Find the product of roots of quadratic equation if |x1-x2|=1 and x^2=2x-m.
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The roots of the quadratic equation are x1 and x2. |x1 - x2| = 1 and x^2 = 2x - m
As x^2 = 2x - m
x1^2 = 2* x1 - m
x2^2 = 2*x2 - m
Subtracting the two we get
x1^2 - x2^2 = 2( x1 - x2)
=> (x1 - x2)(x1+ x2) = 2(x1 - x2)
=> x1 + x2 = 2
|x1 - x2| = 1
=> x1 - x2 = 1 or x1 - x2 = -1
=> (x1 - x2)^2 = 1
=> x1^2 + x2^2 - 2x1*x2 = 1...(1)
x1 + x2 = 2
=> x1^2 + x2^2 + 2x1*x2 = 4...(2)
(2) - (1)
=> 4x1*x2 = 3
=> x1* x2 = 3/4
The required product is 3/4
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We'll re-write the constraint x^2=2x-m moving all terms to the left side.
x^2-2x+m = 0
Since we know that m represents the value of the product of the roots, we'll write the Viete's 2nd relation:
x1*x2 = m
From the module definition we'll write the given constraint:
-1<(x1-x2)<1
If we are square raising, we'll get:
(x1-x2)^2=1
x1^2-2x1*x2+x2^2=1
x1^2+x2^2=1+2x1*x2 (1)
We know that if we are substituting the roots into the equation, they are verifying it. So, we'll introduce the roots, one by one, into the equation.
x1^2-2x1+m=0 (2)
x2^2-2x2+m=0 (3)
Now, we'll add (2) and (3):
x1^2-2x1+m+x2^2-2x2+m=0
We'll combine like terms:
(x1^2+x2^2)-2(x1+x2)+2m=0
From the Viete's relations, we'll have:
x1+x2= - (-2)/1
x1+x2= 2
So, x1^2+x2^2= 2(x1+x2)-2m, where x1+x2= 2
x1^2+x2^2= 2*2 -2m
x1^2+x2^2= 4-2m (4)
But from relation (1), x1^2+x2^2=1+2x1*x2, where x1*x2=m
From (1) and (4):
We'll move the terms in m to the left side and the numbers alone to the right side:
4m=3
The value of the product of the roots is m = 3/4.
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