SOLUTION: please can you show me a method of how to work out this equation x^2+9x+14=0

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Question 122834: please can you show me a method of how to work out this equation x^2+9x+14=0
Found 2 solutions by jim_thompson5910, Earlsdon:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
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%2B9%2Ax%2B14=0 ( notice a=1, b=9, and c=14)




x+=+%28-9+%2B-+sqrt%28+%289%29%5E2-4%2A1%2A14+%29%29%2F%282%2A1%29 Plug in a=1, b=9, and c=14



x+=+%28-9+%2B-+sqrt%28+81-4%2A1%2A14+%29%29%2F%282%2A1%29 Square 9 to get 81



x+=+%28-9+%2B-+sqrt%28+81%2B-56+%29%29%2F%282%2A1%29 Multiply -4%2A14%2A1 to get -56



x+=+%28-9+%2B-+sqrt%28+25+%29%29%2F%282%2A1%29 Combine like terms in the radicand (everything under the square root)



x+=+%28-9+%2B-+5%29%2F%282%2A1%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)



x+=+%28-9+%2B-+5%29%2F2 Multiply 2 and 1 to get 2

So now the expression breaks down into two parts

x+=+%28-9+%2B+5%29%2F2 or x+=+%28-9+-+5%29%2F2

Lets look at the first part:

x=%28-9+%2B+5%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=%28-9+-+5%29%2F2

x=-14%2F2 Subtract the terms in the numerator
x=-7 Divide

So another answer is
x=-7

So our solutions are:
x=-2 or x=-7

Notice when we graph x%5E2%2B9%2Ax%2B14, we get:

+graph%28+500%2C+500%2C+-17%2C+8%2C+-17%2C+8%2C1%2Ax%5E2%2B9%2Ax%2B14%29+

and we can see that the roots are x=-2 and x=-7. This verifies our answer

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for x:
x%5E2%2B9x%2B14+=+0 You can always solve a quadratic equation using the quadratic formula:
x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a and, of course, the a, b, and c come from the standard form of the equation:
ax%5E2%2Bbx%2Bc+=+0
So, in this case, a = 1, b = 9, and c = 14
Making the appropriate substitutions into the quadratic formula, we get:
x+=+%28-9%2B-sqrt%289%5E2-4%281%29%2814%29%29%29%2F2%281%29
x+=+%28-9%2B-sqrt%2881-56%29%29%2F2
x+=+%28-9%2B-sqrt%2825%29%29%2F2
x+=+%28-9%2B5%29%2F2 or x+=+%28-9-5%29%2F2
x+=+-2 or x+=+-7
If the given trinomial can be factored, this offers another way to solve the problem:
x%5E2%2B9x%2B14+=+0 Factor the trinomial.
%28x%2B2%29%28x%2B7%29+=+0 Apply the zero product principle.
x%2B2+=+0 or x%2B7+=+0, then...
If x%2B2+=+0 then x+=+-2
If x%2B7+=+0 then x+=+-7
So in both cases, you get the same solutions:
x+=+-2
x+=+-7
A third way would be to graph the equation (y+=+x%5E2%2B9x%2B14) and identify the x-intercepts (also called "the zeros").
graph%28600%2C400%2C-8%2C5%2C-8%2C5%2Cx%5E2%2B9x%2B14%29