Problem

One root of the equation $-x^{2}-11 x+c=0$ is $\s…

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Problem 28 Hard Difficulty

One root of the equation $-3 x^{2}+9 x+c=0$ is $\sqrt{2}$
a. Find the other root.
b. Find the value of c.
c. Explain why the roots of this equation are not conjugates.

Answer

a) $q=3-\sqrt{2}$ (the second root)
b) $c=6-9 \sqrt{2}$
c) The roots are not congugates because this equation does not have rational coefficients.

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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

we have this quadratic equation where we don't have the c value, but we have that one. Root is the square to to And I'm gonna use the fact that if we add the two routes together, we get negative. Be over a and so I'm gonna have to move this a little bit. So I have room bear with me. We'll just move it and we'll see if that gives us enough room. Maybe, maybe not. So Wonder doesn't swear to to We don't know the other route and the equal Negative B, Which B is nine over a negative three. And so that would give us to negative. So I'm gonna put a positive three. Um, and then nine. Divided by three, of course, is three into. To get that three is equal to the square to two, plus the other route. If we subtract the square to to from both sides of the equation, we're going to get that other route value, and I know it's a little bit messy, but you can see we have three minus square to, To which all rewrite three minus the square root of two is equal to that unknown route very good. And now to find the C value, I'm gonna use this fact the fact that if we multiply the two we get see over a So we know that one route is equal to the square root of two. We're gonna multiply that by this whole thing three minus the square root of two and that is gonna equal. See which we don't know over a, which is negative three. And so, if I do this whole thing, I get three square of two for that first term. And swear to Times Square to negative is going to be a negative too equal to see over negative three. And I guess we can just multiply everything by three to get rid of this fraction. And then we're gonna get it. Because if I most by this, I should say, multiplied by negative three. If I most by this by negative three I get see by itself, which is rape so negative three times a negative. Two is a positive six and negative three times posit three square to is negative. Nine square root of two and so pretty complex answer for see there And for part C, we just have to talk about why these routes are not congregants or something like square to negative Square to, And that really kind of comes down to the B value as long as you have of B value that isn't zero. That's not gonna happen where it's not gonna be like square to negative Square two or four negative for because when the B value is zero, your quadratic equations look something like this. Where there symmetrical about the y axis. But if you have a B value that's non zero. It might be something like this or just anywhere that's not necessarily going to give you the exact opposite on the left and the right. And so there's our roots. Three minus square to to in our see value is negative. Nine for two plus six.

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