Review question

Ref: R7350

## Solution

The equation $x^2+ax+b=0$, where $a$ and $b$ are different, has solutions $x=a$ and $x=b$. How many such equations are there?

As $x=a$ and $x=b$ are the solutions to the equation $x^2+ax+b=0$, we have $x^2+ax+b\equiv(x-a)(x-b).$

Expanding gives $x^2+ax+b\equiv x^2+(-a-b)x+ab$

Comparing coefficients of these two quadratics, we see that we require $a=-a-b$ and $b=ab$.

From the first condition we deduce $a=-b/2$. Substituting this into the second condition gives $b=-\frac{b^2}{2},$ which has solutions $b=0$ and $b=-2$.

As $a$ and $b$ are assumed to be different, we cannot use $b=0$ because this corresponds to $a=0$.

For $b=-2$ we get $a=1$, which is different from $b$.

Alternatively, we could have started with the second condition, $b=ab$. Rearranging this as $ab-b=0$ and factorising to $b(a-1)=0$ shows that $a=1$ or $b=0$, and we obtain the same solutions ($a=b=0$ or $a=1$, $b=-2$) again.

As yet a third approach, we could have substituted $x=a$ and $x=b$ into the original equation to obtain $a^2+a^2+b=0$ and $b^2+ab+b=0$; the first gives $b=-2a^2$ and the second gives $b(a+b+1)=0$. From the second equation, we obtain $b=0$ and hence $a=0$, or $b=-a-1$. Substituting the latter into $b=-2a^2$ gives $2a^2-a-1=0$ or $(2a+1)(a-1)=0$, so $a=1$ or $a=-\frac12$. If $a=1$, then $b=-2$, as before. If $a=-\frac12$, then $b=-\frac12$, so $a=b$.

This is strange: we have obtained a third possible solution using this approach, which did not appear in either of the first two approaches. (Though we have to reject it anyway, as $a=b$, it is still odd that we have found a further solution.)

What’s going on here?

Since the solutions to $x^2+x-2=0$ are indeed $x=1$ and $x=-2$, there is exactly one equation that meets the conditions.