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Equation Problem

This topic has 1 expert reply and 3 member replies

Equation Problem

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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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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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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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[quote="Ian Stewart"]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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Re:[quote="Ian Stewart"]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 Vite'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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