Solve identity (a - b) ^ 2 = a ^ 2 + 2ab + b ^ 2 the 982 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

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

Solve for a (complex solution)

$\left\{\begin{matrix}\\a=0\text{, }&\text{unconditionally}\\a\in \mathrm{C}\text{, }&b=0\end{matrix}\right.$

Steps for Solving Linear Equation

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

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

Subtract $a^{2}$ from both sides.

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

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

$$-2ab+b^{2}=2ab+b^{2}$$

Subtract $2ab$ from both sides.

$$-2ab+b^{2}-2ab=b^{2}$$

Combine $-2ab$ and $-2ab$ to get $-4ab$.

$$-4ab+b^{2}=b^{2}$$

Subtract $b^{2}$ from both sides.

$$-4ab=b^{2}-b^{2}$$

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

$$-4ab=0$$

The equation is in standard form.

$$\left(-4b\right)a=0$$

Divide $0$ by $-4b$.

$$a=0$$

Solve for b (complex solution)

$\left\{\begin{matrix}\\b=0\text{, }&\text{unconditionally}\\b\in \mathrm{C}\text{, }&a=0\end{matrix}\right.$

Steps for Solving Linear Equation

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

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

Subtract $2ab$ from both sides.

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

Combine $-2ab$ and $-2ab$ to get $-4ab$.

$$a^{2}-4ab+b^{2}=a^{2}+b^{2}$$

Subtract $b^{2}$ from both sides.

$$a^{2}-4ab+b^{2}-b^{2}=a^{2}$$

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

$$a^{2}-4ab=a^{2}$$

Subtract $a^{2}$ from both sides.

$$-4ab=a^{2}-a^{2}$$

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

$$-4ab=0$$

The equation is in standard form.

$$\left(-4a\right)b=0$$

Divide $0$ by $-4a$.

$$b=0$$

Solve for a

$\left\{\begin{matrix}\\a=0\text{, }&\text{unconditionally}\\a\in \mathrm{R}\text{, }&b=0\end{matrix}\right.$

Steps for Solving Linear Equation

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

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

Subtract $a^{2}$ from both sides.

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

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

$$-2ab+b^{2}=2ab+b^{2}$$

Subtract $2ab$ from both sides.

$$-2ab+b^{2}-2ab=b^{2}$$

Combine $-2ab$ and $-2ab$ to get $-4ab$.

$$-4ab+b^{2}=b^{2}$$

Subtract $b^{2}$ from both sides.

$$-4ab=b^{2}-b^{2}$$

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

$$-4ab=0$$

The equation is in standard form.

$$\left(-4b\right)a=0$$

Divide $0$ by $-4b$.

$$a=0$$

Solve for b

$\left\{\begin{matrix}\\b=0\text{, }&\text{unconditionally}\\b\in \mathrm{R}\text{, }&a=0\end{matrix}\right.$

Steps for Solving Linear Equation

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

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

Subtract $2ab$ from both sides.

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

Combine $-2ab$ and $-2ab$ to get $-4ab$.

$$a^{2}-4ab+b^{2}=a^{2}+b^{2}$$

Subtract $b^{2}$ from both sides.

$$a^{2}-4ab+b^{2}-b^{2}=a^{2}$$

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

$$a^{2}-4ab=a^{2}$$

Subtract $a^{2}$ from both sides.

$$-4ab=a^{2}-a^{2}$$

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

$$-4ab=0$$

The equation is in standard form.

$$\left(-4a\right)b=0$$

Divide $0$ by $-4a$.

$$b=0$$