# Find the product of roots of quadratic equation if |x1-x2|=1 and x^2=2x-m.

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### 2 Answers

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**

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.**