Solve for x: a) (x+1)^{2}=16 b) (3x-1)^{2}=0 c) (x+8)^{2}=-9

Anne Wacker 2021-12-14 Answered
Solve for x:
a) (x+1)2=16
b) (3x1)2=0
c) (x+8)2=9
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Expert Answer

lovagwb
Answered 2021-12-15 Author has 50 answers
Step 1
a) (x+1)2=16
(x+1)=±4
x=3, 5
No of solutions is 2
Step 2
b) (3x1)2=0
x=13
No of solutions =1
c) (x+8)2=9
x+89
x=±3i8
x=3i8, 3i8
Here no real solution exist for x.
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raefx88y
Answered 2021-12-16 Author has 26 answers
Step 1
a) Given: (x+1)2=16
Take the square root of both sides of the equation.
(x+1)2=16
Simplify.
x+1=4
x+1=4
Subtract 1 from both sides of the equation.
x=3
x=5
Step 2
b) Given: (3x1)2=0
Use binomial theorem (ab)2=a22ab+b2 to expand (3x1)2
9x26x+1=0
This equation is in standard form: ax2+bx+c=0 NSk Substitute 9 for a, -6 for b, and 1 for c in the quadratic formula b±b24ac2a
x=(6)±(6)24×92×9
Square -6.
x=(6)±364×92×9
Multiply -4 times 9.
x=(6)±36362×9
Add 36 to -36.
x=(6)±02×9
Take the square root of 0.
x=62×9
The opposite of -6 is 6.
x=62×9
Multiply 2 times 9.
x=618
Reduce the fraction 618 to lowest terms by extracting and canceling out 6.
x=13
Step 3
c) Given: (x+8)2=9
Subtract 8 from both sides of the equation.
x+88=3i8
x+88=3i8
Subtracting 8 from itself leaves 0.
x=3i8
x=3i8
Subtract 8 from 3i.
x=8+3i
Subtract 8 from -3i.
x=83i
The equation is now solved.
x=8+3i
x=83i
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