SOLUTION: Which is the solution to this equation? x^2+3x-10=0 a.0 b.-2 c.-5 d.-10

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Question 901085: Which is the solution to this equation?
x^2+3x-10=0
a.0
b.-2
c.-5
d.-10
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)


Looking at the expression x%5E2%2B3x-10, we can see that the first coefficient is 1, the second coefficient is 3, and the last term is -10.



Now multiply the first coefficient 1 by the last term -10 to get %281%29%28-10%29=-10.



Now the question is: what two whole numbers multiply to -10 (the previous product) and add to the second coefficient 3?



To find these two numbers, we need to list all of the factors of -10 (the previous product).



Factors of -10:

1,2,5,10

-1,-2,-5,-10



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to -10.

1*(-10) = -10
2*(-5) = -10
(-1)*(10) = -10
(-2)*(5) = -10


Now let's add up each pair of factors to see if one pair adds to the middle coefficient 3:



First NumberSecond NumberSum
1-101+(-10)=-9
2-52+(-5)=-3
-110-1+10=9
-25-2+5=3




From the table, we can see that the two numbers -2 and 5 add to 3 (the middle coefficient).



So the two numbers -2 and 5 both multiply to -10 and add to 3



Now replace the middle term 3x with -2x%2B5x. Remember, -2 and 5 add to 3. So this shows us that -2x%2B5x=3x.



x%5E2%2Bhighlight%28-2x%2B5x%29-10 Replace the second term 3x with -2x%2B5x.



%28x%5E2-2x%29%2B%285x-10%29 Group the terms into two pairs.



x%28x-2%29%2B%285x-10%29 Factor out the GCF x from the first group.



x%28x-2%29%2B5%28x-2%29 Factor out 5 from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.



%28x%2B5%29%28x-2%29 Combine like terms. Or factor out the common term x-2



===============================================================



Answer:



So x%5E2%2B3%2Ax-10 factors to %28x%2B5%29%28x-2%29.



In other words, x%5E2%2B3%2Ax-10=%28x%2B5%29%28x-2%29.



Note: you can check the answer by expanding %28x%2B5%29%28x-2%29 to get x%5E2%2B3%2Ax-10 or by graphing the original expression and the answer (the two graphs should be identical).


Solved by pluggable solver: Quadratic Formula
Let's use the quadratic formula to solve for x:


Starting with the general quadratic


ax%5E2%2Bbx%2Bc=0


the general solution using the quadratic equation is:


x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29




So lets solve x%5E2%2B3%2Ax-10=0 ( notice a=1, b=3, and c=-10)





x+=+%28-3+%2B-+sqrt%28+%283%29%5E2-4%2A1%2A-10+%29%29%2F%282%2A1%29 Plug in a=1, b=3, and c=-10




x+=+%28-3+%2B-+sqrt%28+9-4%2A1%2A-10+%29%29%2F%282%2A1%29 Square 3 to get 9




x+=+%28-3+%2B-+sqrt%28+9%2B40+%29%29%2F%282%2A1%29 Multiply -4%2A-10%2A1 to get 40




x+=+%28-3+%2B-+sqrt%28+49+%29%29%2F%282%2A1%29 Combine like terms in the radicand (everything under the square root)




x+=+%28-3+%2B-+7%29%2F%282%2A1%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)




x+=+%28-3+%2B-+7%29%2F2 Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


x+=+%28-3+%2B+7%29%2F2 or x+=+%28-3+-+7%29%2F2


Lets look at the first part:


x=%28-3+%2B+7%29%2F2


x=4%2F2 Add the terms in the numerator

x=2 Divide


So one answer is

x=2




Now lets look at the second part:


x=%28-3+-+7%29%2F2


x=-10%2F2 Subtract the terms in the numerator

x=-5 Divide


So another answer is

x=-5


So our solutions are:

x=2 or x=-5