make c the subject $c+1=\sqrt{a-ac^2}$

Question

Transpose the formula to c the subject,

$$c+1=\sqrt{a-ac^2}\tag1$$

A possible method

$$(c+1)^2=a-ac^2$$

$$c^2+2c+1+ac^2=a$$

$$c^2(1+a)+2c+1=a$$

I can't seem to make c the subject.

I could use the quadratic formula, but my teacher said you can't use it.

So there must be another way of making c the subject.

Can anyone help me. Thank.

Answer 1

Continuing from your second step,

$$(c+1)^2=a(1-c^2)=a(1-c)(c+1)$$

It is obvious that $c = -1$ is a solution

When $c \neq -1$,

$$c+1=a(1-c)=a-ac$$

$$c+ac=a-1$$

$$c(1+a)=a-1$$

$$c=\frac{a-1}{a+1}$$

Answer 2

Set $c+1=x^2$, with $x\ge0$, which is possible because $\sqrt{a-ac^2}\ge0$; then note that $$ a-ac^2=a(1-c)(1+c)=a(2-x^2)x^2 $$ Thus the equation reads $$ x^2=x\sqrt{a(2-x^2)} $$ Hence $x=0$, that is $c=-1$, or $x=\sqrt{a(2-x^2)}$.

In the latter case we can square both sides (because $x\ge0$ by assumption), leading to $$ x^2=a(2-x^2) $$ hence $$ x^2=\frac{2a}{1+a} $$ and $$ c=x^2-1=\frac{2a}{1+a}-1=\frac{a-1}{a+1} $$ Note that $$ 2-x^2=2-\frac{2a}{1+a}=\frac{2}{1+a} $$ leading to some limitations for $a$, because we need $$ a(2-x^2)=\frac{2a}{1+a}\ge0 $$ so $a\ge0$ or $a<-1$.

Hence the complete answer is

$c=-1$ or $c=\dfrac{a-1}{a+1}$ provided $a\ge0$ or $a<-1$. The equation has no solution for $-1\le a<0$.