SOLUTION: SOLVE THE EQUATION ALGEBRAICALLY. CHECK SOLUTION BY GRAPHING. #38 X^2-11=14 #42 2X^2-89=9 #43 2X^2+8=16 #44 3X^2+5=32

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: SOLVE THE EQUATION ALGEBRAICALLY. CHECK SOLUTION BY GRAPHING. #38 X^2-11=14 #42 2X^2-89=9 #43 2X^2+8=16 #44 3X^2+5=32      Log On


   

Question 118484This question is from textbook ALGEBRA 1 CONCEPTS AND SKILLS
: SOLVE THE EQUATION ALGEBRAICALLY. CHECK SOLUTION BY GRAPHING.
#38 X^2-11=14
#42 2X^2-89=9
#43 2X^2+8=16
#44 3X^2+5=32 This question is from textbook ALGEBRA 1 CONCEPTS AND SKILLS

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

SOLVE THE EQUATION ALGEBRAICALLY. CHECK SOLUTION BY GRAPHING.
#38
x%5E2-11=14
x%5E2+=14+%2B+11
x%5E2+=+25

x+=+sqrt%2825%29
x%5B1%5D+=+5
x%5B2%5D+=+-+5

#42
2x%5E2+-+89+=+9
2x%5E2+=+9+%2B+89
2x%5E2+=+98
x%5E2+=+49
x+=+sqrt%2849%29
x%5B1%5D+=+7
x%5B2%5D+=+-+7

#43
2x%5E2%2B8=16
2x%5E2+=16+-+8
2x%5E2+=+8
x%5E2+=+4
x+=+sqrt%284%29
x%5B1%5D+=+2
x%5B2%5D+=+-+2
#44
3x%5E2+%2B+5+=+32
3x%5E2+=+32+-+5
3x%5E2+=+27
x%5E2+=+9
x+=+sqrt%289%29
x%5B1%5D+=+3
x%5B2%5D+=+-+3
Check solution by graphing:
#38
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B0x%2B-25+=+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=%280%29%5E2-4%2A1%2A-25=100.

Discriminant d=100 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-0%2B-sqrt%28+100+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%280%29%2Bsqrt%28+100+%29%29%2F2%5C1+=+5
x%5B2%5D+=+%28-%280%29-sqrt%28+100+%29%29%2F2%5C1+=+-5

Quadratic expression 1x%5E2%2B0x%2B-25 can be factored:
1x%5E2%2B0x%2B-25+=+%28x-5%29%2A%28x--5%29
Again, the answer is: 5, -5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B0%2Ax%2B-25+%29




#42

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B0x%2B-49+=+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=%280%29%5E2-4%2A1%2A-49=196.

Discriminant d=196 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-0%2B-sqrt%28+196+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%280%29%2Bsqrt%28+196+%29%29%2F2%5C1+=+7
x%5B2%5D+=+%28-%280%29-sqrt%28+196+%29%29%2F2%5C1+=+-7

Quadratic expression 1x%5E2%2B0x%2B-49 can be factored:
1x%5E2%2B0x%2B-49+=+%28x-7%29%2A%28x--7%29
Again, the answer is: 7, -7. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B0%2Ax%2B-49+%29



#43
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B0x%2B-4+=+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=%280%29%5E2-4%2A1%2A-4=16.

Discriminant d=16 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-0%2B-sqrt%28+16+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%280%29%2Bsqrt%28+16+%29%29%2F2%5C1+=+2
x%5B2%5D+=+%28-%280%29-sqrt%28+16+%29%29%2F2%5C1+=+-2

Quadratic expression 1x%5E2%2B0x%2B-4 can be factored:
1x%5E2%2B0x%2B-4+=+%28x-2%29%2A%28x--2%29
Again, the answer is: 2, -2. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B0%2Ax%2B-4+%29



#44

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B0x%2B-9+=+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=%280%29%5E2-4%2A1%2A-9=36.

Discriminant d=36 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-0%2B-sqrt%28+36+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%280%29%2Bsqrt%28+36+%29%29%2F2%5C1+=+3
x%5B2%5D+=+%28-%280%29-sqrt%28+36+%29%29%2F2%5C1+=+-3

Quadratic expression 1x%5E2%2B0x%2B-9 can be factored:
1x%5E2%2B0x%2B-9+=+%28x-3%29%2A%28x--3%29
Again, the answer is: 3, -3. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B0%2Ax%2B-9+%29