# NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.2

## Chapter 2 Ex.2.2 Question 1

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i)  $$x^{2} -2x-8$$

(ii) $$4{{s}^{2}}-4s+1$$

(iii) $$6{{x}^{2}}-3\text{ }-7x$$

(iv) $$4{{u}^{2}}+\text{ }8u$$

(v) $${t^{2}}-15$$

(vi) $$3{{x}^{2}}-x-4$$

### Solution

What is known?

What is unknown?

• The zeroes of the given quadratic polynomials
• Verification of the relationship between the zeroes and the coefficients.

Reasoning:

You can solve this question by following the steps given below

We know that the standard form of the quadratic equation is

$a{x^2} + bx+c = 0$

Simplify the quadratic polynomial by factorisation and find the zeroes of the polynomial.

Now you have to find the relation between the zeroes and the coefficients.

For find out the sum of zeroes and product of zeroes.

We know that

Sum of zeroes

\begin{align}= \frac{-\, \text{coefficient of}\;x}{{\text{coefficientof}\;{x^2}}} \end{align}

\begin{align}\alpha + \beta = \frac{{ - b}}{a}\end{align}

Product of  Zeroes

\begin{align} = \frac{{\text{constant}\;{\text{term}}}}{{\text{coefficient of}\;\;{x^2}}} \end{align}

\begin{align}\alpha .\beta &= \frac{c}{a}\end{align}

Put the values in the above formula and find out the relation between the zeroes and the coefficients.

Steps:

(i) \begin{align}{x^{2}-2 x-8}\end{align}

\begin{align}x^{2}-4 x+2 x-8&=0 \\ x(x-4)+2(x-4)&=0 \\ (x-4) (x+2)&=0 \\ (x-4)&=0,\x+2)&=0\end{align} \(x=4, x=-2 are the zeroes of the polynomial

Relationship between the zeroes and the coefficients

Sum of zeroes

\begin{align}= \frac{{ -\, \text{coefficient of}\;x}}{{\text{coefficient of}\;{x^2}}}\end{align}

\begin{align}\alpha \, + \beta &= \frac{{ - b}}{a}\\ - 2 + 4 &= \frac{{ - \left( { - 2} \right)}}{1}\\2 &= 2\end{align}

Product of zeroes

\begin{align}= \frac{{\text{constant}\;\text{term}}}{{\text{coefficient}\;\text{of}\;{x^2}}}\end{align}

\begin{align}\alpha \,.\,\beta &= \frac{c}{a}\\ - 2 \times 4 &= \frac{{ - 8}}{1}\\-8 &= -8\end{align}

(ii) \begin{align}{4{s^2} - 4s + 1}\end{align}

\begin{align}4{s^2} - 2s - 2s + 1 &= 0\\2s\left( {2s - 1} \right) - \left( {2s - 1} \right) &= 0\\\left( {2s - 1} \right)\left( {2s - 1} \right)&= 0\\2s - 1&= 0,\\2s - 1&= 0\end{align}

\begin{align}{\text{s}} = \frac{1}{2},\,\,{\text{s}} = \frac{1}{2}\end{align} are the zeroes of the polynomial.

Relationship between the zeroes and the coefficients

Sum of zeroes

\begin{align} = \frac{{ - \,\text{coefficient of}\;\;x}}{{\text{coefficient of}\;\;{x^2}}}\end{align}

\begin{align}{\alpha + \beta} &= {\frac{{ - b}}{a}}\\{\frac{1}{2} + \frac{1}{2}}&= {\frac{{ - \left( { - 4} \right)}}{4}}\\{\,\,\,\,\,\,\,\,\,\,\,1}&= {1}\end{align}

Product of zeroes

\begin{align}= \frac{{\text{constant}\;\text{term}}}{{\text{coefficient}\;\text{of}\;{x^2}}}\end{align}

\begin{align}\alpha \,.\,\beta &= \frac{c}{a}\\\frac{1}{2} \times \frac{1}{2} &= \frac{1}{4}\\\end{align}

(iii) \begin{align}{6{x^2} - 3 - 7x}\end{align}

\begin{align}6{x^2}-7x-3&=0\\6{x^2} - 9x + 2x - 3 &= 0\\3x(2x - 3) + (2x - 3) &= 0\2x - 3)&= 0,\\(3x + 1) &= 0\end{align} \(\begin{align} x = \frac{3}{2},\,\,\,x = \frac{{ - 1}}{3}\end{align} are the zeroes of the polynomial

