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 $m\cdot n?$

Guest Feb 11, 2018

#1**-2 **

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**

Guest Feb 11, 2018

#1**-2 **

Best Answer

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**

Guest Feb 11, 2018