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655-705 Level|   Algebra|      
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gmatophobia
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If a and b are constants and the equation x^2 + ax + b = 0 has 03 Mar 2024, 12:18
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55% (hard)

Question Stats:

64% (01:41) correct 36% (01:59) wrong based on 824 sessions

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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

A. III only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III

­
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1709492768309 (004).jpg [ 68.52 KiB | Viewed 11287 times ]
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C
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If a and b are constants and the equation x^2 + ax + b = 0 has 03 Mar 2024, 13:19
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gmatophobia
­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 

    A. III only

    B. I and II only

    C. I and III only

    D. II and III only

    E. I, II, and III 
­ ­
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­

­
    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.­
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If a and b are constants and the equation x^2 + ax + b = 0 has 30 Aug 2024, 07:25
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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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If a and b are constants and the equation x^2 + ax + b = 0 has 03 Mar 2024, 22:06
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­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
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If a and b are constants and the equation x^2 + ax + b = 0 has 05 Jan 2025, 09:17
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So here if I did not know the Vieta's formula, How would I have approached ?
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If a and b are constants and the equation x^2 + ax + b = 0 has 09 Jan 2025, 01:53
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Pif96
So here if I did not know the Vieta's formula, How would I have approached ?

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
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If a and b are constants and the equation x^2 + ax + b = 0 has 19 May 2025, 04:08
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Can’t this question be solve via discriminant as roots are unequal then b^-4ac must be greater then zero.
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If a and b are constants and the equation x^2 + ax + b = 0 has 19 May 2025, 05:09
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Pradhumansingh1
Can’t this question be solve via discriminant as roots are unequal then b^-4ac must be greater then zero.

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.
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