Relationship between the zeroes and the coefficients

Sum of zeroes

\begin{align} = \frac{{ -\,\text{ coefficient of}\;x}}{{\text{coefficient of}\;{x^2}}}\end{align}

\begin{align}{\,\,\,\alpha + \beta} &= {\frac{{ - \left( { - 7} \right)}}{6}}\\{\frac{3}{2} + \frac{{ - 1}}{3}} &={ \frac{{\left( 7 \right)}}{6}}\\{\,\,\,\,\,\,\frac{{\left( 7 \right)}}{6}}&= {\frac{{\left( 7 \right)}}{6}}\end{align}

\begin{align}{\alpha \,.\,\beta }&= {\frac{c}{a}}\\{\frac{3}{2} \times \frac{{ - 1}}{3} }&={ \frac{{\left( { - 3} \right)}}{6}}\\{\,\,\,\frac{{\left( { - 3} \right)}}{6} }&={ \frac{{\left( { - 3} \right)}}{6}}\\{\,\,\,\frac{{\left( { - 1} \right)}}{2} }&={ \frac{{\left( { - 1} \right)}}{2}}\end{align}

(iv) \begin{align}{4{u^2} + 8u}\end{align}

\begin{align}{4u\,\left( {u + 2} \right) }&={ 0}\\{4u = 0{\text{ or }}u + 2 }&= {0}\\u &= 0\\{\text{ or }}\;u &={ - 2}\end{align}

$$u = 0,{\text{ }}u{\text{ }} = - 2$$ are the zeroes of the polynomial

Relationship between the zeroes and the coefficients

Sum of zeroes

\begin{align} = \frac{{ - \;\text{coefficient of}\;u}}{{\text{coefficient of}\;{u^2}}}\end{align}

\begin{align}{\alpha + \beta}&={ \frac{{ - \left( 8 \right)}}{4}}\\{0 + \left( { - 2} \right) }&={ - 2}\\{ - 2} &={ - 2}\end{align}

Product of zeroes

\begin{align} = \frac{{\text{constant term}}}{{\text{coefficient of}\;{u^2}}}\end{align}

\begin{align}{\alpha \,.\,\beta }&={ \frac{c}{a}}\\{0 \times - 2 }&={ \frac{0}{4}}\\{0 }&={ 0}\\\end{align}

(v) \begin{align}{t^{2}}-15\end{align}

\begin{align}{{t^2}- 15} &={ 0}\\{{t^2} - 15 }&= {0}\\{\,\,\,\,\,\,\,\,\,\,\,\,\,t }&={ \sqrt {15} }\end{align}

$$-\sqrt{15},\text{ }t\text{ }+\sqrt{15}$$ are the zeroes of the polynomial

Relationship between the zeroes and the coefficients

Sum of zeroes

\begin{align} = \frac{{ - \;\text{coefficient of}\;t}}{{\text{coefficient of}\;{t^2}}}\end{align}

\begin{align}\alpha + \beta &= \frac{0}{1}\\& = - \sqrt {15} + \sqrt {15} = 0\\0 &= 0\end{align}

Product of zeroes

\begin{align} = \frac{{\text{constant term}}}{{\text{coefficient of}\;{t^2}}}\end{align}

\begin{align}\alpha \,.\,\beta &= \frac{c}{a}\\ - \sqrt {15} {\text{ }} \times \sqrt {15} &= - \frac{{15}}{1}\\- 15 &= - 15\end{align}

(vi) \begin{align}{3{x^2} - x - 4}\end{align}

\begin{align}3{x^2} - x - 4 &= 0\\3{x^2} - 4x + 3x - 4 &= 0\\x\left( {3x - 4} \right) + \left( {3x - 4} \right) &= 0\\\left( {x + 1} \right)\left( {3x - 4} \right) &= 0\\\left( {x + 1} \right) &= 0\\ {\text{ or }}\quad\left( {3x - 4} \right) &= 0\end{align}

\begin{align}x = - 1\,\,{\text{or}}\,\,x = \frac{4}{3}\end{align} are the zeroes of the polynomial

Relationship between the zeroes and the coefficients

Sum of zeroes

\begin{align} = \frac{{ - \;\text{coefficient of}\;x}}{{\text{coefficient of}\;{x^2}}}\end{align}

\begin{align}{\alpha + \beta} &= {\frac{{ - 1}}{3}}\\{ - 1 + \frac{4}{3} }&= {\frac{{ - 1}}{3}}\\{\frac{1}{3} }&= {\frac{1}{3}}\end{align}

