SOLUTION: Use the quadratic formula to solve each of the following quadratic equations... 1. 2x^2-5x=3 2. 3x^2-2x+1=0 3. x^2+4x+4=7 (Hint: Factor the left hand side)

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Use the quadratic formula to solve each of the following quadratic equations... 1. 2x^2-5x=3 2. 3x^2-2x+1=0 3. x^2+4x+4=7 (Hint: Factor the left hand side)      Log On


   

Question 85397: Use the quadratic formula to solve each of the following quadratic equations...
1. 2x^2-5x=3
2. 3x^2-2x+1=0
3. x^2+4x+4=7 (Hint: Factor the left hand side)
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
1.
2x%5E2-5x=3
2x%5E2-5x-3=0Subtract 3 from both sides
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B-5x%2B-3+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-5%29%5E2-4%2A2%2A-3=49.

Discriminant d=49 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--5%2B-sqrt%28+49+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-5%29%2Bsqrt%28+49+%29%29%2F2%5C2+=+3
x%5B2%5D+=+%28-%28-5%29-sqrt%28+49+%29%29%2F2%5C2+=+-0.5

Quadratic expression 2x%5E2%2B-5x%2B-3 can be factored:
2x%5E2%2B-5x%2B-3+=+2%28x-3%29%2A%28x--0.5%29
Again, the answer is: 3, -0.5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-5%2Ax%2B-3+%29


2.
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 3x%5E2%2B-2x%2B1+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-2%29%5E2-4%2A3%2A1=-8.

The discriminant -8 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -8 is + or - sqrt%28+8%29+=+2.82842712474619.

The solution is

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+3%2Ax%5E2%2B-2%2Ax%2B1+%29



3.
x%5E2%2B4x%2B4=7
x%5E2%2B4x%2B4-7=0 Subtract 7 from both sides
x%5E2%2B4x-3=0 Combine like terms
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B4x%2B-3+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%284%29%5E2-4%2A1%2A-3=28.

Discriminant d=28 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-4%2B-sqrt%28+28+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%284%29%2Bsqrt%28+28+%29%29%2F2%5C1+=+0.645751311064591
x%5B2%5D+=+%28-%284%29-sqrt%28+28+%29%29%2F2%5C1+=+-4.64575131106459

Quadratic expression 1x%5E2%2B4x%2B-3 can be factored:
1x%5E2%2B4x%2B-3+=+1%28x-0.645751311064591%29%2A%28x--4.64575131106459%29
Again, the answer is: 0.645751311064591, -4.64575131106459. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B4%2Ax%2B-3+%29