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# help

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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 $a + b + c$?

May 31, 2019

#1
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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?

May 31, 2019
#2
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ummm so i got

a=4

b=88

c=4

i got 96. it should be right.....

ProffesorNobody  May 31, 2019
#3
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ProffesorNobody  Jun 1, 2019
#6
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$$\sqrt{88}$$   can be simplified to   $$2\sqrt{22}$$   because   $$\sqrt{88}\ =\ \sqrt{4\cdot22}\ =\ \sqrt4\cdot\sqrt{22}\ =\ 2\sqrt{22}$$

But it is true that    x = $$\frac{4+\sqrt{88}}{4}$$ hectictar  Jun 1, 2019
#8
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Thank you. I understnad why i was wrong...

ProffesorNobody  Jun 1, 2019
#5
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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

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

Jun 1, 2019
#7
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i used the quadratic formula..... what did i do wrong

ProffesorNobody  Jun 1, 2019
edited by ProffesorNobody  Jun 1, 2019
#9
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$$\begin{array}{rcll} 2x^2 &=& 4x + 9\\ 2x^2 - 4x - 9 &=& 0\\ x &=&\dfrac{4\pm\sqrt{16+4\cdot2\cdot9}}{4}\\ x &=& \dfrac{4\pm2\sqrt{22}}4\\ x &=& \dfrac{2\pm\sqrt{22}}2\\ a + b + c &=&2 + 22 + 2 \\ a + b + c &=& 26 \end{array}$$

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Jun 1, 2019