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

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Answer 1 · 2015-04-16T11:57:11

We can square both sides to get

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

#26x-39 = x^2 + 10x + 25#
(Used the Identity #color(blue)((a + b)^2 = a^2 + 2ab + b^2#

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

#0 = x^2 + 10x - 26x + 25 + 39#

#0 = x^2 - 16x + 64#

#x^2 - 16x + 64 = 0#

#(x - 8) ^2 = 0#

# color(green)(x = 8#

Verfiy your answer :

Left Hand Side = #sqrt(26*8 - 39) = sqrt 169 = 13#

Right Hand Side = #8 + 5 = 13#

Hence our solution is correct.

Answer 2 · 2015-04-16T12:00:48

Have a look:
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