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Equation Problem

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by ritula » Fri Sep 05, 2008 1:10 am
What is the value of K if y ^2 – 12y + k = 0?

(1) One root of the equation is 3.
(2) One root of the equation is 6 more than the other.
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Re: Equation Problem

by parallel_chase » Fri Sep 05, 2008 2:19 am
ritula wrote:What is the value of K if y ^2 – 12y + k = 0?

(1) One root of the equation is 3.
(2) One root of the equation is 6 more than the other.
y^2 - 12y +k = 0

Statement I

one root is 3

3^2 - 12*3 +k = 0

9 - 36 + k = 0

k = 36-9 = 27

Sufficient.

Statement II

One root is 6 more than the other.

We can have infinite values for k.

Therefore insufficient.

Hence A is the answer.
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by Ian Stewart » Fri Sep 05, 2008 4:16 am
If a and b are the roots of the equation, then

(y-a)(y-b) = y^2 - 12y + k

so -a - b = - 12, and (-a)(-b) = k. Rewriting these, we have:

a + b = 12
ab = k

From 1), we can plug in b = 3 to find that a = 9, k = 27. From 2) we can plug in b = a - 6 to find that a = 9, b = 3, and k = 27. D.
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by smclean23 » Fri Sep 05, 2008 11:09 am
Ian Stewart wrote:
so -a - b = - 12, and (-a)(-b) = k. Rewriting these, we have:

a + b = 12
ab = k
quote]

How did you go from -a-b=-12 & (-a)(-b)=k?

Thanks.
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by mayur00 » Fri Sep 05, 2008 1:44 pm
Re:
Ian Stewart wrote:
so -a - b = - 12, and (-a)(-b) = k. Rewriting these, we have:

a + b = 12
ab = k
quote]

How did you go from -a-b=-12 & (-a)(-b)=k?

Thanks.]

These are standard Viète's formulas.

x1+x2 = -b/a

x1.x2 = c/a

where x1,x2 are roots of euqations ax^2+bx+c=0
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