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Earn +20 ptsQ: How do you show an equation has 3 real roots?
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What is the root of equation?
roots of equation are x values when y = 0
If a quadratic equation has roots at 3 and -5 which factors could be multiplied together to obtain an original equation?
(x - 3) and (x - (-5)).
What are the roots of the equation x squared -3x-2 equals 0?
I will use the quadratic equation here. By inspection the discriminant shows 2 real roots. X = -b +/- sqrt(b^2 - 4ac)/2a a = 1 b = - 3 c = - 2 X = -(-3) +/- sqrt[(-3)^2 - 4(1)(-2)]2(1) X = 3 +/- sqrt(17)/2 X = [3 +/- sqrt(17]/2 exact answer.
What are the roots of the quadratic equation 2X2-3X-3 equals 0?
Use the quadratic equation to find the roots of 2x2-3x-3=0.a=2, b=-3, c=-3x =[ -b +- SQRT(b2-4ac)]/2a=[--3 +- SQRT((-3)2 - 4(2)(-3))]/2*2= [3+-SQRT(9--24)]/4so the roots are x = [3+SQRT(33)]/4 and x = [3-SQRT(33)]/4
Complex roots occur in conjugate pairs?
Not for all types of equations. But always in second degree equations they do. Consider a third degree equation with 3 different roots. Obviously, one of the roots can not be in a pair.
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