Answer :
To find the value of [tex]\( a \)[/tex] such that [tex]\( f(a) = 2 \)[/tex] for the function [tex]\( f(x) = x^2 + 16x + 62 \)[/tex], we need to solve the equation:
[tex]\[ f(a) = a^2 + 16a + 62 = 2 \][/tex]
Let's go step-by-step:
1. Start by setting the function equal to 2:
[tex]\[
a^2 + 16a + 62 = 2
\][/tex]
2. Subtract 2 from both sides to set the equation to zero:
[tex]\[
a^2 + 16a + 62 - 2 = 0
\][/tex]
[tex]\[
a^2 + 16a + 60 = 0
\][/tex]
3. Now, you have a quadratic equation in the form of [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 16 \)[/tex], and [tex]\( c = 60 \)[/tex].
4. To solve the quadratic equation, you can use the quadratic formula:
[tex]\[
a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
5. Plug in the values for [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
- [tex]\( b^2 - 4ac = 16^2 - 4 \times 1 \times 60 \)[/tex]
- [tex]\( b^2 - 4ac = 256 - 240 = 16 \)[/tex]
6. Solve using the quadratic formula:
[tex]\[
a = \frac{-16 \pm \sqrt{16}}{2}
\][/tex]
[tex]\[
a = \frac{-16 \pm 4}{2}
\][/tex]
7. Calculate the two possible solutions:
- [tex]\( a = \frac{-16 + 4}{2} = \frac{-12}{2} = -6 \)[/tex]
- [tex]\( a = \frac{-16 - 4}{2} = \frac{-20}{2} = -10 \)[/tex]
Therefore, the values of [tex]\( a \)[/tex] that satisfy [tex]\( f(a) = 2 \)[/tex] are [tex]\( a = -10 \)[/tex] and [tex]\( a = -6 \)[/tex].
[tex]\[ f(a) = a^2 + 16a + 62 = 2 \][/tex]
Let's go step-by-step:
1. Start by setting the function equal to 2:
[tex]\[
a^2 + 16a + 62 = 2
\][/tex]
2. Subtract 2 from both sides to set the equation to zero:
[tex]\[
a^2 + 16a + 62 - 2 = 0
\][/tex]
[tex]\[
a^2 + 16a + 60 = 0
\][/tex]
3. Now, you have a quadratic equation in the form of [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 16 \)[/tex], and [tex]\( c = 60 \)[/tex].
4. To solve the quadratic equation, you can use the quadratic formula:
[tex]\[
a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
5. Plug in the values for [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
- [tex]\( b^2 - 4ac = 16^2 - 4 \times 1 \times 60 \)[/tex]
- [tex]\( b^2 - 4ac = 256 - 240 = 16 \)[/tex]
6. Solve using the quadratic formula:
[tex]\[
a = \frac{-16 \pm \sqrt{16}}{2}
\][/tex]
[tex]\[
a = \frac{-16 \pm 4}{2}
\][/tex]
7. Calculate the two possible solutions:
- [tex]\( a = \frac{-16 + 4}{2} = \frac{-12}{2} = -6 \)[/tex]
- [tex]\( a = \frac{-16 - 4}{2} = \frac{-20}{2} = -10 \)[/tex]
Therefore, the values of [tex]\( a \)[/tex] that satisfy [tex]\( f(a) = 2 \)[/tex] are [tex]\( a = -10 \)[/tex] and [tex]\( a = -6 \)[/tex].
