Question 120709
what is the answer step by step to the following quadratic equations ?
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These need to be solved using the quadratic formula, which is:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
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x^2 - 4x - 15 = 0
In this problem a = 1; b =-4; c =-15; substitute these value in the quad formula
{{{x = (-(-4) +- sqrt(-4^2 - 4 * 1 * -15 ))/(2*1) }}}
{{{x = (4 +- sqrt(16 - (-60) ))/(2) }}}
{{{x = (4 +- sqrt(16 + 60 ))/(2) }}}; remember minus a minus is a plus
{{{x = (4 +- sqrt(76))/(2) }}}
{{{x = (4 +- 8.7178)/2 }}}; find the square root of 76 with a calc
Two solutions
Solution 1:
{{{x = (4 + 8.7178)/2 }}}
{{{x = 12.7178/2}}}
x = 6.3589 is solution 1
Solution 2:
{{{x = (4 - 8.7178)/2 }}}
{{{x = -4.7178/2}}}
x = -2.3589 is solution 2
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:
x^2 + 8x - 33 = 0
In this problem a = 1; b = 8; c =-33, substitute these in the quad formula
{{{x = (-8 +- sqrt(8^2- 4 * 1 * -33 ))/(2*1) }}}
{{{x = (-8 +- sqrt(64 - (-132) ))/(2) }}}
{{{x = (-8 +- sqrt(64 + 132 ))/(2) }}}
{{{x = (-8 +- sqrt(196 ))/(2) }}}
{{{x = (-8 +- 14)/2 }}}
Solution 1
{{{x = (-8 + 14)/2 }}}
{{{x = (+6)/2 }}}
x = 3 is solution 1
Solution 2
{{{x = (-8 - 14)/2}}}
{{{x = (-22)/2}}}
x = -11 is solution 2
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Note that these solutions are both integers, so we could have just factored
(x+11)(x-3)= 0 would have saved a lot of time
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:
x^2 + 5x + 2 = 0
You should be getting the hang of it now
{{{x = (-5 +- sqrt(5^2 - 4 * 1 * 2 ))/(2*1) }}}
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Do the math on this one just like we did on the 1st two
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You should get: x = -4.56 and x = -.438, (neither comes out even)
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This one is a special case, does not have any real roots 
we shall see why as we proceed with the quad formula
3x^2 - 3x + 6 = 0
{{{x = (-(-3) +- sqrt(-3^2- 4 * 3 * 6 ))/(2*3) }}}
{{{x = (3 +- sqrt(9 - 72 ))/(6) }}}
{{{x = (3 +- sqrt(-63 ))/(6) }}}
we have a square root of a negative number so there is no real roots
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Look up the word "discriminant" in the index of your algebra book.
This is related to what we just did.
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Did this make sense to you. It's very important that you understand exactly what's going on here. Do you have any questions?