2 years ago

Please help me with this geometry question and show work

Mathematics

1 answer:

soldi70 [24.7K]2 years ago

8 0

(3) 8

Step-by-step explanation:

Note that ∆ADE is similar to ∆ABC. As such, the ratios of their legs are equal. Also note that AD = AB + BD.

BC/AB = DE/AD

BC/10 = 12/(10 + 5) = 12/15 = 4/5

BC = 10(4/5) = 8

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Answer:

$x=4\sqrt{3},\:x=-4\sqrt{3}$ are the roots.

Step-by-step explanation:

$x=4\sqrt{3},\:x=-4\sqrt{3}$

Considering the expression

$x^2\:-\:48$

Solving

$x^2-48=0$

$x^2-48+48=0+48$

$x^2=48$

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=\sqrt{48},\:x=-\sqrt{48}$

Solving

$x=\sqrt{48}$

= $\sqrt{2^4\cdot \:3}$

$\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$

= $\sqrt{3}\sqrt{2^4}$

$\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a^m}=a^{\frac{m}{n}}$

$\sqrt{2^4}=2^{\frac{4}{2}}=2^2$

$=2^2\sqrt{3}$

$=4\sqrt{3}$

So,

$x=4\sqrt{3}$

Similarly,

$x=-\sqrt{48}=-4\sqrt{3}$

Therefore, $x=4\sqrt{3},\:x=-4\sqrt{3}$ are the roots.

Keywords: roots, expression

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