How do you simplify $\sqrt{x - 1} + \sqrt{2 x} = 3$ $sqrt(x-1) + sqrt( 2x) = 3$ ?

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Answer 1 · 2018-05-06T16:13:34

$\to x = 2$ $rarrx=2$

$\to \sqrt{x - 1} + \sqrt{2 x} = 3$ $rarrsqrt(x-1)+sqrt(2x)=3$

$\to \sqrt{x - 1} = 3 - \sqrt{2 x}$ $rarrsqrt(x-1)=3-sqrt(2x)$

$\to {[\sqrt{x - 1}]}^{2} = {[3 - \sqrt{2 x}]}^{2}$ $rarr[sqrt(x-1)]^2=[3-sqrt(2x)]^2$

$\to x - 1 = 9 - 6 \sqrt{2 x} + 2 x$ $rarrx-1=9-6sqrt(2x)+2x$

$\to 6 \sqrt{2 x} = x + 10$ $rarr6sqrt(2x)=x+10$

$\to {[6 \sqrt{2 x}]}^{2} = {[x + 10]}^{2}$ $rarr[6sqrt(2x)]^2=[x+10]^2$

$\to 36 \cdot (2 x) = x^{2} + 20 x + 100$ $rarr36*(2x)=x^2+20x+100$

$\to x^{2} - 52 x + 100 = 0$ $rarrx^2-52x+100=0$

$\to x^{2} - 2 \cdot x \cdot 26 + 26^{2} - 26^{2} + 100 = 0$ $rarrx^2-2*x*26+26^2-26^2+100=0$

$\to {(x - 26)}^{2} = 26^{2} - 100 = 576$ $rarr(x-26)^2=26^2-100=576$

$\to x - 26 = \sqrt{576} = \pm 24$ $rarrx-26=sqrt(576)=+-24$

$\to x = 26 + 24, 26 - 24 = 50 or 2$ $rarrx=26+24,26-24=50 or 2$

Putting $x = 50$ $x=50$ in given equation, we get,

$\to \sqrt{50 - 1} + \sqrt{2 \cdot 50} = 17 (r e j e c t e d)$ $rarrsqrt(50-1)+sqrt(2*50)=17(rejected)$

Putting $x = 2$ $x=2$ in given equation, we get,

$rarrsqrt(2-1)+sqrt(2*2)=3(accepted)$

So, the required value of x is $2.$

How do you simplify sqrt(x-1) + sqrt( 2x) = 3√x−1+√2x=3sqrt(x-1) + sqrt( 2x) = 3?