Let #h(x)=12x+x^2#, how do you find a such that h(a)=-27?

2 Answers
Nov 14, 2016

#a= -9 or a= -3#

Explanation:

#h(a) = 12a +a^2 = -27 or a^2 +12a +27 =0 or (a +9)(a+3)=0#. Either #a+9=0 or a+3=0 :. a= -9 or a= -3# [Ans]

Nov 14, 2016

#a=-3 , a=-9#

Explanation:

Express h(x) in terms of a.

That is #h(color(red)(a))=12color(red)(a)+(color(red)(a))^2=12a+a^2#

#h(a)=-27" and " h(a)=12a+a^2#

#"solve " 12a+a^2=-27" to find a"#

since this is a quadratic function, equate to zero.

#rArra^2+12a+27=0#

using the a-c method, we require the product of factors of 27 that also sum to + 12. These are +3 and +9.

#rArr(a+3)(a+9)=0#

solve : #a+3=0rArra=-3#

solve : #a+9=0rArra=-9#

Check :

#a=-3rArr12xx(-3)+(-3)^2=-36+9=-27color(white)(x)#

#a=-9rArr12xx(-9)+(-9)^2=-108+81=-27#

#rArra=-3,a=-9" are the solutions"#