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Which of these are possible rational roots of the above polynomial:

1) +/- 7
2) +/- 1/2
3) +/- 1/7
4) +/- 1
5) +/- 2

2007-01-29 06:53:20 · 6 answers · asked by Random G 3 in Science & Mathematics Mathematics

You guys are saying none, but according to the question before me, I must choose at least two of them for my answer to be correct.

2007-01-29 07:08:07 · update #1

6 answers

If you do the quadratic equation you'll find that this has no real numbers as roots. You'll have imaginary numbers as roots.

Rational numbers are a subcategory of real numbers. So there is no rational root of the polynomial.

you may want to check and see if you wrote the problem correctly.

2007-01-29 07:03:23 · answer #1 · answered by Zonte 2 · 0 0

information on a thanks to remedy that is first to work out that this equation is a quadratic: A x X^2 + B x X + C the position A=2, B=3, and C=2. you go with to memorize the equation for the discriminant, it is B^2 - 4AC. (Spoken "b squared minus 4 cases a cases c"). B^2 = 3 * 3 = 9 4 x A x C = 4 x 2 x 2 = 16. So the discriminant is 9-16 or -7. A discriminant less than 0 ability no authentic roots, so the reply is B. (A, C, and D ought to all require a minimum of one authentic root.)

2016-10-16 06:33:31 · answer #2 · answered by Anonymous · 0 0

If you have a polynome ax^n+..........+b=0 if there is a rational root

x=p/q p must divide b and q must divide a

Here a= 2 so q can be +-1 or +-2

b=7 so p can be -+1 or+-7

so you get +-7 ,+-1/2,+-1 but not +-2 and +-1/7

2007-01-29 07:50:15 · answer #3 · answered by santmann2002 7 · 1 0

None of the above.

===

You either put in the question wrong or the answers wrong. For the two real roots of a quadratic equation to be of the +/- variety, the equation has to boil down to (x² - a = 0), and then x=± √a. This equation is not that kind of equation. That's how I answered so quickly.

2007-01-29 06:57:20 · answer #4 · answered by bequalming 5 · 1 0

Solve this equation using the quadratic equation:

ƒ(x) = ax² + bx + c

so in your equation ...

ƒ(x) = 2x² - 3x + 7

a = 2
b = - 3
c = 7

Quadratic equation tells us:

x = [ -b +- sqrt( 4ac) ] / 2a

plug in: ........... x = [-(2) +- sqrt( 4*2*7 ) ] / 2(2)
simplify .......... x = [ -2 +- sqrt( 56 ) ] / 4

.............sqrt(56) appoximately equals 7.48

Roots are then x1 = -2 + 7.48 / 4 = 1.37
and ................. x2 = -2 - 7.48 / 4 = -2.37

2007-01-29 07:10:18 · answer #5 · answered by Razor 2 · 0 0

none of the above!! good luck

2007-01-29 06:58:36 · answer #6 · answered by Anonymous · 0 0

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