Solve (alpha^{v}+2)^2-(alpha^v-3)^2-10(alpha^v-1)=5

$$( \alpha ^ { v } + 2 ) ^ { 2 } - ( \alpha ^ { v } - 3 ) ^ { 2 } - 10 ( \alpha ^ { v } - 1 ) = 5$$

Solve for v (complex solution)

$$v\in \mathrm{C}$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\alpha ^{v}+2\right)^{2}.

$$\left(\alpha ^{v}\right)^{2}+4\alpha ^{v}+4-\left(\alpha ^{v}-3\right)^{2}-10\left(\alpha ^{v}-1\right)=5$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\alpha ^{v}-3\right)^{2}.

$$\left(\alpha ^{v}\right)^{2}+4\alpha ^{v}+4-\left(\left(\alpha ^{v}\right)^{2}-6\alpha ^{v}+9\right)-10\left(\alpha ^{v}-1\right)=5$$

To find the opposite of \left(\alpha ^{v}\right)^{2}-6\alpha ^{v}+9, find the opposite of each term.

$$\left(\alpha ^{v}\right)^{2}+4\alpha ^{v}+4-\left(\alpha ^{v}\right)^{2}+6\alpha ^{v}-9-10\left(\alpha ^{v}-1\right)=5$$

Combine \left(\alpha ^{v}\right)^{2} and -\left(\alpha ^{v}\right)^{2} to get 0.

$$4\alpha ^{v}+4+6\alpha ^{v}-9-10\left(\alpha ^{v}-1\right)=5$$

Combine 4\alpha ^{v} and 6\alpha ^{v} to get 10\alpha ^{v}.

$$10\alpha ^{v}+4-9-10\left(\alpha ^{v}-1\right)=5$$

Subtract 9 from 4 to get -5.

$$10\alpha ^{v}-5-10\left(\alpha ^{v}-1\right)=5$$

Use the distributive property to multiply -10 by \alpha ^{v}-1.

$$10\alpha ^{v}-5-10\alpha ^{v}+10=5$$

Combine 10\alpha ^{v} and -10\alpha ^{v} to get 0.

$$-5+10=5$$

Add -5 and 10 to get 5.

$$5=5$$

Compare 5 and 5.

$$\text{true}$$

This is true for any v.

$$v\in \mathrm{C}$$

Solve for α (complex solution)

$$\alpha \in \mathrm{C}$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\alpha ^{v}+2\right)^{2}.

$$\left(\alpha ^{v}\right)^{2}+4\alpha ^{v}+4-\left(\alpha ^{v}-3\right)^{2}-10\left(\alpha ^{v}-1\right)=5$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\alpha ^{v}-3\right)^{2}.

$$\left(\alpha ^{v}\right)^{2}+4\alpha ^{v}+4-\left(\left(\alpha ^{v}\right)^{2}-6\alpha ^{v}+9\right)-10\left(\alpha ^{v}-1\right)=5$$

To find the opposite of \left(\alpha ^{v}\right)^{2}-6\alpha ^{v}+9, find the opposite of each term.

$$\left(\alpha ^{v}\right)^{2}+4\alpha ^{v}+4-\left(\alpha ^{v}\right)^{2}+6\alpha ^{v}-9-10\left(\alpha ^{v}-1\right)=5$$

Combine \left(\alpha ^{v}\right)^{2} and -\left(\alpha ^{v}\right)^{2} to get 0.

$$4\alpha ^{v}+4+6\alpha ^{v}-9-10\left(\alpha ^{v}-1\right)=5$$

Combine 4\alpha ^{v} and 6\alpha ^{v} to get 10\alpha ^{v}.

$$10\alpha ^{v}+4-9-10\left(\alpha ^{v}-1\right)=5$$

Subtract 9 from 4 to get -5.

$$10\alpha ^{v}-5-10\left(\alpha ^{v}-1\right)=5$$

Use the distributive property to multiply -10 by \alpha ^{v}-1.

$$10\alpha ^{v}-5-10\alpha ^{v}+10=5$$

Combine 10\alpha ^{v} and -10\alpha ^{v} to get 0.

$$-5+10=5$$

Add -5 and 10 to get 5.

$$5=5$$

Compare 5 and 5.

$$\text{true}$$

This is true for any \alpha .

