The Purplemath Forums

by **pistolpete** on Sun May 31, 2009 1:07 am

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

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...
answer:
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

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

Is this correct? Thanks in advance.

**Sponsor**

by **stapel_eliz** on Sun May 31, 2009 11:56 am

pistolpete wrote:Problem #1
5,-3

...answer:
x^2 - 2x - 15=0

If you're not sure, check! :wink:

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

Factor the quadratic:

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

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! :thumb:

pistolpete wrote:Problem #2
6, -1/2

...Answer:
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. :mrgreen:

by **pistolpete** on Sun May 31, 2009 2:46 pm

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. :oops:

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.. :confused:

Stapel please?

by **stapel_eliz** on Sun May 31, 2009 7:41 pm

pistolpete wrote:I forgot how to add and multiply fractions.

You can try using the "fraction" key on your calculator, if it's got one. :wink:

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.

pistolpete wrote: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!)

:wink:

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

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

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