Question

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

Answer 1

`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]

Answer 2

`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"`