3(x - 5)^2 = 147
(x - 5)^2 = 49
x - 5 = ± 7
x = 5 ± 7
x = 12 or x = - 2
check in the original equation.
id est
To solve this, you need to isolate the variable, x.
First, divide both sides by 3.
3(x-5)^2 / (3) = 147 / (3)
The 3 on the left will divide out.
(x-5)^2 = 49
Now, take the square root of both sides.
â (x-5)^2 = â49
x - 5 = ±7
I will write these separately, although you use the same steps to solve.
x - 5 = 7
x - 5 + 5 = 7 + 5
x = 12
AND
x - 5 = -7
x - 5 + 5 = -7 + 5
x = -2
147/3=49
square root of 49 is 7
x-5=7
7+5=12
x=12
Start by dividing by 3
3(x-5)^2 = 147 DIVIDE BY 3
(X-5)^2 = 49 the sqrt of 49 is
(x-5) = 7 add 5
147/3=49...
(x-5)²=49...
(x-5)²=(+7 or -7)²...
therefore, x-5=+7 or x-5= -7......
so, x=12 or -2...
hope this help...
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Verified answer
3(x - 5)^2 = 147
(x - 5)^2 = 49
x - 5 = ± 7
x = 5 ± 7
x = 12 or x = - 2
check in the original equation.
id est
To solve this, you need to isolate the variable, x.
First, divide both sides by 3.
3(x-5)^2 / (3) = 147 / (3)
The 3 on the left will divide out.
(x-5)^2 = 49
Now, take the square root of both sides.
â (x-5)^2 = â49
x - 5 = ±7
I will write these separately, although you use the same steps to solve.
x - 5 = 7
x - 5 + 5 = 7 + 5
x = 12
AND
x - 5 = -7
x - 5 + 5 = -7 + 5
x = -2
147/3=49
square root of 49 is 7
x-5=7
7+5=12
x=12
Start by dividing by 3
3(x-5)^2 = 147 DIVIDE BY 3
(X-5)^2 = 49 the sqrt of 49 is
(x-5) = 7 add 5
x = 12
147/3=49...
(x-5)²=49...
(x-5)²=(+7 or -7)²...
therefore, x-5=+7 or x-5= -7......
so, x=12 or -2...
hope this help...