Question

Rearrange the top equation to match the bottom equation.

Answer 1

See below for proof.

Explanation:

Rearrange the given equation into a quadratic with terms in y:

`\implies x^2=y^2+a\left((x-y)/(a)\right)^2`

`\implies y^2+a\left(((x-y)^2)/(a^2)\right)-x^2=0`

`\implies y^2+((x-y)^2)/(a)-x^2=0`

`\implies y^2+(x^2)/(a)-(2xy)/(a)+(y^2)/(a)-x^2=0`

`\implies y^2+(y^2)/(a)-(2xy)/(a)+(x^2)/(a)-x^2=0`

`\implies \left(1+(1)/(a)\right)y^2-\left((2x)/(a)\right)y+\left((x^2)/(a)-x^2\right)=0`

`\boxed{\begin{minipage}{3.6 cm}\underline{Quadratic Formula}\\\\$x=(-b \pm √(b^2-4ac))/(2a)$\\\\when $ax^2+bx+c=0$ \\\end{minipage}}`

Therefore, comparing the quadratic:

`a=\left(1+(1)/(a)\right)`

`b=\left(-(2x)/(a)\right)`

`c=\left((x^2)/(a)-x^2\right)`

Substitute the values of a, b and c into the formula:

`\implies y=\frac{-\left(-(2x)/(a)\right) \pm \sqrt{\left(-(2x)/(a)\right)^2-4\left(1+(1)/(a)\right)\left((x^2)/(a)-x^2\right)}}{2\left(1+(1)/(a)\right)}`

`\implies y=\frac{(2x)/(a) \pm \sqrt{(4x^2)/(a^2)-4\left((a+1)/(a)\right)\left((x^2-ax^2)/(a)\right)}}{2\left(1+(1)/(a)\right)}`

`\implies y=\frac{(2x)/(a) \pm \sqrt{(4x^2)/(a^2)-(4(a+1)(x^2-ax^2))/(a^2)}}{2\left(1+(1)/(a)\right)}`

`\implies y=\frac{(2x)/(a) \pm \sqrt{(4x^2)/(a^2)-(4x^2-4a^2x^2)/(a^2)}}{2\left(1+(1)/(a)\right)}`

`\implies y=\frac{(2x)/(a) \pm \sqrt{(4x^2)/(a^2)-(4x^2)/(a^2)+(4a^2x^2)/(a^2)}}{2\left(1+(1)/(a)\right)}`

`\implies y=((2x)/(a) \pm √(4x^2))/(2\left(1+(1)/(a)\right))`

`\implies y=((2x)/(a) \pm2x)/(2\left(1+(1)/(a)\right))`

`\implies y=((x)/(a) \pm x)/(1+(1)/(a))`

`\implies y=x\left[((1)/(a) \pm 1)/(1+(1)/(a))\right]`

Case 1

`\implies y=x\left[((1)/(a)-1 )/(1+(1)/(a))\right]`

Add and subtract 1 from the numerator:

`\implies y=x\left[(1+(1)/(a)-1-1 )/(1+(1)/(a))\right]`

`\implies y=x\left[(1+(1)/(a)-2 )/(1+(1)/(a))\right]`

`\implies y=x\left[(1+(1)/(a))/(1+(1)/(a)) -(2)/(1+(1)/(a))\right]`

`\implies y=x\left[1 -(2)/(1+(1)/(a))\right]`

Case 2

`\implies y=x\left[((1)/(a)+1 )/(1+(1)/(a))\right]`

`\implies y=x`