Question

Q23. Find the zeroes of the polynomial : 4x2 +5V2x-3​

Answer 1

Answer:

$\huge\underline\bold\red{AnswEr}$

$4x {}^{2} + 5 \sqrt{2} x - 3$

$= 4x {}^{2} + 6 \sqrt{2} x - \sqrt{2} x - 3$

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

$= ( \sqrt{2} x + 3)( 2\sqrt{2} x - 1)$

So , the zeros of the polynomial are -----

$( \sqrt{2} x + 3) = 0 \: and \: (2 \sqrt{2} x - 1) = 0$

$x = \frac{ - 3}{ \sqrt{2} } \: and \: x = \frac{1}{2 \sqrt{2} }$

$x = \frac{ - 3 \sqrt{2} }{2} \: and \: x = \frac{ \sqrt{2} }{4}$

Now the polynomial is 4x² + 5√2x - 3

Comparing it with ax² + bx + c = 0 , we get a = 4 , b = 5√2 and c = -3

$sum \: of \: the \: zeros \: = \frac{ - 3 \sqrt{2} }{2} + \frac{ \sqrt{2} }{4}$

$= \frac{ - 6 \sqrt{2} + \sqrt{2} }{4}$

$= \frac{ - 5 \sqrt{2} }{4}$

$= \frac{ - b}{a}$

$product \: of \: the \: zeros \: = ( \frac{ - 3 \sqrt{2} }{2} ) + ( \frac{ \sqrt{2} }{4} )$

$= ( \frac{ - 3}{2} )( \frac{2}{4} )$

$= \frac{ - 3}{4}$

$= \frac{c}{a}$