Math problems from the *free* official practice tests and
problems from mba.com
pmvartanian
Forum Guests
 
Posts: 1
Joined: Thu Jun 28, 2012 2:32 pm
 

Quadratic Equation Help

by pmvartanian Mon Jul 16, 2012 9:44 pm

I have been going through the flash cards and came across this one for EIVs.

Is the statement sufficient?

What are the solutions to the equation x^2 + kx - 10 = 0, where k is a constant?

(1) One of the solutions is -5.

I was able to find the other solution plugging the numbers in.

(x+5) (x-2) since the two numbers in the parentheses need to equal -10 when multiplying.

But I am scratching my head when plugging them into the equation to solve.
RonPurewal
Students
 
Posts: 19744
Joined: Tue Aug 14, 2007 8:23 am
 

Re: Quadratic Equation Help

by RonPurewal Thu Jul 19, 2012 7:31 am

pmvartanian Wrote:I have been going through the flash cards and came across this one for EIVs.

Is the statement sufficient?

What are the solutions to the equation x^2 + kx - 10 = 0, where k is a constant?

(1) One of the solutions is -5.

I was able to find the other solution plugging the numbers in.

(x+5) (x-2) since the two numbers in the parentheses need to equal -10 when multiplying.

But I am scratching my head when plugging them into the equation to solve.


you've actually got the more sophisticated solution here, so, nice job.

you can also solve by just plugging -5 in for x, since it's a solution of the equation (according to what you're told). if you plug that number in, you get 25 - 5k - 10 = 0, or 15 - 5k = 0. therefore, k = 3.
this gives you the equation x^2 + 3x - 10 = 0, which can be factored as you've shown above.

the best thing, of course, is to get both of these solution methods under your belt.