Answer:
Part 1) [tex]2x^{2}-32=0----> (-4,4)[/tex]
Part 2) [tex]4x^{2}-100=0----> (-5,5)[/tex]
Part 3) [tex]x^{2}-55=9----> (-8,8)[/tex]
Part 4) [tex]x^{2}-140=-19----> (-11,11)[/tex]
Part 5) [tex]2x^{2}-18=0----> (-3,3)[/tex]
see the attached figure
Step-by-step explanation:
Solve each quadratic equation
Part 1) we have
[tex]2x^{2}-32=0[/tex]
Adds 32 both sides
[tex]2x^{2}=32[/tex]
Divide by 2 both sides
[tex]x^{2}=16[/tex]
take square root both sides
[tex]x=\pm4[/tex]
therefore
The solutions are (-4,4)
Part 2) we have
[tex]4x^{2}-100=0[/tex]
Adds 100 both sides
[tex]4x^{2}=100[/tex]
Divide by 4 both sides
[tex]x^{2}=25[/tex]
take square root both sides
[tex]x=\pm5[/tex]
therefore
The solutions are (-5,5)
Part 3) we have
[tex]x^{2}-55=9[/tex]
Adds 55 both sides
[tex]x^{2}=9+55[/tex]
[tex]x^{2}=64[/tex]
take square root both sides
[tex]x=\pm8[/tex]
therefore
The solutions are (-8,8)
Part 4) we have
[tex]x^{2}-140=-19[/tex]
Adds 140 both sides
[tex]x^{2}=-19+140[/tex]
[tex]x^{2}=121[/tex]
take square root both sides
[tex]x=\pm11[/tex]
therefore
The solutions are (-11,11)
Part 5) we have
[tex]2x^{2}-18=0[/tex]
Adds 18 both sides
[tex]2x^{2}=18[/tex]
Divide by 2 both sides
[tex]x^{2}=9[/tex]
take square root both sides
[tex]x=\pm3[/tex]
therefore
The solutions are (-3,3)
