Let x be the positive number such that 2x^2=4x+9
If x can be written in simplified form as \(\dfrac{a + \sqrt{b}}{c}\) such that a, b, and c are positive integers, what is a+b+c?
+364 \(2x^2=4x+9\)
\(2x^2-9=4x\)
\(2x^2-4x-9=0\)
\(x_{1,\:2}=\frac{-\left(-4\right)\pm \sqrt{\left(-4\right)^2-4\cdot \:2\left(-9\right)}}{2\cdot \:2}\)
\(x_{1,\:2}=\frac{-\left(-4\right)\pm \:2\sqrt{22}}{2\cdot \:2}\)
\(x_1=\frac{-\left(-4\right)+2\sqrt{22}}{2\cdot \:2},\:x_2=\frac{-\left(-4\right)-2\sqrt{22}}{2\cdot \:2}\)
\(x=\frac{2+\sqrt{22}}{2},\:x=\frac{2-\sqrt{22}}{2}\)
In this case, it is
\(x=\frac{2+\sqrt{22}}{2}\)
2+22+2=26
so 26 is your answer!
