Simple Quadratic Equation(s)

**Raj** · October 12th 2007, 08:59 AM

Determine the roots of the equations, use any method (complete sq, factor, quad. formula) :

A) $50x^2-2x-77=0$

I came up with the below using the quadratic formula. Just need to know how to get it to its simplest form.

$(2+{\sqrt{15404}})/(100)$

B) $(x-1/x+4) - (x/x-3) = 9$

$(x-1)(x-3)-(x)(x+4) = 9(x+4)(x-3)$
$(x^2-4x+3)-(x^2+4x) = 9(x^2+x-12)$
$(x^2-4x+3)-(x^2+4x) = (9x^2+9x-108)$
New Equation: $9x^2+9x-111 = 0$ ?

If somone could check if i did the algebra right for the new equation i would be thankfull.

**earboth** · October 12th 2007, 09:44 AM

Originally Posted by Raj

Determine the roots of the equations, use any method (complete sq, factor, quad. formula) :

A) $50x^2-2x-77=0$

I came up with the below using the quadratic formula. Just need to know how to get it to its simplest form.

$(2+{\sqrt{15404}})/(100)$

B) $(x-1/x+4) - (x/x-3) = 9$

$(x-1)(x-3)-(x)(x+4) = 9(x+4)(x-3)$
$(x^2-4x+3)-(x^2+4x) = 9(x^2+x-12)$
$(x^2-4x+3)-(x^2+4x) = (9x^2+9x-108)$
New Equation: $9x^2+9x-111 = 0$ ?

If somone could check if i did the algebra right for the new equation i would be thankfull.

Hello,

A) there is missing the solution $x = (2-{\sqrt{15404}})/(100)$

B) You did all the calculations really fine except the last line:

$(x^2-4x+3)-(x^2+4x) = (9x^2+9x-108)~\iff~-8x+3 = 9x^2+9x-108$ . that means you have to solve for x:

$9x^2+17x-111 = 0$ . I leave this for you

**topsquark** · October 12th 2007, 10:13 AM

Originally Posted by Raj

$(2+{\sqrt{15404}})/(100)$

As earboth said you are missing the other solution.

The question remains though, if this is in its simplest form.

$\frac{2 \pm \sqrt{15404}}{100} = \frac{2 \pm \sqrt{4 \cdot 3851}}{100}$

$= \frac{2 \pm 2 \sqrt{3851}}{100} = \frac{1 \pm \sqrt{3851}}{50}$

By the way. A quick way to note that 15404 is divisible by 4 is to look at the last two digits. If this number is divisible by 4 then the whole number is divisible by 4.

-Dan

**Raj** · October 12th 2007, 12:37 PM

@Earboth

Thankyou, i wasn't sure if i needed to distribute the (-) to the +4x.

@Topsquark

Yea i still dont know all of the math tags so i left out the $\pm$ . Thanks for confirming the answer, and the tip

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