How to solve 3√(2x-3) + 2√(7-x) =11 and

Question

also √(7-2x) - √(5+x) = √(4+3x) ?

Answer 1

Given that 3√(2x - 3) + 2√(7 - x) = 11

 By squaring on both sides we get,        

9(2x - 3) + 12*√((2x - 3)(7 - x)) + 4(7 - x) = 121              [Since (a+b)^2 = a^2+2ab+b^2]

18x - 27 + 12*√(14x - 2x^2 - 21 + 3x) + 28 - 4x = 121

12*√(-2x^2 + 17x - 21) = 121 - 18x + 27 - 28 + 4x

12*√(-2x^2 + 17x - 21) = 120 - 14x

By squaring both sides we get,

144(-2x^2 + 17x - 21) = 14400 - 3360x + 196x^2

-288x^2 + 2448x - 3024 = 14400 - 3360x + 196x^2

-288x^2 + 2448x - 3024 - 14400 + 3360x - 196x^2 = 0

-484x^2 + 5808x - 17424 = 0

x = (-5808 +/- √(5808^2 - 4(-484)(-17424))) / (2(-484))

x = (-5808 +/- √(33732864 - 33732864)) / (-968)

x = (-5808 +/- √(0)) / (-968)

x = (-5808 +/- 0) / (-968)

x = (-5808) / (-968)

x = 6

Therefore x = 6.

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2)

Given that √(7 - 2x) - √(5 + x) = √(4 + 3x)

 By squaring on both sides we get,  

7 - 2x - 2*√((7 - 2x)(5 + x)) + 5 + x = 4 + 3x                     [Since (a-b)^2 = a^2-2ab+b^2]

-2*√(35 + 7x - 10x - 2x^2) = 4 + 3x - 7 + 2x - 5 - x

-2*√(35 - 3x - 2x^2) = 4x - 8

√(35 - 3x - 2x^2) = (4x - 8) / (-2)

√(35 - 3x - 2x^2) = 4 - 2x

 By squaring on both sides we get,     

35 - 3x - 2x^2 = 16 - 16x + 4x^2

35 - 3x - 2x^2 - 16 + 16x - 4x^2 = 0

-6x^2 + 13x + 19 = 0

6x^2 - 13x - 19 = 0

6x^2 - 19x + 6x - 19 = 0

x(6x - 19) + (6x - 19) = 0

(x + 1)(6x - 19) = 0

x + 1 = 0 or 6x - 19 = 0

x = -1 or 6x = 19

x = -1 or x = 19/6

But if we substitute 19/6 in the given equation, we will get a negative number on the left, and a positive number on the right, so that doesn't work.

Hence the only solution is x = -1