Positive root of the following equation:

34*x^2 + 68*x - 510

Recall:

a*x^2 + b*x + c

x1 = ( - b + sqrt ( b*b - 4*a*c ) ) / ( 2*a)

I have no idea how to go about doing this. Can someone help me out?

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- Jun 27th 2012, 07:56 AMgregzaj1Find the Positive Root Of The Following EquationPositive root of the following equation:

34*x^2 + 68*x - 510

Recall:

a*x^2 + b*x + c

x1 = ( - b + sqrt ( b*b - 4*a*c ) ) / ( 2*a)

I have no idea how to go about doing this. Can someone help me out?

- Jun 27th 2012, 08:02 AMrichard1234Re: Find the Positive Root Of The Following Equation
Divide both sides of the equation $\displaystyle 34x^2 + 68x - 510 = 0$ by 34.

$\displaystyle x^2 + 2x - 15 = 0$

By the quadratic formula, $\displaystyle x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-15)}}{2(1)}$

$\displaystyle =\frac{-2 \pm \sqrt{64}}{2} = \frac{-2 \pm 8}{2}$

The positive root is $\displaystyle \frac{-2 + 8}{2} = 3$.

Alternatively, $\displaystyle x^2 + 2x - 15 = 0$ may be factored: $\displaystyle (x+5)(x-3) = 0$. The roots are -5 and 3. - Jun 27th 2012, 09:25 AMHallsofIvyRe: Find the Positive Root Of The Following Equation

First, you understand that this is NOT an equation, don't you? I think you meant $\displaystyle 34x^2+ 68x- 510= 0$ which is an equation because it has "=" but I had to guess at the "0"

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Recall:

a*x^2 + b*x + c

Now, compare that with your $\displaystyle 34x^2+ 68x- 510= 0$. You should see that a= 34, b= 68, and c= -510

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x1 = ( - b + sqrt ( b*b - 4*a*c ) ) / ( 2*a)

So "b*b- 4*a*c" is (68)(68)- 4(34)(-510)= 4624+ 69360= 73984 and the square root is 272.

-b+sqrt(b*b- 4*a*c)= -68+ 272= 204. 2*a= 2*34= 68 so x= 204/68= 3.

That wasn't so hard was it?

(If you are very clever, like richard1234, you might recognise that 68= 2(34) and 510= 15(34) so dividing through by 34 at the start simplifies the numbers.)

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I have no idea how to go about doing this. Can someone help me out?