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PS Fourth Degree Equations

by rodalvarezz » Mon Nov 14, 2011 9:49 pm
This is from a Kaplan practice test:

If 4y^4 − 41y^2 + 100 = 0, then what is the sum of the two greatest possible values of y ?
a) 4
b) 9/2
c) 7
d) 41/4
e) 25

OA is b.
[/url]

Can anyone please help and explain how to solve 4th degree equations.
Say z=y^2

Then we can rewrite the equation as:
4z^2-41z+100=0

So I am stuck here, what is the best method/technique to solve this quadratic equation. The quadratic term has a coefficient and I don't even know if I can solve it by looking at two numbers that multiplied together give 100 and whose sum give -41. Can anyone please explain how to solve these equations?

Thank you!
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by user123321 » Mon Nov 14, 2011 9:55 pm
4z^2-41z+100=0
we can factorize this one as
4z^2 - 16z - 25z +100 = 0
(4z-25)(z-4) = 0
so y = 5/2,-5/2,2,-2
adding greatest among them will give 5/2 + 2 = 9/2

IMO B

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by neelgandham » Tue Nov 15, 2011 1:56 am
user123321 wrote:4z^2-41z+100=0
IMO B
user123321
At times, When your eyes/brain don't appear to be focused on the problem after fighting to find the factors, you might want to use the formula Roots of the equation
a*z^2 + b*z + c = 0 are z = (-b+ Square root((b^2)-(4*a*c)))/2a

Here
z = (41+Square root((41^2)-(4*100*4)))/(2*4)
z = (41+Square root((41^2)-(40^2)))/(2*4)
z = (41+Square root(81*1))/(2*4)
z = (41+9)/(2*4)
z = 50/8 or 4
Implies the values of y are 5/2,-5/2,2,-2.
Adding greatest among them will give you 9/2(5/2 + 2)
Note:
For a given quadratic equation a*z^2 + b*z + c = 0
Sum of roots = -b/a and product of roots = c/a
My 2 cents !
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by lunarpower » Tue Nov 15, 2011 1:58 am
looks like the user above got here before me; that explanation is good. basically, this equation is really just a quadratic in disguise, with x^2 as the variable instead of x.

NOTE:
i don't think i've ever seen an official problem that requires the student to factor an expression with a coefficient in front of the square term ... so you might not necessarily have to be able to do this.
there are all kinds of factoring problems without a coefficient in front of the square term, but i don't think i've seen one like this.
the only exception is certain differences of squares (e.g., 4x^2 - 25y^2), but those aren't very hard to factor.

rodalvarezz wrote:This is from a Kaplan practice test:

If 4y^4 − 41y^2 + 100 = 0, then what is the sum of the two greatest possible values of y ?
a) 4
b) 9/2
c) 7
d) 41/4
e) 25

OA is b.
[/url]

Can anyone please help and explain how to solve 4th degree equations.
Say z=y^2

Then we can rewrite the equation as:
4z^2-41z+100=0

So I am stuck here, what is the best method/technique to solve this quadratic equation. The quadratic term has a coefficient and I don't even know if I can solve it by looking at two numbers that multiplied together give 100 and whose sum give -41. Can anyone please explain how to solve these equations?

Thank you!
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by GMATGuruNY » Tue Nov 15, 2011 4:52 am
rodalvarezz wrote:This is from a Kaplan practice test:

If 4y^4 − 41y^2 + 100 = 0, then what is the sum of the two greatest possible values of y ?
a) 4
b) 9/2
c) 7
d) 41/4
e) 25

OA is b.
[/url]

Can anyone please help and explain how to solve 4th degree equations.
Say z=y^2

Then we can rewrite the equation as:
4z^2-41z+100=0

So I am stuck here, what is the best method/technique to solve this quadratic equation. The quadratic term has a coefficient and I don't even know if I can solve it by looking at two numbers that multiplied together give 100 and whose sum give -41. Can anyone please explain how to solve these equations?

Thank you!
I received a PM asking me to comment.

Given any quadratic in the form ax² + bx + c = 0:
The sum of the roots = -b/a.
The product of the roots = c/a.

As noted above, if we let z=y², then the equation becomes:
4z² - 41z + 100 = 0.
The sum of the roots = -b/a = 41/4.

Since y² is equal to each of the roots, and there is no √ in the answer choices, it is virtually guaranteed that each root = (perfect square)/4.
Perfect squares less than 41 are 1,4,9,16,25,36.
Since 16+25 = 41, the two values of y² must be 16/4 and 25/4, yielding a sum of 16/4 + 25/4 = 41/4.

If y² = 16/4 = 4, then y = ±2.
If y² = 25/4, then y = ± 5/2.
Thus, the sum of the two greatest possible values of y = 2 + 5/2 = 9/2.

The correct answer is B.

I agree with Ron: this question seems beyond the scope of the GMAT. Quadratics on the GMAT typically do not include a coefficient in front of the squared term.
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