How do solve $x^2+21>10x$ and write the answer as a inequality and interval notation?

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Answer 1 · 2016-11-06T22:59:30

Rewrite as a quadratic equation, solve, and use test points to see in what interval the solution lies.

$x^2 + 21 = 10x$

$x^2 - 10x + 21 = 0$

$(x- 7)( x - 3) = 0$

$x = 7 and 3$

TEST POINT 1: $x= 2$

$2^2 + 21 >^? 10(2)$

$25 > 20" "color(green)(√)$

TEST POINT 2: $x = 4$

$4^2 + 21 >^? 10(4)$

$37 >^O/ 40" "color(red)(xx)$

TEST POINT 3: $x =8$

$8^2 + 21>^?10(8)$

$85 > 80" "color(green)(√)$

So, the solution is $x < 3 and x > 7$ .

In interval notation, this would be $( -oo, 3)$ and $(7, oo)$ .

Hopefully this helps!

How do solve x^2+21>10xx^2+21>10x and write the answer as a inequality and interval notation?