# the roots of 2x^2-3x-2=0 are alpha and beta find an equation with integral coefficients whose roots are 2alpha+beta and 2beta+alpha

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2x^2-3x-2=0

First let us find the zeros of the equation:

(2x+1)(x-2)=0

==> x= -1/2 and x=2

Or alpha=-1/2 and beta=2

Now we need to find the equation whose roots are:

2alpha+beta or 2(-1/2)+2=1

and 2beta+alpha OR 2(2)-1/2 = 7/2

Then the factors are:

(x-1)(x-7/2)=0

x^2-(9/2)x+7/2=0

Multiply by 2:

2x^2+9x+7=0

Then the function is:

f(x) = 2x^2+9x+7

The roots of 2x^2-3x-2 = 0 are alpha =m and betan. To fibd the equation whose roots are 2m+n and and 2n+m.

Solution:

By the relation between the roots and coefficients,

Sum of the roota = m+n = -coefficient of x/ coeefficient of x^2 = -(-3)/2 = 3/2......(1)

Product of the roots. = mn = constant terrm /coefficient of x^2 = -2/2 = -1............(2).

We use the above results to obtain the valuee of the sum of the new roots 2m+n and 2n+m and their product:

The sum of the roots (2m+n)+(2n+m) = 3(m+n) = 3(3/2) =

9/2

Product of the roota (2m+n)*(2n+m) = 4mn+2(m^2+n^2)+mn = 5mn +2[(m+n)^2)-2mn] = 5(-1)+[(3/2)^2-2(-1)] =

-5+2[(3/2)^2-2(-1)] = -5+2[9/4+2] = -5+9/2+4 = 9/2-1 = 7/2.

Therefore the required equation is:

x^2 - (9/2)x+ (7/2) = 0. Making the coefficients integral by multiplying by 2, we get:

2x^2-9x+7 = 0.