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!