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) \(4s^2 – 4s + 1\) (iii) \(6x^2 – 3 – 7x\) (iv) \(4u^2 + 8u\) (v) \( t^2 – 15\) (vi) \(3x^2 – x – 4\)

Answer 1

(i) \(x^2-2x-8\)
\(=(x-4)(x+2)\)

The value of \(x^2-2x-8 \) is zero when \(x - 4 = 0\) or \(x + 2 = 0\) , 

i.e., when \(x = 4\) or \(x = -2\)
Therefore, the zeroes of \(x^2-2x-8\) are \(4\) and \(-2.\)
Sum of zeroes \(= 4-2=2=\) \(\dfrac{-(-2)}{1} \)=\(-(\)Coefficient of \(x)\) Coefficient of \(x^2\)
Product of zeroes \(= 4x(-2)=-8= \dfrac{(-8)}{1} = \dfrac{\text{ Constant term}}{\text{Coefficient of }x^2}\)

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(ii)\( 4s^2-4s+1\)
\(=(2s-1)^2\)

The value of \( 4s^2-4s+1\) is zero when \(2s - 1 = 0,\)

 i.e.,\( s = \dfrac{1}{2} \) Therefore, the zeros of \(4s^2 - 4s + 1\) are \(\dfrac{1}{2}\) and  \(\dfrac{1}{2} .\)
Therefore,
The sum of zeroes\( = \dfrac{1}{2} + \dfrac{1}{2} = \dfrac{1}{1}= \dfrac{-(-4)}{ 4}\) [Multiply by \(4\) in Numerator and Denominator]\(= \dfrac{-\text{(Coefficient of s)}}{\text{(Coefficient of } s^2)}\)
Product of zeroes \(= \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4} = \dfrac{\text{ Constant term}}{\text{Coefficient of }S^2}\)

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(iii) \(6x^2-3-7x\)
\(= 6x^2-7x-3=(3x+1)(2x-3)\)

The value of \(6x^2 - 3 - 7x\) is zero when \(3x + 1 = 0\) or \(2x -3 = 0\), i.e.,
\(x = \dfrac{-1}{3}\) or \(x = \dfrac{3}{2} \)

Therefore, the zeroes of \(6 x^2 − 3 − 7 x\) are \(6x^2 -3 -3 -7x\) are \(\dfrac{-1}{3}\) and \(\dfrac{3}{2}\).

Sum of Zeros \(= \dfrac{-1}{3} + \dfrac{3}{2} = \dfrac{7}{6} = \dfrac{-(-7)}{6} = \dfrac{-(\text{Coefficient of } x)}{ \text{Coefficient of  }  x^2}\)

Product of zeroes \(= \dfrac{-1}{3} \times \dfrac{3}{2} = \dfrac{-1}{ 2} = \dfrac{-3}{6} = \dfrac{\text{ Constant term}}{\text{Coefficient of }x^2}\)

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(iv)\( 4u^2 + 8u \)
\(= 4u^2 + 8u +0\)
\(=4u(u+2 )\)

The value of \(4u^2 + 8u\) is zero when \(4u = 0\) or \(u + 2 = 0\), i.e., 
\(u = 0\) or \(u = −2\) 

Therefore, the zeroes of \(4u^2 + 8u\) are \(0\) and \(−2.\)

The sum of Zeros \(=0 + (-2) = -2 = \dfrac{-(8)}{4}\) [Multiply by \(4\) in Numerator and Denominator]\(= \dfrac{-\text{(Coefficient of u)}}{\text{(Coefficient of } u^2)}\)
Product of zeroes \(= 0 \times (-2) =0 = \dfrac{0}{4} = \)\(\dfrac{\text{ Constant term}}{\text{Coefficient of }u^2}\)

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(v) \(t^2 -15\)
\(= t^2 -0.t -15\)
\(=(t-\sqrt{15}) (t + \sqrt {15})\)

The value of \(t^2 − 15\) is zero when \(t-\sqrt{15} = 0\) or \(t + \sqrt{15} =0\),

 i.e., when\(t = \sqrt{15}\) or\(t = -\sqrt{15}\)

Therefore, the zeroes of \(t^2 − 15\) are\(\sqrt{15} \text{  and } -\sqrt{15}.\) 

Sum of Zeros\(= \sqrt{15} + (-\sqrt{15}) = 0 =-0/1 = -\)\(\dfrac{-\text{(Coefficient of t)}}{\text{(Coefficient of } t^2)}\)

Product of zeroes \(= \sqrt{15} \times (-sqrt{15}) = -15 = -15/1=\) \(\dfrac{\text{ Constant term}}{\text{Coefficient of }t^2}\)

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(vi) \(3x^2 − x − 4\)
\(= (3x-4)(x+1)\)

The value of \(3x^2 − x − 4\) is zero when \(3x − 4\) \(= 0\) or \(x + 1 = 0\),

 i.e., when\(x= \dfrac{4}{3} \text{ or } x = −1\)

Therefore, the zeroes of \(3x^2 − x − 4\) are \(\dfrac{4}{3}\)  and \(−1\).

Sum of Zeros \(= \dfrac{4}{3} + (-1) = \dfrac{1}{3} = \dfrac{-(-1)}{3} =\)\( \dfrac{-\text{(Coefficient of x)}}{\text{(Coefficient of } x^2)}\)
Product of zeroes \(= \dfrac{4}{3} \times (-1)\) = \(= \dfrac{4}{3}\)= \(\dfrac{\text{ Constant term}}{\text{Coefficient of }x^2}\)