Question

8. Find the zeros of the quadratic polynomial (8x2 - 4) and verify the
relation between the zeros and the coefficients.
find the zeros of the quadratic polynomial (51,2
10 and​

Answer 1

Correct Question:

Find the zeros of the quadratic polynomial $\sf{8x^{2} - 4}$ and verify the relation between the zeros and the coefficients.

Answer:-

$\sf{\frac{-1}{\sqrt2} \ and \ \frac{1}{\sqrt2} \ are \ the \ roots}$

$\sf{of \ the \ given \ quadratic \ polynomial. }$

Given:

• The given quadratic polynomial is $\sf{8x^{2}-4.}$

To find:

• Zeroes of the polynomial.

Solution:-

$\sf{The \ given \ quadratic \ polynomial \ is}$

$\sf{\implies{8x^{2}-4}}$

$\sf{\implies{(2\sqrt2x)^{2}-2^{2}}}$

$\sf{According \ to \ the \ identity}$

$\boxed{\sf{a^{2}-b^{2}=(a+b)(a-b)}}$

$\sf{\implies{(2\sqrt2x+2)(2\sqrt2-2)}}$

$\sf{\implies{x=\frac{-2}{2\sqrt2} \ or \ x=\frac{2}{2\sqrt2}}}$

$\sf{\implies{x=\frac{-1}{\sqrt2} \ or \frac{1}{\sqrt2}}}$

$\sf\purple{\tt{\therefore{\frac{-1}{\sqrt2} \ and \ \frac{1}{\sqrt2} \ are \ the \ roots}}}$

$\sf\purple{\tt{of \ the \ given \ quadratic \ polynomial. }}$

_______________________________________

$\sf\blue{Verification:}$

$\sf{Let \ \alpha \ be \ \frac{-1}{\sqrt2} \ and \ \beta \ be \ \frac{1}{\sqrt2}}$

$\sf{The \ given \ quadratic \ polynomial \ is}$

$\sf{\implies{8x^{2}-4}}$

$\sf{Here, \ a=8, \ b=0 \ and \ c=-4}$

$\sf{\alpha+\beta=\frac{-1}{\sqrt2}+\frac{1}{\sqrt2}}$

$\sf{\therefore{\alpha+\beta=0...(1)}}$

$\sf{\frac{-b}{a}=0...(2)}$

$\sf{...from \ (1) \ and \ (2)}$

$\boxed{\sf{Sum \ of \ zeroes=\frac{-b}{a}}}$

$\sf{\alpha\beta=\frac{-1}{\sqrt2}\times\frac{1}{\sqrt2}}$

$\sf{\therefore{\alpha\beta=\frac{-1}{2}...(3)}}$

$\sf{\frac{c}{a}=\frac{-4}{8}}$

$\sf{\therefore{\frac{c}{a}=\frac{-1}{2}...(4)}}$

$\sf{...from \ (3) \ and \ (4)}$

$\boxed{\sf{Product \ of \ zeroes=\frac{c}{a}}}$

$\sf{Hence, \ verified. }$