Solving the Quadratic Equation: y = x^2-8x+7

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Joebird
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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! :)
 
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Joebird said:
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! :)

Try searching on google.

Do you know what the discriminant is?

Do you know how to factor?

Do you know what the quadratic equation is?

I can help you, but any type of help would be giving the answer.

http://home.alltel.net/okrebs/page6.html

Here's a link if you don't feel like searching.
 
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Tryed myself, the discriminant is "x^2 - 8x"?

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

Not sure about #3.

Thanks for the help! :)

PS: Yes, this is in my Algebra II class.
 
Do you know the quadratic formula? I'll bet it's in your book. I'll also bet they define "discriminant" in the same section. (No, it's not "x^2- 8x". The discriminant is a number not an expression in x.)
 
HallsofIvy said:
Do you know the quadratic formula? I'll bet it's in your book. I'll also bet they define "discriminant" in the same section. (No, it's not "x^2- 8x". The discriminant is a number not an expression in x.)

Why look in the book when people online will answer it? :rolleyes:
 
It probably is in my book, but I'm sick and my book was left in my locker as I thought I was going to school today. I'm looking it up on how to do it, although I can find examples I can't figure out how they change it.

Thanks :)

PS: The roots are x = {7,1} I believe?
 
The quadratic formula is:

{-8 ± Sq Root (64 - 28)} ÷ 2

That means the discriminant is:

64 - 28

I hope that's right.

Thanks! :blushing:
 
Thanks for all the help. :)

On the question:

Find the roots by using the quadratic equation.

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

Thanks :)
 
Joebird said:
Thanks for all the help. :)

On the question:

Find the roots by using the quadratic equation.

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

Thanks :)

Substitute your answers in the original equation. You'll see immediately if they're correct.
 
Okay, thanks. :)

How would I find the 'x =' in the equation? (y = x^2-8x+7)

THANKS! :)
 
High school and college level homework goes into the Science Education Zone, please.

Joebird said:
How would I find the 'x =' in the equation? (y = x^2-8x+7)

You have to show some work first.
 
Joebird said:
Okay, thanks. :)

How would I find the 'x =' in the equation? (y = x^2-8x+7)

THANKS! :)

Hint: Take a closer look at the Quadratic Formula.
 
Joebird said:
Okay, thanks. :)

How would I find the 'x =' in the equation? (y = x^2-8x+7)

THANKS! :)

The quadratic formula is "overkill" for this one. How can you factor 7?

Edit: I just noticed, this is not x^2- 8x+ 7= 0, which would give x= 7, x= 1 as roots, but y= x^2 - 8x+ 7. To solve that for x, in terms of y, you would need to use the quadratic formula (or complete the square).
 
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discriminant = #'s inside the radical.

to solve the factoring problem you must know :
F-irst
O-utside
I-nside
L-ast

For example if we had [tex]x^2+7x+12=0[/tex] we would factor it light this
[tex](x+3)(x+4)[/tex]

To know if you have done your work right use FOIL
First: [tex]x \times x[/tex]
Outside: [tex]4 \times x[/tex]
Inside: [tex]3 \times x[/tex]
Last: [tex]3 \times 4[/tex]To find the zeros (or answers) of x for : [tex](x+3)(x+4)[/tex] we would simply take:
[tex](x+3)=0[/tex] and solve for x and [tex](x+4)=0[/tex]