How do you solve $\sqrt{10 x + 6} - \sqrt{x + 1} = \sqrt{8 x - 8}$ $sqrt (10x+6) - sqrt (x+1) = sqrt (8x-8)$ ?

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Answer 1 · 2018-07-20T12:11:41

The solution is $S = {3}$ $S={3}$

The equation is

$\sqrt{10 x + 6} - \sqrt{x + 1} = \sqrt{8 x - 8}$ $sqrt(10x+6)-sqrt(x+1)=sqrt(8x-8)$

Squaring both sides

${(\sqrt{10 x + 6} - \sqrt{x + 1})}^{2} = (8 x - 8)$ $(sqrt(10x+6)-sqrt(x+1))^2=(8x-8)$

$10 x + 6 + x + 1 - 2 \sqrt{(10 x + 6) (x + 1)} = 8 x - 8$ $10x+6+x+1-2sqrt((10x+6)(x+1))=8x-8$

$11 x + 7 - 2 \sqrt{(10 x + 6) (x + 1)} = 8 x - 8$ $11x+7-2sqrt((10x+6)(x+1))=8x-8$

$3 x + 15 = 2 \sqrt{(10 x + 6) (x + 1)}$ $3x+15=2sqrt((10x+6)(x+1))$

Squaring both sides

${(3 x + 15)}^{2} = 4 ((10 x + 6) (x + 1))$ $(3x+15)^2=4((10x+6)(x+1))$

$9 x^{2} + 90 x + 225 = 4 (10 x^{2} + 16 x + 6)$ $9x^2+90x+225=4(10x^2+16x+6)$

$9 x^{2} + 90 x + 225 = 40 x^{2} + 64 x + 24$ $9x^2+90x+225=40x^2+64x+24$

$31 x^{2} - 26 x - 201 = 0$ $31x^2-26x-201=0$

Solving this quadratic equation

$x = \frac{26 \pm \sqrt{26^{2} + 4 \cdot 31 \cdot 201}}{2 \cdot 31}$ $x=(26+-sqrt(26^2+4*31*201))/(2*31)$

$= \frac{26 \pm 160}{62}$ $=(26+-160)/(62)$

$x_{1} = 3$ $x_1=3$

$x_{2} = - 2.16$ $x_2=-2.16$

We keep only $x_{1} = 3$ $x_1=3$

How do you solve √10x+6−√x+1=√8x−8sqrt (10x+6) - sqrt (x+1) = sqrt (8x-8) ?