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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

Best Answer 

 #1
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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
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1+0 Answers

 #1
avatar
-1
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

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