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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?

 Feb 21, 2022
 #1
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\(2x^2=4x+9\)

\(2x^2-9=4x\)

\(2x^2-4x-9=0\)

\(x_{1,\:2}=\frac{-\left(-4\right)\pm \sqrt{\left(-4\right)^2-4\cdot \:2\left(-9\right)}}{2\cdot \:2}\)

\(x_{1,\:2}=\frac{-\left(-4\right)\pm \:2\sqrt{22}}{2\cdot \:2}\)

\(x_1=\frac{-\left(-4\right)+2\sqrt{22}}{2\cdot \:2},\:x_2=\frac{-\left(-4\right)-2\sqrt{22}}{2\cdot \:2}\)

\(x=\frac{2+\sqrt{22}}{2},\:x=\frac{2-\sqrt{22}}{2}\)

In this case, it is

\(x=\frac{2+\sqrt{22}}{2}\)

2+22+2=26

so 26 is your answer!

 Feb 21, 2022
 #2
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Thank you!

Guest Feb 24, 2022

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