Question

Find x Value.
under square root of 4x-3 + under square root of 2x+3 = 6

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Answer 1

Answer:

3

Step-by-step explanation:

$\sqrt{(4x-3)}$ + $\sqrt{2x+3}$ = 6

Squaring on both sides and using $(a+b)^{2}$ formula on LHS we have

(4x-3)+(2x+3)+2$\sqrt{(4x-3)*(2x+3)}$ = 36

6x+2$\sqrt{8x^{2}+6x-9}$ = 36

2$\sqrt{8x^{2}+6x-9}$ = 36-6x

taking squares again on both sides

4($8x^{2}$+6x-9) = $36x^{2}$-432x+1296

$36x^{2}$ - $32x^{2}$ -432x-24x+1296+36 = 0

$4x^{2}$-456x+1332 = 0

Dividing by 4 on both sides

$x^{2}$-114x+333 = 0

Solving the quadratic equation we have

(x-111)*(x-3) = 0

So the possible values for x are 111 and 3

Putting 111 in the original equation and solving we have

$\sqrt{441}$+$\sqrt{228}$ ≠ 6

Putting 3 in the original equation we have

$\sqrt{9} +\sqrt{9} = 3+3 = 6$

Hence value of x = 3