If the product of the roots of the equation (a+1) x2+(2 a+3) x+(3 a+4)=0 be 2, then the sum of roots is

Question

If the product of the roots of the equation (a+1) x2+(2 a+3) x+(3 a+4)=0 be 2, then the sum of roots is (A) 1 (B) -1 (C) 2 (D) -2

Tardigrade Admin · Accepted Answer

It is given that: α β=2 ⇒ (3 a+4/a+1)=2 ⇒ 3 a+4=2 a+2 ⇒ a=-2 Also, α+β=-(2 a+3/a+1) Putting this value of a, we get sum of roots =-(2 a+3/a+1)=-(-4+3/-2+1)=-1

Q. If the product of the roots of the equation $(a + 1) x^{2} + (2 a + 3) x + (3 a + 4) = 0$ be $2$ , then the sum of roots is

1

-1

2

-2

It is given that: $α β = 2$
$\Rightarrow \frac{3 a + 4}{a + 1} = 2$
$\Rightarrow 3 a + 4 = 2 a + 2$
$\Rightarrow a = - 2$
Also, $α + β = - \frac{2 a + 3}{a + 1}$
Putting this value of a, we get sum of roots
$= - \frac{2 a + 3}{a + 1} = - \frac{- 4 + 3}{- 2 + 1} = - 1$

Q. If the product of the roots of the equation (a+1)x2+(2a+3)x+(3a+4)=0 be 2, then the sum of roots is

1

-1

2

-2

It is given that: αβ=2 ⇒a+13a+4​=2 ⇒3a+4=2a+2 ⇒a=−2 Also, α+β=−a+12a+3​ Putting this value of a, we get sum of roots =−a+12a+3​=−−2+1−4+3​=−1

Q. If the product of the roots of the equation $(a + 1) x^{2} + (2 a + 3) x + (3 a + 4) = 0$ be $2$ , then the sum of roots is

It is given that: $α β = 2$
$\Rightarrow \frac{3 a + 4}{a + 1} = 2$
$\Rightarrow 3 a + 4 = 2 a + 2$
$\Rightarrow a = - 2$
Also, $α + β = - \frac{2 a + 3}{a + 1}$
Putting this value of a, we get sum of roots
$= - \frac{2 a + 3}{a + 1} = - \frac{- 4 + 3}{- 2 + 1} = - 1$