SOLUTION: How do I find a root of the equation 2x^2-x-15=0 ALSO How do I find the x coordinate of the vertex for the quadratic equation x^2+6x+5=0 Thanks(:

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Question 273005: How do I find a root of the equation 2x^2-x-15=0
ALSO
How do I find the x coordinate of the vertex for the quadratic equation x^2+6x+5=0

Thanks(:
Found 2 solutions by stanbon, dabanfield:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
How do I find a root of the equation 2x^2-x-15=0
Think of two numbers whose product is -15*2 = -30
and whose sum is -1
The numbers are -6 and +5
Replace -x by -6x+5x and factor:
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2x^2 -6x + 5x - 15 = 0
2x(x-3) + 5(x-3) = 0
(x-3)(2x+5) = 0
x = 3 or x = -5/2
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ALSO
How do I find the x coordinate of the vertex for the quadratic equation x^2+6x+5=0
The x-value of the vertex is -b/2a = -6/(2*1) = -3
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Cheers,
Stan H.

Answer by dabanfield(803) (Show Source):
You can put this solution on YOUR website!
How do I find a root of the equation 2x^2-x-15=0
ALSO
How do I find the x coordinate of the vertex for the quadratic equation x^2+6x+5=0
2x^2 - x - 15 = 0
(2x+5)*(x-3) = 0
Zeros are solutions for:
2x+5 = 0
x=-5/2 and
x-3 = 0
x=3
The vertex form of an equation of a parabola is:
y = a*(x-h)^+ k
In this form the vertex is at point (h,k)
We need to rewrite equation y = 2x^2-x-15 in this form. Let's start with:
y = 2(x^2-1/2x) - 15
We can "complete the square" in the parentheses by adding 2*1/16. To keep the equation balanced we also need to substract 2*1/16 = 1/8:
y = 2(x^2 -1/2x + 1/16) - 1/8 - 15
y = 2*(x - 1/4)^2 - 1/8 - 15
y = 2*(x - 1/4)^2 - 121/8
In this case then a = 2 and the vertex (h,k) = (1/4,-121/8)