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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.

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    $\begingroup$ 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? $\endgroup$ Commented Jul 17, 2023 at 11:12

2 Answers 2

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

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

This leads to :

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

Thus, $x_1=\alpha=3$ .

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

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

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