$$\alpha \in \mathrm{C}$$

Solve for v

$$\left\{\begin{matrix}v\in \mathrm{R}\text{, }&\alpha >0\text{ or }\left(\alpha <0\text{ and }Denominator(v)\text{bmod}2=1\right)\\v>0\text{, }&\alpha =0\end{matrix}\right.$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\alpha ^{v}+2\right)^{2}.

$$\left(\alpha ^{v}\right)^{2}+4\alpha ^{v}+4-\left(\alpha ^{v}-3\right)^{2}-10\left(\alpha ^{v}-1\right)=5$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\alpha ^{v}-3\right)^{2}.

$$\left(\alpha ^{v}\right)^{2}+4\alpha ^{v}+4-\left(\left(\alpha ^{v}\right)^{2}-6\alpha ^{v}+9\right)-10\left(\alpha ^{v}-1\right)=5$$

To find the opposite of \left(\alpha ^{v}\right)^{2}-6\alpha ^{v}+9, find the opposite of each term.

$$\left(\alpha ^{v}\right)^{2}+4\alpha ^{v}+4-\left(\alpha ^{v}\right)^{2}+6\alpha ^{v}-9-10\left(\alpha ^{v}-1\right)=5$$

Combine \left(\alpha ^{v}\right)^{2} and -\left(\alpha ^{v}\right)^{2} to get 0.

$$4\alpha ^{v}+4+6\alpha ^{v}-9-10\left(\alpha ^{v}-1\right)=5$$

Combine 4\alpha ^{v} and 6\alpha ^{v} to get 10\alpha ^{v}.

$$10\alpha ^{v}+4-9-10\left(\alpha ^{v}-1\right)=5$$

Subtract 9 from 4 to get -5.

$$10\alpha ^{v}-5-10\left(\alpha ^{v}-1\right)=5$$

Use the distributive property to multiply -10 by \alpha ^{v}-1.

$$10\alpha ^{v}-5-10\alpha ^{v}+10=5$$

Combine 10\alpha ^{v} and -10\alpha ^{v} to get 0.

$$-5+10=5$$

Add -5 and 10 to get 5.

$$5=5$$

Compare 5 and 5.

$$\text{true}$$

This is true for any v.

$$v\in \mathrm{R}$$

Solve for α

$$\left\{\begin{matrix}\\\alpha >0\text{, }&\text{unconditionally}\\\alpha <0\text{, }&Denominator(v)\text{bmod}2=1\\\alpha =0\text{, }&v>0\end{matrix}\right.$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\alpha ^{v}+2\right)^{2}.

$$\left(\alpha ^{v}\right)^{2}+4\alpha ^{v}+4-\left(\alpha ^{v}-3\right)^{2}-10\left(\alpha ^{v}-1\right)=5$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\alpha ^{v}-3\right)^{2}.

$$\left(\alpha ^{v}\right)^{2}+4\alpha ^{v}+4-\left(\left(\alpha ^{v}\right)^{2}-6\alpha ^{v}+9\right)-10\left(\alpha ^{v}-1\right)=5$$

To find the opposite of \left(\alpha ^{v}\right)^{2}-6\alpha ^{v}+9, find the opposite of each term.

$$\left(\alpha ^{v}\right)^{2}+4\alpha ^{v}+4-\left(\alpha ^{v}\right)^{2}+6\alpha ^{v}-9-10\left(\alpha ^{v}-1\right)=5$$

Combine \left(\alpha ^{v}\right)^{2} and -\left(\alpha ^{v}\right)^{2} to get 0.

$$4\alpha ^{v}+4+6\alpha ^{v}-9-10\left(\alpha ^{v}-1\right)=5$$

Combine 4\alpha ^{v} and 6\alpha ^{v} to get 10\alpha ^{v}.

$$10\alpha ^{v}+4-9-10\left(\alpha ^{v}-1\right)=5$$

Subtract 9 from 4 to get -5.

$$10\alpha ^{v}-5-10\left(\alpha ^{v}-1\right)=5$$

Use the distributive property to multiply -10 by \alpha ^{v}-1.

$$10\alpha ^{v}-5-10\alpha ^{v}+10=5$$

Combine 10\alpha ^{v} and -10\alpha ^{v} to get 0.

$$-5+10=5$$

Add -5 and 10 to get 5.

$$5=5$$

Compare 5 and 5.

$$\text{true}$$

This is true for any \alpha .

$$\alpha \in \mathrm{R}$$