How do you solve $\sqrt{4 x + 8} - 1 = x - 2$ $sqrt (4x + 8) - 1 = x - 2$ ?

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Answer 1 · 2016-01-26T04:11:25

1) leave the square root alone on 1 side:
$\sqrt{4 x + 8} = x - 2 + 1 = x - 1$ $sqrt(4x+8) = x-2+1 = x-1$

2) square both sides:
$4 x + 8 = {(x - 1)}^{2}$ $4x+8 = (x-1)^2$

3) Expand the right side: ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $(a+b)^2=a^2+2ab+b^2$
$4 x - 8 = x^{2} - 2 x + 1$ $4x-8 = x^2 -2x+1$

4) Move everything to one side and you get:
$x^{2} - 6 x + 9 = 0$ $x^2 -6x +9 = 0$

5) This. luckily factors out to:
${(x - 3)}^{2} = 0$ $(x-3)^2=0$
so $x = 3$ $x=3$

Answer 2 · 2016-01-31T07:08:42

$x = 7, - 1$ $x=7,-1$

$\sqrt{4 x + 8} - 1 = x - 2$ $sqrt(4x+8)-1=x-2$

$\to \sqrt{4 x + 8} = x - 2 + 1$ $rarrsqrt(4x+8)=x-2+1$

$\to \sqrt{4 x + 8} = x - 1$ $rarrsqrt(4x+8)=x-1$

Square both sides:

$\to {(\sqrt{4 x - 8})}^{2} = {(x - 1)}^{2}$ $rarr(sqrt(4x-8))^2=(x-1)^2$

Use the formula ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $(a+b)^2=a^2+2ab+b^2$ for the equation on the R.h.s

$\to 4 x - 8 = x^{2} - 2 x + 1$ $rarr4x-8=x^2-2x+1$

$\to 0 = x^{2} - 2 x + 1 - 4 x 8$ $rarr0=x^2-2x+1-4x8$

$\to 0 = x^{2} - 6 x - 7$ $rarr0=x^2-6x-7$

Now this is a quadratic equation:Solve using quadratic formula:

$\to x = 7, - 1$ $rarrx=7,-1$

How do you solve √4x+8−1=x−2sqrt (4x + 8) - 1 = x - 2?