Solve (2a+3b)^2+(2a-3b)^2=8a^2+18b^2

$$( 2 a + 3 b ) ^ { 2 } + ( 2 a - 3 b ) ^ { 2 } = 8 a ^ { 2 } + 18 b ^ { 2 }$$

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(2a+3b\right)^{2}.

$$4a^{2}+12ab+9b^{2}+\left(2a-3b\right)^{2}=8a^{2}+18b^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(2a-3b\right)^{2}.

$$4a^{2}+12ab+9b^{2}+4a^{2}-12ab+9b^{2}=8a^{2}+18b^{2}$$

Combine 4a^{2} and 4a^{2} to get 8a^{2}.

$$8a^{2}+12ab+9b^{2}-12ab+9b^{2}=8a^{2}+18b^{2}$$

Combine 12ab and -12ab to get 0.

$$8a^{2}+9b^{2}+9b^{2}=8a^{2}+18b^{2}$$

Combine 9b^{2} and 9b^{2} to get 18b^{2}.

$$8a^{2}+18b^{2}=8a^{2}+18b^{2}$$

Subtract 8a^{2} from both sides.

$$8a^{2}+18b^{2}-8a^{2}=18b^{2}$$

Combine 8a^{2} and -8a^{2} to get 0.

$$18b^{2}=18b^{2}$$

Cancel out 18 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(2a+3b\right)^{2}.

$$4a^{2}+12ab+9b^{2}+\left(2a-3b\right)^{2}=8a^{2}+18b^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(2a-3b\right)^{2}.

$$4a^{2}+12ab+9b^{2}+4a^{2}-12ab+9b^{2}=8a^{2}+18b^{2}$$

Combine 4a^{2} and 4a^{2} to get 8a^{2}.

$$8a^{2}+12ab+9b^{2}-12ab+9b^{2}=8a^{2}+18b^{2}$$

Combine 12ab and -12ab to get 0.

$$8a^{2}+9b^{2}+9b^{2}=8a^{2}+18b^{2}$$

Combine 9b^{2} and 9b^{2} to get 18b^{2}.

$$8a^{2}+18b^{2}=8a^{2}+18b^{2}$$

Subtract 18b^{2} from both sides.

$$8a^{2}+18b^{2}-18b^{2}=8a^{2}$$

Combine 18b^{2} and -18b^{2} to get 0.

$$8a^{2}=8a^{2}$$

Cancel out 8 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(2a+3b\right)^{2}.

$$4a^{2}+12ab+9b^{2}+\left(2a-3b\right)^{2}=8a^{2}+18b^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(2a-3b\right)^{2}.

$$4a^{2}+12ab+9b^{2}+4a^{2}-12ab+9b^{2}=8a^{2}+18b^{2}$$

Combine 4a^{2} and 4a^{2} to get 8a^{2}.

$$8a^{2}+12ab+9b^{2}-12ab+9b^{2}=8a^{2}+18b^{2}$$

Combine 12ab and -12ab to get 0.

$$8a^{2}+9b^{2}+9b^{2}=8a^{2}+18b^{2}$$

Combine 9b^{2} and 9b^{2} to get 18b^{2}.

$$8a^{2}+18b^{2}=8a^{2}+18b^{2}$$

Subtract 8a^{2} from both sides.

$$8a^{2}+18b^{2}-8a^{2}=18b^{2}$$

Combine 8a^{2} and -8a^{2} to get 0.

$$18b^{2}=18b^{2}$$

Cancel out 18 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(2a+3b\right)^{2}.

$$4a^{2}+12ab+9b^{2}+\left(2a-3b\right)^{2}=8a^{2}+18b^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(2a-3b\right)^{2}.

$$4a^{2}+12ab+9b^{2}+4a^{2}-12ab+9b^{2}=8a^{2}+18b^{2}$$

Combine 4a^{2} and 4a^{2} to get 8a^{2}.

$$8a^{2}+12ab+9b^{2}-12ab+9b^{2}=8a^{2}+18b^{2}$$

Combine 12ab and -12ab to get 0.

$$8a^{2}+9b^{2}+9b^{2}=8a^{2}+18b^{2}$$

Combine 9b^{2} and 9b^{2} to get 18b^{2}.

$$8a^{2}+18b^{2}=8a^{2}+18b^{2}$$

Subtract 18b^{2} from both sides.

$$8a^{2}+18b^{2}-18b^{2}=8a^{2}$$

Combine 18b^{2} and -18b^{2} to get 0.

$$8a^{2}=8a^{2}$$

Cancel out 8 on both sides.

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

Reorder the terms.

$$\text{true}$$

This is true for any b.

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