Solve for b 4^2+b^2=( square root of 69)^2

Math
42+b2=(69)2
Raise 4 to the power of 2.
16+b2=(69)2
Rewrite 692 as 69.
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Use axn=axn to rewrite 69 as 6912.
16+b2=(6912)2
Apply the power rule and multiply exponents, (am)n=amn.
16+b2=6912⋅2
Combine 12 and 2.
16+b2=6922
Cancel the common factor of 2.
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Cancel the common factor.
16+b2=6922
Divide 1 by 1.
16+b2=691
16+b2=691
Evaluate the exponent.
16+b2=69
16+b2=69
Move all terms not containing b to the right side of the equation.
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Subtract 16 from both sides of the equation.
b2=69-16
Subtract 16 from 69.
b2=53
b2=53
Take the square root of both sides of the equation to eliminate the exponent on the left side.
b=±53
The complete solution is the result of both the positive and negative portions of the solution.
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First, use the positive value of the ± to find the first solution.
b=53
Next, use the negative value of the ± to find the second solution.
b=-53
The complete solution is the result of both the positive and negative portions of the solution.
b=53,-53
b=53,-53
The result can be shown in multiple forms.
Exact Form:
b=53,-53
Decimal Form:
b=7.28010988…,-7.28010988…
Solve for b 4^2+b^2=( square root of 69)^2

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