MHB Proving - properties of quadratic roots

AI Thread Summary
The discussion focuses on proving properties of quadratic roots using the discriminant. When the discriminant is negative, the quadratic equation has two imaginary solutions, expressed as x = (-b ± i√(4ac - b²)) / 2a. Conversely, when the discriminant is zero, there is one real solution, represented as x = -b / 2a, indicating a double root. Participants emphasize the importance of understanding the quadratic formula and how to manipulate the discriminant to derive these results. Overall, the conversation clarifies the relationship between the discriminant and the nature of the roots in quadratic equations.
Drain Brain
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This is one of my weakness in Math, to prove an existing fact. please Tell how to go about doing these problem.

1. Prove that when the discriminant of a quadratic equation with
real coefficients is negative, the equation has two imaginary
solutions.

2. Prove that when the discriminant of a quadratic equation with
real coefficients is zero, the equation has one real solution.

regards!:)
 
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Re: Proving

Here again, you want to look at the quadratic formula, particularly the discriminant. Answering this question will help you answer your other question. Tell me what you see when you look at the quadratic formula...
 
Re: Proving

MarkFL said:
Here again, you want to look at the quadratic formula, particularly the discriminant. Answering this question will help you answer your other question. Tell me what you see when you look at the quadratic formula...
when the $b^2-4ac<0$

the roots are

$x=\frac{-b+\sqrt{b^-4ac}i}{2a}$ and $x=\frac{-b-\sqrt{b^-4ac}i}{2a}$ they are both imaginary.

for the 2nd question

when $b^2-4ac=0$

the roots are

$x=\frac{-b}{2a}$ and $x=\frac{-b}{2a}$

root of multiplicity two(double root)
 
Your reasoning is sound, but for the imaginary case, you want to express the roots as:

$$x=\frac{-b\pm i\sqrt{4ac-b^2}}{2a}$$

Do you see how we negated the discriminant to pull $i$ out front?
 
do you mean $\sqrt{-(4ac-b^2)}=i\sqrt{4ac-b^2}$

but why do we have to negate the discriminant if it is assumed to be $b^2-4ac<0$ already?
 
Drain Brain said:
do you mean $\sqrt{-(4ac-b^2)}=i\sqrt{4ac-b^2}$

but why do we have to negate the discriminant if it is assumed to be $b^2-4ac<0$ already?

We want what's under the radical to be positive, so if it is initially negative, we pull $i$ out, and then negate the radicand so that it is now positive.
 
MarkFL said:
We want what's under the radical to be positive, so if it is initially negative, we pull $i$ out, and then negate the radicand so that it is now positive.

Thanks! now it's clear!
 
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