How do you solve $\sqrt{2 x - 2} - \sqrt{x} + 3 = 4$ $sqrt(2x-2) - sqrtx + 3 = 4$ ?

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Answer 1 · 2016-02-10T21:42:07

$x = 9$ $x=9$

First thing, determine the dominion:

$2 x - 2 > 0 and x \geq 0$ $2x-2>0 and x>=0$

$x \geq 1 and x \geq 0$ $x>=1 and x>=0$

$x \geq 1$ $x>=1$

The standard way is to put one root in each side of the equality and calculate the squares:

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

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

squaring:

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

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

Now, you have only one root. Isolate it and square it again:

$x - 3 = 2 \sqrt{x}$ $x-3=2sqrt(x)$ ,
We must remember that $2 \sqrt{x} \geq 0$ $2sqrt(x)>=0$ then $x - 3 \geq 0$ $x-3>=0$ also.

This means the dominion has changed to $x \geq 3$ $x>=3$

squaring:

$x^{2} - 6 x + 9 = 4 x$ $x^2-6x+9=4x$

$x^{2} - 10 x + 9 = 0$ $x^2-10x+9=0$

$x = \frac{10 \pm \sqrt{10^{2} - 4 \cdot 9}}{2}$ $x=(10+-sqrt(10^2-4*9))/2$

$x = \frac{10 \pm \sqrt{64}}{2}$ $x=(10+-sqrt(64))/2$

$x = \frac{10 \pm 8}{2}$ $x=(10+-8)/2$

$x = 5 \pm 4$ $x=5+-4$

$x = 9 or x = 1$ $x=9 or x=1$ ,

Only the solution $x = 9$ $x=9$ is valid.

How do you solve √2x−2−√x+3=4sqrt(2x-2) - sqrtx + 3 = 4?