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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parallel_chase
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y^2 - 12y +k = 0ritula 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.
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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Ian Stewart
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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.
(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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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
