It has to do with Parabola .
I have to Find The Solution to these problems using Factoring and Square Roots
My algebra teacher is a bit old,64, he is kinda hard to learn from . help
i think there are 3 possible answers . No solution X<0 , you have 2 answers x>0, or 1 answer x=0
Here are problems i have trouble with and thank you
3x^2-60=87 and 1/2x^2-1/2=0 also 2x^2-33=17
I have to Find The Solution to these problems using Factoring and Square Roots
My algebra teacher is a bit old,64, he is kinda hard to learn from . help
i think there are 3 possible answers . No solution X<0 , you have 2 answers x>0, or 1 answer x=0
Here are problems i have trouble with and thank you
3x^2-60=87 and 1/2x^2-1/2=0 also 2x^2-33=17
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3x^2-60=87
Combine like terms:
3x^2 = 27
Divide by 3:
x^2 = 9
Take the square root:
x = ± 3
1/2x^2-1/2=0
Multiply by 2 to get rid of the fractions:
x^2 - 1 = 0
Factor the difference of squares:
(x + 1) (x - 1) = 0
Use the zero product principle twice:
x = -1, 1
2x^2-33=17
Combine like terms:
2x^2 - 50 = 0
Divide by 2:
x^2 - 25 = 0
Factor the difference of squares:
(x + 5) (x - 5) = 0
Use the zero product principle twice:
x = -5, 5
Combine like terms:
3x^2 = 27
Divide by 3:
x^2 = 9
Take the square root:
x = ± 3
1/2x^2-1/2=0
Multiply by 2 to get rid of the fractions:
x^2 - 1 = 0
Factor the difference of squares:
(x + 1) (x - 1) = 0
Use the zero product principle twice:
x = -1, 1
2x^2-33=17
Combine like terms:
2x^2 - 50 = 0
Divide by 2:
x^2 - 25 = 0
Factor the difference of squares:
(x + 5) (x - 5) = 0
Use the zero product principle twice:
x = -5, 5