Question

If 'a' and 'b' are the roots of the equation $3m^2=6m+5$, find the value of $(a+2b)(2a+b)$

Answer 1

We know that if $a$ and $b$ are the roots of a quadratic equation $a{x}^2+bx+c=0$, the sum of the roots is $a+b=-\frac{b}{a}$ and the product of the roots is $ab=\frac{c}{a}$.

Here, the given quadratic equation $3m^2=6m+5$ can be rewritten as $3m^2-6m-5=0$ is in the form $p{x}^2+qx+r=0$ where $p=3,q=-6$ and $r=-5$.

The sum of the roots is:

$a+b=-\frac{q}{p}=-\frac{(-6)}{3}=2$

The product of the roots is $\frac{r}{p}$ that is:

$ab=\frac{r}{p}=\frac{-5}{3}$

Now, we find $(a^2+b^2)$ as follows:

$(a+b{)}^2=a^2+b^2+2ab\Rightarrow 2^2=a^2+b^2+(2\times -\frac{5}{3})\Rightarrow 4=a^2+b^2-\frac{10}{3}\Rightarrow a^2+b^2=\frac{10}{3}+4\Rightarrow a^2+b^2=\frac{10+12}{3}\Rightarrow a^2+b^2=\frac{22}{3}$

Therefore,

$(a+2b)(2a+b)=a(2a+b)+2b(2a+b)=2a^2+ab+4ab+2b^2=2(a^2+b^2)+5ab=(2\times \frac{22}{3})+(5\times -\frac{5}{3})$

$=\frac{44}{3}-\frac{25}{3}=\frac{19}{3}$

Hence, the value of $(a+2b)(2a+b)=\frac{19}{3}$.