Solve (a+2b)^2+(2a-b)^2=5(a^2+b^2)

$$( a + 2 b ) ^ { 2 } + ( 2 a - b ) ^ { 2 } = 5 ( a ^ { 2 } + b ^ { 2 } )$$

Solve for a (complex solution)

$$a\in \mathrm{C}$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+2b\right)^{2}.

$$a^{2}+4ab+4b^{2}+\left(2a-b\right)^{2}=5\left(a^{2}+b^{2}\right)$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(2a-b\right)^{2}.

$$a^{2}+4ab+4b^{2}+4a^{2}-4ab+b^{2}=5\left(a^{2}+b^{2}\right)$$

Combine a^{2} and 4a^{2} to get 5a^{2}.

$$5a^{2}+4ab+4b^{2}-4ab+b^{2}=5\left(a^{2}+b^{2}\right)$$

Combine 4ab and -4ab to get 0.

$$5a^{2}+4b^{2}+b^{2}=5\left(a^{2}+b^{2}\right)$$

Combine 4b^{2} and b^{2} to get 5b^{2}.

$$5a^{2}+5b^{2}=5\left(a^{2}+b^{2}\right)$$

Use the distributive property to multiply 5 by a^{2}+b^{2}.

$$5a^{2}+5b^{2}=5a^{2}+5b^{2}$$

Subtract 5a^{2} from both sides.

$$5a^{2}+5b^{2}-5a^{2}=5b^{2}$$

Combine 5a^{2} and -5a^{2} to get 0.

$$5b^{2}=5b^{2}$$

Cancel out 5 on both sides.

$$b^{2}=b^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any a.

$$a\in \mathrm{C}$$

Solve for b (complex solution)

$$b\in \mathrm{C}$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+2b\right)^{2}.

$$a^{2}+4ab+4b^{2}+\left(2a-b\right)^{2}=5\left(a^{2}+b^{2}\right)$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(2a-b\right)^{2}.

$$a^{2}+4ab+4b^{2}+4a^{2}-4ab+b^{2}=5\left(a^{2}+b^{2}\right)$$

Combine a^{2} and 4a^{2} to get 5a^{2}.

$$5a^{2}+4ab+4b^{2}-4ab+b^{2}=5\left(a^{2}+b^{2}\right)$$

Combine 4ab and -4ab to get 0.

$$5a^{2}+4b^{2}+b^{2}=5\left(a^{2}+b^{2}\right)$$

Combine 4b^{2} and b^{2} to get 5b^{2}.

$$5a^{2}+5b^{2}=5\left(a^{2}+b^{2}\right)$$

Use the distributive property to multiply 5 by a^{2}+b^{2}.

$$5a^{2}+5b^{2}=5a^{2}+5b^{2}$$

Subtract 5b^{2} from both sides.

$$5a^{2}+5b^{2}-5b^{2}=5a^{2}$$

Combine 5b^{2} and -5b^{2} to get 0.

$$5a^{2}=5a^{2}$$

Cancel out 5 on both sides.

$$a^{2}=a^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any b.

$$b\in \mathrm{C}$$

Solve for a

$$a\in \mathrm{R}$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+2b\right)^{2}.

$$a^{2}+4ab+4b^{2}+\left(2a-b\right)^{2}=5\left(a^{2}+b^{2}\right)$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(2a-b\right)^{2}.

$$a^{2}+4ab+4b^{2}+4a^{2}-4ab+b^{2}=5\left(a^{2}+b^{2}\right)$$

Combine a^{2} and 4a^{2} to get 5a^{2}.

$$5a^{2}+4ab+4b^{2}-4ab+b^{2}=5\left(a^{2}+b^{2}\right)$$

Combine 4ab and -4ab to get 0.

$$5a^{2}+4b^{2}+b^{2}=5\left(a^{2}+b^{2}\right)$$

Combine 4b^{2} and b^{2} to get 5b^{2}.

$$5a^{2}+5b^{2}=5\left(a^{2}+b^{2}\right)$$

Use the distributive property to multiply 5 by a^{2}+b^{2}.

$$5a^{2}+5b^{2}=5a^{2}+5b^{2}$$

Subtract 5a^{2} from both sides.

$$5a^{2}+5b^{2}-5a^{2}=5b^{2}$$

Combine 5a^{2} and -5a^{2} to get 0.

$$5b^{2}=5b^{2}$$

Cancel out 5 on both sides.

$$b^{2}=b^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any a.

$$a\in \mathrm{R}$$

Solve for b

$$b\in \mathrm{R}$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+2b\right)^{2}.

$$a^{2}+4ab+4b^{2}+\left(2a-b\right)^{2}=5\left(a^{2}+b^{2}\right)$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(2a-b\right)^{2}.

$$a^{2}+4ab+4b^{2}+4a^{2}-4ab+b^{2}=5\left(a^{2}+b^{2}\right)$$

Combine a^{2} and 4a^{2} to get 5a^{2}.

$$5a^{2}+4ab+4b^{2}-4ab+b^{2}=5\left(a^{2}+b^{2}\right)$$

Combine 4ab and -4ab to get 0.

$$5a^{2}+4b^{2}+b^{2}=5\left(a^{2}+b^{2}\right)$$

Combine 4b^{2} and b^{2} to get 5b^{2}.

$$5a^{2}+5b^{2}=5\left(a^{2}+b^{2}\right)$$

Use the distributive property to multiply 5 by a^{2}+b^{2}.

$$5a^{2}+5b^{2}=5a^{2}+5b^{2}$$

Subtract 5b^{2} from both sides.

$$5a^{2}+5b^{2}-5b^{2}=5a^{2}$$

Combine 5b^{2} and -5b^{2} to get 0.

$$5a^{2}=5a^{2}$$

Cancel out 5 on both sides.

$$a^{2}=a^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any b.

$$b\in \mathrm{R}$$