How do you solve sqrt(26x-39)=x+5√26x−39=x+5?

2 Answers
Don't Memorise
Apr 16, 2015

We can square both sides to get

(sqrt(26x-39))^2=(x+5)^2(√26x−39)2=(x+5)2

26x-39 = x^2 + 10x + 2526x−39=x2+10x+25
(Used the Identity color(blue)((a + b)^2 = a^2 + 2ab + b^2(a+b)2=a2+2ab+b2

Transposing the terms from the left hand side to the right, we get:

0 = x^2 + 10x - 26x + 25 + 390=x2+10x−26x+25+39

0 = x^2 - 16x + 640=x2−16x+64

x^2 - 16x + 64 = 0x2−16x+64=0

(x - 8) ^2 = 0(x−8)2=0

color(green)(x = 8x=8

Verfiy your answer :

Left Hand Side = sqrt(26*8 - 39) = sqrt 169 = 13√26⋅8−39=√169=13

Right Hand Side = 8 + 5 = 138+5=13

Hence our solution is correct.

Gió
Apr 16, 2015

Have a look:
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