# Ex.2.2 Q1 Polynomials Solution - NCERT Maths Class 10

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## Question

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)  $$3{{x}^{2}}-x-4$$

Video Solution
Polynomials
Ex 2.2 | Question 1

## Text 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}

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