Solving Quadratic Equations

[pistolpete]

I am having some trouble with these 3 quadratic equations. I need to solve for x.

x^4+2x^2-15=0

2(x-1)^2+7(x-1)+6=0

2x-9sqrt(x)+4=0

[stapel_eliz]

x^4+2x^2-15=0

This one factors, to give you (x² + 5)(x² - 3) = 0. Then solve the two quadratic factors by taking square roots.

2(x-1)^2+7(x-1)+6=0

This is just 2Y² + 7Y + 6 = 0, with Y = x - 1. (You can find examples of this here.) Factor as usual, and solve the factors.

2x-9sqrt(x)+4=0

This is a quadratic in sqrt[x], so you've got 2Y² - 9Y + 4 = 0. Factor as usual, and then solve the factors.

If you get stuck, please reply showing how far you have gotten. Thank you!

[pistolpete]

Thanks for your help, stapel..

Here's what I got for the first one,

x= i sqrt(5) and x=sqrt(3)

i being the imaginary number of sqrt(-1)
I'm not sure it is correct though

For the second one, I can't seem to get the factors to add up to 7y (the middle term) so I'm stuck here
I want to do (2z+1) (z+6) but doing the "smiley face" check it does not match to the 7y in 2y^2+7y+6=0

For the last one, I also get stuck at factoring

2x-9sqrt(x)+4=0
2y^2-9y+4=0
(2y )(y )=0

1 and 4 or 2 and 2 as the factorals don't match the middle term once again... can u tell me what I'm doing wrong? Thanks

[stapel_eliz]

Here's what I got for the first one,

x= i sqrt(5) and x=sqrt(3)

You should have gotten a "plus minus" sign somewhere...? To learn how this works, try here.

For the second one, I can't seem to get the factors to add up to 7y (the middle term) so I'm stuck here

Factors of 12 that add to 7 are 3 and 4. To learn how to do the rest of the factorization, try here.

For the last one, I also get stuck at factoring

Factors of 8 that add to -9 are -1 and -8.

[pistolpete]

Wait I'm confused. For 2(x-1)^2+7(x-1)+6=0 shouldn't I be looking for factors of 6 and not 12?

And for, 2x-9sqrt(x)+4=0, don't i need factors of 4 and not 8 ?

Thanks..

[stapel_eliz]

Wait I'm confused. For 2(x-1)^2+7(x-1)+6=0 shouldn't I be looking for factors of 6 and not 12?

And for, 2x-9sqrt(x)+4=0, don't i need factors of 4 and not 8 ?

Ah; so you haven't yet studied the lesson on how to factor this sort of quadratic....

As you'll learn when you read the lesson, no, when the leading coefficient is something other than 1, the rules are changed a bit. For the second exercise, you need to find factors of (2)(6) = 12, not just of 6; for the third, you need factors of (2)(4) = 8, not just of 4.