Question

find the range of values for rhe the following quadratic equation:-
3-(4x-3)²​

Answer 1

Answer:

3-(4x_3)2

3- (4x)2 -2×4x×3 +(3)2

3-2x2 -24x+9

Answer 2

$(\sqrt{3} - 4x - 3 )= 0 \\and (\sqrt{3} + 4x + 3 ) = 0$Answer:

Quadratic equation:-

Step-by-step explanation:

   3 - $(4x -3)^{2}$

 = $\sqrt({3)} ^{2} - (4x - 3)^{2}$                                ( $\sqrt{3} * \sqrt{3} = 3$

 = $[\sqrt{3} - (4x+3)] [\sqrt{3} - (4x -3)]$             ( $using identity: a^{2} - b^{2} = ( a-b) (a+b)$

 = $(\sqrt{3} - 4x - 3 ) (\sqrt{3} + 4x + 3 )$

     $(\sqrt{3} - 4x - 3 ) (\sqrt{3} + 4x + 3 )$ = 0  ( to find zeros)

 $(\sqrt{3} - 4x - 3 ) = 0 - and - (\sqrt{3} + 4x + 3 ) = 0$

$-4x = 3 - \sqrt{3} \\4x = -3 - \sqrt{3}$

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