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Algebra Examples
a=πd24a=πd24
Step 1
Rewrite the equation as πd24=aπd24=a.
πd24=a
Step 2
Multiply both sides of the equation by 4π.
4π⋅πd24=4πa
Step 3
Step 3.1
Simplify the left side.
Step 3.1.1
Simplify 4π⋅πd24.
Step 3.1.1.1
Combine.
4(πd2)π⋅4=4πa
Step 3.1.1.2
Cancel the common factor of 4.
Step 3.1.1.2.1
Cancel the common factor.
4(πd2)π⋅4=4πa
Step 3.1.1.2.2
Rewrite the expression.
πd2π=4πa
πd2π=4πa
Step 3.1.1.3
Cancel the common factor of π.
Step 3.1.1.3.1
Cancel the common factor.
πd2π=4πa
Step 3.1.1.3.2
Divide d2 by 1.
d2=4πa
d2=4πa
d2=4πa
d2=4πa
Step 3.2
Simplify the right side.
Step 3.2.1
Combine 4π and a.
d2=4aπ
d2=4aπ
d2=4aπ
Step 4
Take the specified root of both sides of the equation to eliminate the exponent on the left side.
d=±√4aπ
Step 5
Step 5.1
Rewrite √4aπ as √4a√π.
d=±√4a√π
Step 5.2
Simplify the numerator.
Step 5.2.1
Rewrite 4 as 22.
d=±√22a√π
Step 5.2.2
Pull terms out from under the radical.
d=±2√a√π
d=±2√a√π
Step 5.3
Multiply 2√a√π by √π√π.
d=±2√a√π⋅√π√π
Step 5.4
Combine and simplify the denominator.
Step 5.4.1
Multiply 2√a√π by √π√π.
d=±2√a√π√π√π
Step 5.4.2
Raise √π to the power of 1.
d=±2√a√π√π1√π
Step 5.4.3
Raise √π to the power of 1.
d=±2√a√π√π1√π1
Step 5.4.4
Use the power rule aman=am+n to combine exponents.
d=±2√a√π√π1+1
Step 5.4.5
Add 1 and 1.
d=±2√a√π√π2
Step 5.4.6
Rewrite √π2 as π.
Step 5.4.6.1
Use n√ax=axn to rewrite √π as π12.
d=±2√a√π(π12)2
Step 5.4.6.2
Apply the power rule and multiply exponents, (am)n=amn.
d=±2√a√ππ12⋅2
Step 5.4.6.3
Combine 12 and 2.
d=±2√a√ππ22
Step 5.4.6.4
Cancel the common factor of 2.
Step 5.4.6.4.1
Cancel the common factor.
d=±2√a√ππ22
Step 5.4.6.4.2
Rewrite the expression.
d=±2√a√ππ1
d=±2√a√ππ1
Step 5.4.6.5
Simplify.
d=±2√a√ππ
d=±2√a√ππ
d=±2√a√ππ
Step 5.5
Combine using the product rule for radicals.
d=±2√πaπ
d=±2√πaπ
Step 6
Step 6.1
First, use the positive value of the ± to find the first solution.
d=2√πaπ
Step 6.2
Next, use the negative value of the ± to find the second solution.
d=-2√πaπ
Step 6.3
The complete solution is the result of both the positive and negative portions of the solution.
d=2√πaπ
d=-2√πaπ
d=2√πaπ
d=-2√πaπ