Solve x=2√(x-4) + 4.

Mathematics

1 answer:

stich3 [128]3 years ago

8 0

Answer:x=4,8

Step-by-step explanation:

Given

$x=2\sqrt{x-4}+4$

$x-4=2\sqrt{x-4}$

Squaring both sides

$(x-4)^2=4(x-4)$

$x^2+16-8x=4x-16$

$x^2-12x+32=0$

$x=\frac{12\pm \sqrt{12^2-4\times 32}}{2}$

$x=\frac{12\pm 4}{2}$

$x=4,8$

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Sedaia [141]

Check the picture below.

so let's find the lengths of those two sides in red, since are the length and width of the rectangle.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-6}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{-3}~,~\stackrel{y_2}{6})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d = \sqrt{[-3-(-6)]^2+[6-3]^2}\implies d=\sqrt{(-3+6)^2+(6-3)^2} \\\\\\ d=\sqrt{9+9}\implies \boxed{d=\sqrt{18}} \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-6}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-1})~\hfill d=\sqrt{[-2-(-6)]^2+[-1-3]^2} \\\\\\ d=\sqrt{(-2+6)^2+(-1-3)^2}\implies d=\sqrt{16+16}\implies \boxed{d=\sqrt{32}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of the rectangle}}{(\sqrt{18})(\sqrt{32})}\implies \sqrt{18\cdot 32}\implies \sqrt{576}\implies 24$