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Math - OG 38.

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by odod » Mon Nov 09, 2009 3:34 pm
If 4 is one solution of the equation X^2 +3x +k =10, where k is constant, what is the other solution?

A)-7
B) -4
C) -3
D) 1
E)6

I took out the page that explains the answer. The answer is A. PLEASE explain
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Re: Math - OG 38.

by life is a test » Mon Nov 09, 2009 6:02 pm
odod wrote:If 4 is one solution of the equation X^2 +3x +k =10, where k is constant, what is the other solution?

A)-7
B) -4
C) -3
D) 1
E)6

I took out the page that explains the answer. The answer is A. PLEASE explain
try to factor the eqn to see the two solutions to the eqns. You know 4 is one solution hence one of the brackets will have (x-4) and you also know that 3x is part of the equation so you need another bracket which will multiply out to give 3x; only (x+7) will work since (x-4) (x+7) = x^2 + 3x... hence the other solution must be x=-7.

hope that helps.
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by papgust » Mon Nov 09, 2009 11:33 pm
You could also substitue 4 in the equation as you already know that 4 is one solution.

You get k=-18. Sub k in eqn, you get this - x^2+3x-28=0.

Factorize and you will find that -7 is another solution.
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by Scott@TargetTestPrep » Wed May 20, 2015 4:49 am
odod wrote:If 4 is one solution of the equation X^2 +3x +k =10, where k is constant, what is the other solution?

A)-7
B) -4
C) -3
D) 1
E)6

I took out the page that explains the answer. The answer is A. PLEASE explain
Solution:

The phrase “4 is one solution of the equation” means that one value of x is 4. Thus, we first must plug 4 for x into the given equation to determine the value of k. So we have

4^2 + (3)(4) + k = 10

16 + 12 + k = 10

28 + k = 10

k = -18

Next we plug -18 into the given equation for k and then solve for x.

x^2 + 3x – 18 = 10

x^2 + 3x – 28 = 0

(x+7)(x-4) = 0

x = -7 or x = 4

Thus, -7 is the other solution.

The answer is A

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