

$15-12=3$, the coefficient of the middle term.

The intermediate stage of the quadratic polynomial can be expressed in factorized form as,

$2x^2+3x-90=(2x+p)(x+q)=0$, where $p$ and $q$ are real numbers.

The multiplicative factor of 2, the coefficient of $x$ in first linear factor is $q$, the constant term of the second linear factor.

$q$ must equal $-6$ to generate $2q=-12$ and $p$ must equal $15$.

The quadratic polynomial thus can be factorized as,

$2x^2+3x-90=(2x+15)(x-6)=0$.

The two roots of the quadratic equation will then be given by,

$(2x+15)=0$, or, $x=-\displaystyle\frac{15}{2}$, an invalid value, and

$(x-6)=0$, or, $x=6$.

Number of articles produced being 6, cost of each article is,

$2\times{6}+3=\text{Rs.}15$.

Answer: Number of articles produced 6 and cost of each article Rs.15.


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 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