Let , be the roots of the equation x^2-3x+r=0, and ( )/(2),2 be the roots of the equation …

Question

Let $\alpha ,\beta $ be the roots of the equation ${x}^{2}-3x+r=0$, and $\frac{\alpha }{2},2\beta $ be the roots of the equation ${x}^{2}+3x+r=0$.
If the roots of the equation ${x}^{2}+6x=m$ are $2\alpha +\beta +2r$ and $\alpha -2\beta -\frac{r}{2}$, then $m$ is equal to :

Accepted Answer

For the equation ${x}^{2}-3x+r=0$ with roots $\alpha ,\beta $, by using the standard results:

$\begin{gathered} \alpha +\beta =3 \ \frac{\alpha }{2}+2\beta =-3 \end{gathered}$

The first line comes from ``sum of roots'' $=-\frac{x}{{x}^{2}}=-\frac{-3}{1}=3$.

The second line comes from the equation ${x}^{2}+3x+r=0$, whose roots are $\frac{\alpha }{2}$ and $2\beta $. So, their sum is:

$\frac{\alpha }{2}+2\beta =-\frac{3}{1}=-3$

Now solve the two equations:

$\alpha +\beta =3\Rightarrow \alpha =3-\beta $

Substitute into $\frac{\alpha }{2}+2\beta =-3$:

$\frac{3-\beta }{2}+2\beta =-3$

Multiply by $2$:

$3-\beta +4\beta =-6\Rightarrow 3+3\beta =-6\Rightarrow 3\beta =-9\Rightarrow \beta =-3$

Then $\alpha =3-(-3)=6$.

Now use ``product of roots'' for ${x}^{2}-3x+r=0$:

$\alpha \beta =r\Rightarrow r=6	imes (-3)=-18$

Now consider the equation ${x}^{2}+6x=m$. Rewrite it in standard form:

${x}^{2}+6x-m=0$

So, product of its roots $=\frac{-m}{1}=-m$.

The roots are given as $2\alpha +\beta +2r$ and $\alpha -2\beta -\frac{r}{2}$. Substitute $\alpha =6$, $\beta =-3$, $r=-18$:

First root:

$2\alpha +\beta +2r=2(6)+(-3)+2(-18)=12-3-36=-27$

Second root:

$\alpha -2\beta -\frac{r}{2}=6-2(-3)-\frac{-18}{2}=6+6+9=21$

Now product of roots:

$\begin{gathered} =(2\alpha +\beta +2r)(\alpha -2\beta -\frac{r}{2}) \ \Rightarrow -m=(-27)(21) \ \Rightarrow m=567 \end{gathered}$

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