Let x$ be a 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 $a + b + c$?
+24 uhhh can you put it in latex????????????? wait lemme do that:
Let x be a 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?
+24 ummm so i got
a=4
b=88
c=4
so if you add ummm
aww i cant add in my head..... uhhhhhhhhh 88+8=96?
i got 96. it should be right.....
+9479 2x2 = 4x + 9
Subtract 4x from both sides of the equation.
2x2 - 4x = 9
Subtract 9 from both sides of the equation.
2x2 - 4x - 9 = 0
By the quadratic formula,
x = \({-(-4) \pm \sqrt{(-4)^2-4(2)(-9)} \over 2(2)}\ =\ \frac{4\pm\sqrt{16+72}}{4}\ =\ \frac{4\pm\sqrt{88}}{4}\ =\ \frac{4\pm2\sqrt{22}}{4}\ =\ \frac{2\pm\sqrt{22}}{2}\)
We're given that x is positive so....
x = \(\frac{2+\sqrt{22}}{2}\)
Now it is in simplified form as \(\frac{a+\sqrt{b}}{c}\) where a, b, and c are positive integers.
a + b + c = 2 + 22 + 2 = 26
