If a and b are constants and the equation x^2 + ax + b = 0 has

Question

If a and b are constants and the equation \(x^2 + ax + b = 0\) has two different positive roots, which of the following must be true?

I. a < 0
II. b < 0
III. ab < 0

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Bunuel · Accepted Answer

Vieta's Formula for Quadratic Equation:  if  f(x) = ax^2 + bx + c  is a quadratic equation with roots α and β, then the sum of roots =  α + β = -\frac{b}{a} , and the product of roots =  αβ = \frac{c}{a} .

According to the above, for the equation  x^2 + ax + b = 0  with the roots x1 and x2,  x_1+x_2=-a  and  x_1 * x_2=b . Since both roots must be positive, then  x_1+x_2=-a  must be positive, implying a is negative, and  x_1 * x_2=b  must also be positive, implying b is positive, making ab negative.

Answer: C.

AT1982 · Answer

Ok

(X-2)(X-4) is the equation 
2 x roots = 2 & 4
Both positive

Now 
Eqn x^2 -6x +8 = 0

a = -6<0 (I)
b = 8>0 (II)
ab = -48<0 (III)

Thnx

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sachi-in · Answer

A quadratic equation is of form :

x^2 + -(p + q)x + pq = 0    where p and q are the roots

Hence
a = -(p + q) which is negative
b = pq is positive

a < 0
b > 0
ab < 0

C is correct ans

Pif96 · Answer

So here if I did not know the Vieta's formula, How would I have approached ?

KarishmaB · Answer

Vieta's formula isn't something you need to learn. It is something you need to understand because it forms the very basics of quadratics. If a quadratic has roots p and q, it can be written as (x - p)(x - q) = x^2 - (p+q)x + pq = 0 If p and q are positive, co-efficient of x (which is -(p+q)) is negative and the constant term (which is pq) is positive. Compare this with x^2 + ax + b = 0 So a must be negative and b must be positive. Answer (C) Discussion on Quadratic Equations: https://youtu.be/QOSVZ7JLuH0

Pradhumansingh1 · Answer

Can’t this question be solve via discriminant as roots are unequal then b^-4ac must be greater then zero.

Bunuel · Answer

For the equation in question, using the discriminant gives a^2 - 4a > 0, but that alone isn't enough to solve the problem. It doesn't directly lead to the required conclusions.

vasu1104 · Answer

x^2 + ax+ b = 0 we know quadratic equation, ax^2+bx+c =0 when we compare the given eqn with quadratic one, we get a= 1, b= a , c= b now we know, sum = -b/a we found the value of b and a. so use put into eqn. sum = -a /1 sum = -a so a<0 product= c/a here c = b and a=1 product= b so b>0 s-1 - as we discussed, a<0. so this is correct. s-2- no b>0 as per our calculations. incorrect. s-3 since a<0 and b>0, the product of ab will always be < 0. so correct. option D

vasu1104 · Answer

x^2 + ax+ b = 0 we know quadratic equation, ax^2+bx+c =0 when we compare the given eqn with quadratic one, we get a= 1, b= a , c= b now we know, sum = -b/a we found the value of b and a. so use put into eqn. sum = -a /1 sum = -a so a<0 product= c/a here c = b and a=1 product= b so b>0 s-1 - as we discussed, a<0. so this is correct. s-2- no b>0 as per our calculations. incorrect. s-3 since a<0 and b>0, the product of ab will always be < 0. so correct. option D

If a and b are constants and the equation x^2 + ax + b = 0 has

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If a and b are constants and the equation x^2 + ax + b = 0 has

Prep Toolkit

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