Question

What method would be best to use 53=4(x-3)^2-11

Answer 1

Answer:

x = 7 , -1

Step-by-step explanation:

SOLUTION :-

$4(x-3)^2-11 = 53$

• Add 11 to both the sides.

$=> 4(x-3)^2-11+11=53+11$

$=> 4(x-3)^2 = 64$

• Divide both the sides by 4.

$=> \frac{4(x-3)^2}{4} = \frac{64}{4}$

$=> (x-3)^2 = 16$

• Root square both the sides.

$=> \sqrt{(x-3)^2} = \sqrt{16}$

$=> x-3 = +4 \; or -4$

Here , x will have two values -

1) $x-3 = 4$

$=> x = 4 + 3 = 7$

2) $x - 3 = -4$

$=> x = -4 + 3 = -1$

VERIFICATION :-

When x = 7 ,

$4(x-3)^2 - 11 = 4(7 - 3)^2 - 11$

                       $= 4 \times 4^2 - 11$

                       $= 4 \times 16 - 11$

                       $= 64 - 11$

                       $= 53$

When x = -1 ,

$4(x-3)^2 - 11 = 4(-1 - 3)^2 - 11$

                       $= 4 \times (-4)^2 - 11$

                       $= 4 \times 16 - 11$

                       $= 64 - 11$

                       $= 53$