M30-13

I see what you wrote, but I don't get it. If i substract -9 from 2x²-10x-3=0, I get 2x²-10x-12=-9 i can write it like this: (2x+2)*(x-6)=-9
If I try to insert 5 here, it doesn't work.

How can this be correct then, since my way is a more standard way of doing this, which has worked for me on other GMAT questions with similar difficulty. I usually had to add 1 or 2 to both sides in most cases, but this formula does seem to work for other cases than ax²+bx+c=0.

Do you personally think studying this formula is necessary for the gmat?

SolidGod

Inserting 5 into an equation will not work of course, because 5 is the sum of all possible solutions for 2m^2-10m-3=0, not one particular solution to the equation.

for example, for (x-2)(x+3)=0, sum of all possible solutions is = -3+2 = -1.
if you plug-in either -3 or +2 to the equation, you will get a valid answer. However, if you plug-in -1 to the equation, this will not work. What you are essentially doing is plugging in -1 instead of individual solutions such as -3 or 2.

With that said, i think the theorm may be helpful as it can be applied to other similar topics as well.

Hope it helps!

I hate using the quadratic equation, but I solved it via [-(-10) +- root(10^2- 4*2*-3)]/2*2 = [10 +- root(124)]/4

root(124) ~ 11 so (10+11)/4 + (10-11)/4

(21/4) - (1/4) = 20/4; answer is 5

I have never heard of this theorem before and I have been studying quite a while. Is this common?

Hi Bunuel,

Just to get this straight, I worked the solution up to 2m^2-10m-3

I couldn't work it out as a quadratic from here...does that mean that the use of the theorem was the only way of solving this - if so what conditions have to be met for one to use the viete theorem?

Additionally, was there a more conventional way of solving this question? If so...please how can it be solved more conventionally without reverting to the theorem.

Tosin

You could solve 2m^2-10m-3=0 to get two values of m, and get their sum.

Hi Bunuel

how to solve 2m^2 - 10m - 3 to find the two values of m
product of factors has to equal -6 while sum should equal -10

Does not seem possible at all. Please help

Thanks
Rahul

Factoring Quadratics
Solving Quadratic Equations

Check more links on Algebra above.

I think this is a high-quality question and I agree with explanation. I didn't remember for the formula for the sum of two roots but I remembered the formula for each root lol

I think this is a high-quality question and I agree the with the explanation. I had forgotten the formula for sum of roots so this jogged my memory.

If you are unfamiliar with Viete's theorem and forget the quadratic formula (since there are so many ways you can slip up with such a lengthy formula), you can solve by completing the square.

\(2m^2 - 10m - 3 = 0\)

\(2m^2 - 10m = 3\)

\(m^2 - 5m = \frac{3}{2}\)

\(m^2 - 5m + (\frac{-5}{2})^2 = \frac{3}{2} + (\frac{-5}{2})^2\)

\(m^2 - 5m + \frac{25}{4} = \frac{3}{2} + \frac{25}{4}\)

\((m - \frac{5}{2})^2 = \frac{6}{4} + \frac{25}{4}\)

\(\sqrt{(m - \frac{5}{2})^2} = \sqrt{\frac{31}{4}}\)

\(m - \frac{5}{2} = +/- \frac{\sqrt{31}}{2}\)

\(m = \frac{5}{2} +/- \frac{\sqrt{31}}{2}\)

Now for the sum:

\(\frac{(5 + \sqrt{31}) + (5 - \sqrt{31})}{2}\)

\(\frac{10}{2}\)

\(5\)

The answer must be (D). I hope this solution may prove helpful to some in the community.

- Andrew

hey AndrewN hope you are well can you pls explain how did you get the highlighted part ?

Hello, dave13. Thank you for the well wishes. Remember, when you solve a quadratic, say, using the quadratic formula, you get both a positive and a negative solution. In this problem, those two solutions were given in the step just above the part you highlighted:

\(m = \frac{5}{2} +/- \frac{\sqrt{31}}{2}\)

We can tease the two apart to do away with the +/-:

\(m = \frac{5}{2} + \frac{\sqrt{31}}{2}\) OR

\(m = \frac{5}{2} - \frac{\sqrt{31}}{2}\)

When we sum the two solutions, we can either keep the two fractions apart or combine them, since they have the same denominator. First, separately:

\(m = (\frac{5}{2} + \frac{\sqrt{31}}{2}) + (\frac{5}{2} - \frac{\sqrt{31}}{2})\)

Note that the roots cancel out, so we are left with

\(\frac{5}{2} + \frac{5}{2} = \frac{10}{2} = 5\)

Now, as explained above, we could also add the numerators over the common denominator in one fraction, rather than separate the two, and that is exactly what I did in my earlier post:

\(\frac{(5 + \sqrt{31}) + (5 - \sqrt{31})}{2} = \frac{10}{2} = 5\)

Perhaps that makes more sense now. Thank you for thinking to ask.

- Andrew

Great question, even without the theorem its calm because the (delta)^1/2 of each root equation cancel each other out.

Still gave me a mini panic attack though.

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.