Question 211831This question is from textbook Introductory Algebra
: evaluate the discriminate b^2-4ac, then use the answer to state how many real-number solutions exist for the equation y=x^2+8x+16 This question is from textbook Introductory Algebra
You can put this solution on YOUR website! your equation is y=x^2+8x+16
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this is in the standard form of the equation.
you find the roots when you set y = 0 and solve.
your equation becomes:
x^2 + 8x + 16 = 0
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to find the roots using the quadratic formula, you would get the coefficients and plug them into that equation.
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standard form of a quadratic equation is:
ax^2 + bx + c = 0
your equation makes:
a = 1
b = 8
c = 16
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the discriminant in this equation is b^-4ac
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if b^2-4ac = 0, then you have one real root.
if b^2-4ac = positive you have two real roots.
if b^2-4ac = positive and a perfect square you have two real integer roots. This assumes that a and c are also integers.
if b^2-4ac = negative you have two imaginary roots.
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imaginary root means your graph does not cross the x-axis.
it is either above it or below it but never crosses it.
it would be above it if the x^2 portion of your equation is positive.
it would be below it if the x^2 portion of your equation is negative.
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b^2-4ac = (8^2 - 4*1*16) = 64 - 64 = 0
your equation will have one root.
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the quadratic formula says that this root will be at -8 +/- sqrt(0)/2 which makes it at -4.
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graph of your equation is shown below:
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you have one root at x = -4.
