Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
pistolpete
Posts: 13
Joined: Wed May 20, 2009 1:07 am
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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
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
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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
Posts: 13
Joined: Wed May 20, 2009 1:07 am
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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
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
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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
Posts: 13
Joined: Wed May 20, 2009 1:07 am
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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
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
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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.