# Three equations with a common positive root

If the equations $$x^2+ax+12=0$$, $$x^2+bx+15=0$$ and $$x^2+(a+b)x+36=0$$ have a common positive root, then $$(b-2a)$$ is equal to

What I tried:

Let $$\alpha$$ be common positive root of all equation. Then

$$\alpha^2+a\alpha+12=0 \tag{1}$$

$$\alpha^2+b\alpha+15=0 \tag{2}$$

$$\alpha^2+(a+b)\alpha+36=0 \tag{3}$$

From (1) and (2), we get

$$(a-b)\alpha=3 \tag{4}$$

And doing (3) minus (1) and (3) minus (2), we get

$$b\alpha=-24 \text{ and} \\ a\alpha=-21$$

Now I don’t understand how to find the values of $$b$$ and $$a$$. Help me, please.

Thanks.

• Hint: you've tried simplifying the quadratics by eliminating the terms in $\alpha^2$, which is sensible, but you can also simplify them by eliminating terms in $\alpha$. Can you see how to do that? Commented Jul 17, 2023 at 11:12

Let $$x=\alpha$$ be a common positive root . Then, define :

\begin{align}A(x)&=x^2+ax+12\\ B(x)&=x^2+bx+15\\ C(x)&=x^2+(a+b)x+36\end{align}

\begin{align}A(\alpha)+B(\alpha)-C(\alpha)\\ =\alpha^2-9=0\end{align}

Thus, $$x_1=\alpha=3$$ .

Finally, if you plug $$x=3$$, then you can determine $$a$$ and $$b$$ .

Given that

$$x^2 + ax + 12 = 0 \tag{1}$$ $$x^2 + bx + 15 = 0\tag{2}$$ $$x^2 + (b+a)x + 36 = 0 \tag{3}$$

have a common root, subtracting $$(1)+(2)$$ from $$(3)$$ gives

$$-x^2 + 36 - 12 -15 = -x^2 + 9 = 0,$$

which gives us $$x = 3$$ as a solution.

Thus, we have $$9 + 3a + 12 = 0 \implies a = -1$$ and $$9 + 3b + 15 = 0 \implies b = -2$$.

Thus, $$b - 2a = -2 -2(-1) = 0$$.