## Using the formula x^2 - (r1 + r2)x + (r1 * r2) = 0

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
pistolpete
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Joined: Wed May 20, 2009 1:07 am
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### Using the formula x^2 - (r1 + r2)x + (r1 * r2) = 0

Can anyone who understands this well take a minute and check my work and see if I did this right?

The book says:
Use the form x^2 - (r1 + r2)x + (r1 * r2)=0
Write a quadratic equation having the given roots.

Problem #1
5,-3

My Work:
Sum of roots: 5+(-3)= 2
Product of roots: 5*-3= -15

so...
x^2 - 2x - 15=0

Problem #2
6, -1/2

My Work:
Sum of roots: 6+(-1/2)= 5 1/2
Product of roots: 6*-1/2 = -3

x^2 - 5 1/2x - 3=0

Is this correct? Thanks in advance.

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
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Problem #1
5,-3

x^2 - 2x - 15=0
If you're not sure, check!

. . . . .$x^2\, -\, 2x\, -\, 15\, =\, 0$

. . . . .$(x\, -\, 5)(x\, +\, 3)\, =\, 0$

Then solve the factors:

. . . . .$x\, -\, 5\, =\, 0\, \mbox{ or }\, x\, +\, 3\, =\, 0$

. . . . .$x\, =\, 5\, \mbox{ or }\, x\, =\, -3$

Your quadratic does indeed have the required zeroes, so it must be correct!
Problem #2
6, -1/2

x^2 - 5 1/2x - 3=0
It would almost certainly be better to use an improper fraction, rather than a mixed number, for the coefficient of the linear term. So I would suggest changing the "five and a half" to "eleven halves".

. . . . .$x^2\, -\, \frac{11}{2}\, x\, -\, 3\, =\, 0$

To check, use the Quadratic Formula, or else multiply through to clear the denominator and then factor. I prefer the latter:

. . . . .$2x^2\, -\, 11x\, -\, 6\, =\, 0$

. . . . .$2x^2\, -\, 12x\, +\, 1x\, -\, 6\, =\, 0$

. . . . .$2x(x\, -\, 6)\, +\, 1(x\, -\, 6)\, =\, 0$

. . . . .$(x\, -\, 6)(2x\, +\, 1)\, =\, 0$

Completing the solution, you should get that x = 6 or x = -1/2, as required, so your quadratic is a valid solution.

pistolpete
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Joined: Wed May 20, 2009 1:07 am
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### Re: Using the formula x^2 - (r1 + r2)x + (r1 * r2) = 0

Great! I'm happy I got those right. But, unfortunately they get harder.

#3
2/3, -3/5

I forgot how to add and multiply fractions. I think I need a common denominator to add so that would be 15. And Idk about multiplying. Just multiply the numerator 2 with the other numerator -3. And just multiply the denominators too? Or do i need a common factor for multiplying also?

#4
2+sqrt(2), 2-sqrt(2)

Totally lost here... The product of the roots might be 0 since these cancel out? Not sure..

stapel_eliz
Posts: 1628
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I forgot how to add and multiply fractions.
You can try using the "fraction" key on your calculator, if it's got one. Otherwise, since we really can't teach lessons here, you might need to brush up by studying a few online lessons instead.

Note: The common denominator, for thirds and fifths, is going to be 3*5 = 15.
2+sqrt(2), 2-sqrt(2)

Totally lost here...
The sum shouldn't be a problem, since the radicals will cancel out. To learn how to multiply the radical expressions to get the product, try here. (And yes, I would recommend using the "vertical" method of multiplication!)