Problem

Write a quadratic equation with integer coefficie…

01:06

Problem 29 Hard Difficulty

One root of the equation $-x^{2}-11 x+c=0$ is $\sqrt{3}$
a. Find the other root.
b. Find the value of c.
c. Explain why the roots of this equation are not conjugates.

Answer

a) $q=-11-\sqrt{3}(\text { the second root })$
b) $c=3+11 \sqrt{3}$
c) The roots are not congugates because this equation does not have rational coefficients.

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Algebra

Algebra 2 and Trigonometry

Chapter 5

QUADRATIC FUNCTIONS AND COMPLEX NUMBERS

Section 7

Sum and Product of the Roots of a Quadratic Equation

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Video Transcript

All right, we have this quadratic equation. We know that one of the routes it's call it Route one is equal to square to the street. Yes. We're gonna use this formula to find the other route. So if one of the richest square 23 say are to the second route is the one we don't know is equal to negative. Be over a well, negative B. Let's just keep it as you see, others Two negatives here. So we have a negative 11 over a negative one. It will ultimately cancel out. And let me explain that one more time Negative 11 over negative one is gonna be a positive 11 Number one. But we have a negative here. So that's three negatives leaving this whole thing as negative 11. And so we have negative. 11 is equally square to three. Plus are to. So let's get rid of square to three on this side of leaving our to all by itself. But of course, we have to subtract it from this side. And that actually is it right there? I'll rewrite it in a different location, But that route is negative. 11 minus the square root of three. Very nice. Now let's go ahead and use this. Backed in this fact here that if we multiply them in equals negative, see over a to find that missing see value. So if I take the square 231 of our roots multiplied by this whole thing Negative 11 minus square to three. I should get that equal. See over a We don't know, See? But we do know a is a negative one. And so we can kind of say that whole thing will be negative. See, we'll get there in a bit. Ah, negative. 11 Times Square to three is just kind of rewritten like this square 23 times Square to three is three, but there's a negative there, and this whole thing is equal to a negative. See? So if we multiply everything, everything, everything by negative one will get R C. Answer because negative one times of negative C is a positive. See negative one times negative three is a positive three in a negative one times negative 11 3rd in negative. 11 square three, I should say, is a positive 11 square root free. And so there is a very complex see value. Now I have to talk about why these roots aren't conjure gets and the Kontic it would be like this. If it's a square to three, you would be like negative square to three. And that basically comes down to the B value. When the B value isn't zero are quadratic equations look a little different. For instance, with a B bio zero a quadratic equation like might look like this and your roots might be negative two and positive too. Or it might look like I don't know, like this in your routes could be positive one and negative one. But when you're be value is non zero. It might move it to the right to the left. Um, and it just basically makes it so that it isn't symmetrical about the y axis. And so for that reason, we don't have this like square three Negative Square three situation. We have a route that is quite different. And so for that reason, they're not contra gets

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