If the sum of the roots of (a + 1)x^2 + (2a + 3)x + (3a + 4) = 0 is –1

Question

If the sum of the roots of  (a + 1)x^2 + (2a + 3)x + (3a + 4) = 0  is –1, then what is the product of the roots?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

bhawesh96 · Accepted Answer

Sum of roots = \frac{-b}{a}

For the given equation, sum of roots:   \frac{-(2a+3)}{(a+1)} = -1

On solving, we get  a=-2

Now, product of roots =  \frac{-c}{a}

For the given equation, product of roots:  \frac{-(3a+4)}{(a+1)}

Substituting  a=-2,  we get the product of roots as 2 (option C)

Thanks

umangshah · Answer

Actually, product of the roots is equal to  \frac{c}{a} .

sjmonroe · Answer

You could alternatively recognize the following:

Say the roots of this are r and t. From factoring we know that r*t will be the product coefficients on the squared term and the numerical (leftmost) term and that r+t will be equal to the middle term.

So, r+t = 2a+3 and r*t = (a+1)*(3a+4)
From the problem we are given that the sum of the roots are -1, so r+t = -1 = 2a+3

Using, -1 = 2a+3, we can solve for a, which will be -2.

Finally, we can use a = -2 to solve for r*t: (-2+1)*(3(-2)+4) = 2

The answer is C.

HRX273 · Answer

can any expert explain me this.. there above but it bit vague.....explaination

If the sum of the roots of (a + 1)x^2 + (2a + 3)x + (3a + 4) = 0 is –1

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If the sum of the roots of (a + 1)x^2 + (2a + 3)x + (3a + 4) = 0 is –1

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