The values of $x$ such that $$2x^2 - 6x + 5 = 0$$ are $m+ni$ and $m-ni,$ where $m$ and $n$ are positive. What is $m\cdot n?$
Solve for x:
2 x^2 - 6 x + 5 = 0
Write the quadratic equation in standard form.
Divide both sides by 2:
x^2 - 3 x + 5/2 = 0
Solve the quadratic equation by completing the square.
Subtract 5/2 from both sides:
x^2 - 3 x = -5/2
Take one half of the coefficient of x and square it, then add it to both sides.
Add 9/4 to both sides:
x^2 - 3 x + 9/4 = -1/4
Factor the left hand side.
Write the left hand side as a square:
(x - 3/2)^2 = -1/4
Eliminate the exponent on the left hand side.
Take the square root of both sides:
x - 3/2 = i/2 or x - 3/2 = -i/2
Look at the first equation: Solve for x.
Add 3/2 to both sides:
x = 3/2 + i/2 or x - 3/2 = -i/2
Look at the second equation: Solve for x.
Add 3/2 to both sides:
x = 3/2 + i/2 or x = 3/2 - i/2
So, [ 3/2 + i/2] / [ 3/2 - i/2] =4/5 + (3 i)/5
Solve for x:
2 x^2 - 6 x + 5 = 0
Write the quadratic equation in standard form.
Divide both sides by 2:
x^2 - 3 x + 5/2 = 0
Solve the quadratic equation by completing the square.
Subtract 5/2 from both sides:
x^2 - 3 x = -5/2
Take one half of the coefficient of x and square it, then add it to both sides.
Add 9/4 to both sides:
x^2 - 3 x + 9/4 = -1/4
Factor the left hand side.
Write the left hand side as a square:
(x - 3/2)^2 = -1/4
Eliminate the exponent on the left hand side.
Take the square root of both sides:
x - 3/2 = i/2 or x - 3/2 = -i/2
Look at the first equation: Solve for x.
Add 3/2 to both sides:
x = 3/2 + i/2 or x - 3/2 = -i/2
Look at the second equation: Solve for x.
Add 3/2 to both sides:
x = 3/2 + i/2 or x = 3/2 - i/2
So, [ 3/2 + i/2] / [ 3/2 - i/2] =4/5 + (3 i)/5