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Alternate approach (we don't have to find the value to k)

Given eq. -> x^2 - 5x +2k = 13
We know that sum of roots of an equation is given by -b/a,
Let the two roots be x1 and x2
x1 + x2 = -b/a
a = 1; b = -5; x1 = 3 (given)
3 + x2 = -(-5)/1 ---> x2 = 2

We now know that the values of x1 and x2 are 3 and 2, hence the product = 3*2 = 6

Answer => D
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Bunuel
If 3 is one of the roots of x^2 - 5x +2k = 13, what is the product of the two roots of the equation?

(A) -12
(B) -6
(C) 5
(D) 6
(E) 12


­
x^2 - 5x + 2k = 13

Sum of roots = 5
3 + other root = 5
other root = 2

Product = 3*2 = 6

IMO: D
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Bunuel
If 3 is one of the roots of x^2 - 5x +2k = 13, what is the product of the two roots of the equation?

(A) -12
(B) -6
(C) 5
(D) 6
(E) 12


­
x^2 - 5x +2k = 13 => x^2 - 5x +2k - 13 = 0
The expression we want to factorise is: y = x^2 - 5x + (2k - 13)
Since this has a root 3, (x - 3) is a factor.
If the other root be r, the factor would be (x - r)
=> y = x^2 - 5x + (2k - 13) = (x - 3)(x - r) ... this is true for all values of x
=> x^2 - 5x + (2k - 13) = x^2 - x(r + 3) + 3r

Comparing the coefficient of x on both sides: -5 = -(r + 3) => r = 2
=> Product of roots = 3 * 2 = 6 Answer D

Point to note: For y = ax^2 + bx + c: Sum of roots = -b/a and Product of roots = c/a
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Bunuel
If 3 is one of the roots of x^2 - 5x +2k = 13, what is the product of the two roots of the equation?

(A) -12
(B) -6
(C) 5
(D) 6
(E) 12


­
Sum of Roots = -coeff of x/coeff of x^2. Therefore 3 + Root 2 = 5, Therefore Root 2 = 2. Hence product of roots is 6.
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aX^2 + bx + c = 0;

Have, Sum of Roots = -b/a; and Product of Roots= c/a;

x^2 -5x + 2k = 13;

x^2 -5x + 2k - 13 = 0; a = 1; b = -5; c = 2k-13

-b/a = +5; if one root is three then the other must be 2;

And product of 3 & 2 is 6;

Ans: D
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sum of roots = 5, therefore other root will be 2.

Hence answer, 3*2 =6
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