Question

SAT Math Question (Thanks!)

f(x) = x^2 - 2x + 8
g(x) = -x^2 +18x +4

The two functions defined above are equal to each other when x=5+a or x=5-a, where a is a constant. What is the value of a?

Answer 1

Answer:

$\sqrt{23}$.

Step-by-step explanation:

So we are given the equations are equal for $x=5\pm a$.

If the functions are equal for those values then their difference is zero for those values:

$f-g=0$

$(x^2- 2x+8)-(-x^2 +18x +4)=0$

Combine like terms; keep in mind we are subtracting over the parenthesis:

$2x^2-20x+4=0$

Since all terms have a common factor of 2, then divide both sides by 2:

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

When compared to $ax^2+bx+c=0$ we should see:

$a=1$

$b=-10$

$c=2$

Now use quadratic formula.

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Plug in the values we found:

$x=\frac{10 \pm \sqrt{(-10)^2-4(1)(2)}}{2(1)}$

Let's simplify:

$x=\frac{10 \pm \sqrt{100-8}}{2}$

$x=\frac{10 \pm \sqrt{92}}{2}$

92 isn't a perfect square but contains a factor that is; 92=4(23):

$x=\frac{10 \pm \sqrt{4} \sqrt{23}}{2}$

$x=\frac{10 \pm 2\sqrt{23}}{2}$

Divide top and bottom by since all three terms have a common factor 2:

$x=\frac{10 \pm 2\sqrt{23}}{1}$

$x=5 \pm \sqrt{23}$

So f and g are equal for:

$x=5 \pm \sqrt{23}$

When compared to $x=5\pm a$ we see that $\sqrt{23}$.