Question

FIND THE ROOT OF THE EQUATION :-

(i) (x-4)^2 + 5^2= 13^2​

Answer 1

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Answer 2

Answer:

$(x - 4) {}^{2} + 5 {}^{2} = 13 {}^{2}$

$x {}^{2} + 4 {}^{2} - 2(x)(4) + 25 = 169$

$x {}^{2} + 16 - 8x + 25 = 169$

$x {}^{2} - 8x + 41 = 169$

$x {}^{2} - 8x = 128$

$x {}^{2} - 8x - 128 = 0$

$x {}^{2} - (16 - 8)x - 128 = 0$

$x {}^{2} - 16x + 8x - 128 = 0$

$x (x - 16) + 8(x - 16) = 0$

$(x + 8)(x - 16) = 0$

$either \:$

$x = - 8$

$or$

$x = 16$