Product of zeroes

\begin{align} = \frac{{\text{constant term}}}{{\text{coefficient of}\;{x^2}}}\end{align}

\begin{align}\alpha \,.\,\beta &= \frac{c}{a}\\ - 1 \times \frac{4}{{\frac{4}{3}}} &= - \frac{4}{3}\\ - \frac{4}{3} &= - \frac{4}{3}\end{align}

## Chapter 2 Ex.2.2 Question 2

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) \begin{align}\frac{1}{4},\, - 1\end{align}

(ii) \begin{align}\sqrt 2 ,\;\frac{1}{3}\end{align}

(iii) $${0,\;5}$$

(iv) $$1, \;1$$

(v) \begin{align}\frac{1}{4},\;\frac{1}{4} \end{align}

(vi) $$4,\;1$$

### Solution

What is known?

The sum and product of zeroes of quadratic polynomials.

What is unknown?

A quadratic polynomial with the given numbers as the sum and product of its zeroes respectively.

Reasoning:

This question is straight forward - the value of sum of roots and product of roots is given. You have to form a quadratic polynomial. Put the values in the general equation of the quadratic polynomial i.e.

k\left[ \begin{align}&{{x}^{2}}-\left( \text{sum of roots} \right)x+ \\ &\text{product of roots} \\ \end{align} \right]

Steps:

(i) \begin{align}\,\,\frac{1}{4},\; - 1\end{align}

We know that the general equation of a quadratic polynomial is:

\begin{align}&k\left[ \begin{array}{i}{{x}^{2}}-\left( \text{sum of roots} \right)x+ \\ \text{product of roots} \\ \end{array} \right]\\&{k\left[ {{x^2} - \frac{1}{4}x + \frac{1}{4} \times - 1} \right]}\\&{k\left[ {{x^2} - \frac{1}{4}x - \frac{1}{4}} \right]}\end{align}

(ii)  \begin{align}\sqrt 2,\;\frac{1}{3}\end{align}

We know that the general equation of a quadratic polynomial is:

\begin{align}&k\left[ \begin{array}{i}{{x}^{2}}-\left( \text{sum of roots} \right)x+ \\ \text{product of roots} \\ \end{array} \right]\\&k \left[ {{x^2} - \sqrt 2 x + \frac{1}{3}} \right]\end{align}

(iii)  $$\,0,\;\sqrt 5$$

We know that the general equation of a quadratic polynomial is

\begin{align}&k\left[ \begin{array}{i}{{x}^{2}}-\left( \text{sum of roots} \right)x+ \\\text{product of roots} \\ \end{array} \right]\\&k\left[ {{x^2}-{\text{ }}0.x+\sqrt 5 } \right]\\&k\left[ {{x^2}+ {\text{ }}\sqrt 5 } \right]\end{align}

(iv) $$\,1,\;1$$

We know that the general equation of a quadratic polynomial is

\begin{align}&k\left[ \begin{array}{I}{{x}^{2}}-\left( \text{sum of roots} \right)x+ \\ \text{product of roots} \\ \end{array} \right]\\&k\left[ {{x^2}-{\text{ }}1x{\text{ }} + {\text{ }}1} \right]\\&k\left[ {{x^2}-{\text{ }}x{\text{ }} + {\text{ }}1} \right]\end{align}

(v) \begin{align} - \frac{1}{4},\;\frac{1}{4}\end{align}

We know that the general equation of a quadratic polynomial is:

\begin{align}&k\left[ \begin{array}{I}{{x}^{2}}-\left( \text{sum of roots} \right)x+ \\ \text{product of roots} \\ \end{array} \right]\\&k\left[ {{x^2} - \left( { - \frac{1}{4}x} \right) + \frac{1}{4}} \right] \\& k\left[ {{x^2} + \frac{1}{4}x + \frac{1}{4}} \right] \\ \end{align}

(vi)  $$4,\;1$$

We know that the general equation of a quadratic polynomial is:

\begin{align}&k\left[ \begin{array}&{{x}^{2}}-\left( \text{sum of roots} \right)x+ \\\text{product of roots} \\ \end{array} \right]\\&k\left[ {{x^2}-{\text{ }}4x + {\text{ }}1} \right]\end{align}