SAT Math Question (Thanks!)f(x) = x^2 - 2x + 8g(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: [tex]\sqrt{23}[/tex].Step-by-step explanation:So we are given the equations are equal for [tex]x=5\pm a[/tex].If the functions are equal for those values then their difference is zero for those values:[tex]f-g=0[/tex][tex](x^2- 2x+8)-(-x^2 +18x +4)=0[/tex]Combine like terms; keep in mind we are subtracting over the parenthesis:[tex]2x^2-20x+4=0[/tex]Since all terms have a common factor of 2, then divide both sides by 2:[tex]x^2-10x+2=0[/tex]When compared to [tex]ax^2+bx+c=0[/tex] we should see:[tex]a=1[/tex][tex]b=-10[/tex][tex]c=2[/tex]Now use quadratic formula.[tex]x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]Plug in the values we found:[tex]x=\frac{10 \pm \sqrt{(-10)^2-4(1)(2)}}{2(1)}[/tex]Let's simplify:[tex]x=\frac{10 \pm \sqrt{100-8}}{2}[/tex][tex]x=\frac{10 \pm \sqrt{92}}{2}[/tex]92 isn't a perfect square but contains a factor that is; 92=4(23):[tex]x=\frac{10 \pm \sqrt{4} \sqrt{23}}{2}[/tex][tex]x=\frac{10 \pm 2\sqrt{23}}{2}[/tex]Divide top and bottom by since all three terms have a common factor 2:[tex]x=\frac{10 \pm 2\sqrt{23}}{1}[/tex][tex]x=5 \pm \sqrt{23}[/tex]So f and g are equal for: [tex]x=5 \pm \sqrt{23}[/tex]When compared to [tex]x=5\pm a[/tex] we see that [tex]\sqrt{23}[/tex].