Algebra II

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Joebird

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Jan 29, 2006
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I'm supposed to use this equation:

y = x^2-8x+7

To solve the following questions:

1) What is the value of the discriminant?
2) Find the roots by factoring and solving.
3) Find the roots by using the quadratic equation.

Can anyone give me some help?

Thanks! :)
 
What have you tried?

Please reply showing your steps.

To refresh your memory, here is the quadratic formula:
\(\displaystyle \L\ x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}\)

For part a), determine which part of the formula is the discriminant.
 
Tryed , the discriminant is "x^2 - 8x" right?

The factor is: (x - 7)(x - 1) But are those the roots also?

Not sure about #3.

Thanks for the help! :)
 
Joebird said:
Tryed , the discriminant is "x^2 - 8x" right?

The factor is: (x - 7)(x - 1) But are those the roots also?

Not sure about #3.

Thanks for the help! :)
Click to expand...

The discriminant is the part of the quadratic formula that's under the radical, b² - 4ac.

To find the roots, set the factors = 0. Then use the zero-product property. If (x-7)(x-1) = 0, then at least one of the factors MUST = 0. x - 7 = 0 OR x - 1 = 0
Solve each of those to get the roots (values for x that solve the equation).

For #3, use the quadratic formula and the values for a, b, and c taken from your equation, x² - 8x + 7 = 0, to find the roots by that method.
 
Thanks! :)

On the question:

Find the roots by using the quadratic equation.

Would the answer be x = {7,14}?

Thanks :)
 
On the question:

Find the roots by using the quadratic equation.

Would the answer be x = {7,14}?

The roots better be the same as you got by factoring or algebra theory is in trouble. You got the discriminate = 36 so the roots are (8+6)/2 and (8-6)/2

How would I find the vertex of the equation? (y = x^2-8x+7)
The proper way is to use the vertex equation x=-b/(2a) but since you know the roots, the vertex will be half way between them.
 
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