If JK=20-x^2, KL=2-x, and JL=10 find x

1 answer:

6 0

$|JK|=20-x^2;\ |KL|=2-x;\ |JL|=10\\D:20-x^2 > 0\ and\ 2-x > 0\ and\ 20-x^2 < 10\ and\ 2-x < 10\\x\in(-\sqrt{20};\ \sqrt{20})\ and\ x < 2\ and\\x\in(-\infty; -\sqrt{10})\ \cup\ (\sqrt{10};\ \infty)\ and\ x >-8\\\\x\in(-\sqrt{20};-\sqrt{10})\\\\|JL|=|JK|+|KL|\\\\20-x^2+2-x=10\\\\-x^2-x+22=10\ \ \ \ \ |subrtact\ 10\ from\ both\ sides\\\\-x^2-x+12=0\\\\-(x^2+x-12)=0\\\\x^2+x-12=0\\\\x^2+4x-3x-12=0\\\\x(x+4)-3(x+4)=0\\\\(x+4)(x-3)=0\iff x+4=0\ or\ x-3=0\\\\x=-4\in D\ or\ x=3\notin D\\\\Answer:\boxed{x=-4}$

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The number after the tenths place is 5 (which is either greater than or equal to 5)

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