SOLUTION: i need help solving these quadratic equations using the quadratic formula. 2x^2 - 5x = 3 3x^2 - 2x + 1 = 0

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Question 107041: i need help solving these quadratic equations using the quadratic formula.
2x^2 - 5x = 3
3x^2 - 2x + 1 = 0
Answer by HyperBrain(694) About Me  (Show Source):
You can put this solution on YOUR website!
2x%5E2-5x=3
2x%5E2-5x-3=3-3
2x%5E2-5x-3=0
The quadratic formula for solving the value of x in ax%5E2%2Bbx%2Bc=0 is
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

Here, a=2, b=-5, and c=-3.So,
x=%28-%28-5%29%2B-sqrt%28%28-5%29%5E2-4%2A2%2A%28-3%29%29%29%2F%282%2A2%29
x=%285%2B-sqrt%2825-%28-24%29%29%29%2F%284%29
x=%285%2B-sqrt%2825%2B24%29%29%2F%284%29
x=%285%2B-sqrt%2849%29%29%2F%284%29
x=%285%2B-+7%29%2F%284%29
x=%285%2B7%29%2F%284%29 or x=%285-7%29%2F%284%29
x=12%2F4 or x=-2%2F4
x=3 or x=-1%2F2

For your next problem, 3x%5E2+-2x%2B1+=+0

Here, a=3, b=-2, c=1.

Using the quadratic formula,
x=%28-%28-2%29%2B-sqrt%28%28-2%29%5E2-4%2A3%2A1%29%29%2F%282%2A3%29
x=%282%2B-sqrt%284-12%29%29%2F%286%29
x=%282%2B-sqrt%28-8%29%29%2F%286%29
We know that square root of a negative number does not exist because all squares are positive numbers. In this cases, oftentimes, mathematicians use the symbol i for the square root of -1
Let's continue

x=%282%2B-sqrt%28%28-1%29%284%29%282%29%29%29%2F%286%29
x=%282%2B-+%28sqrt%28-1%29%29%28sqrt%284%29%29%28sqrt%282%29%29%29%2F%286%29
x=%282%2B-+%28i%29%282%29%28sqrt%282%29%29%29%2F%286%29
x=%282%2B-+2i+%28sqrt%282%29%29%29%2F%286%29
Roots involving i are called imaginary roots.

Power up,
HyperBrain!