Question

If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p\left(x\right)=2x^2-5x+7$, find a polynomial whose zeroes are of $(2\alpha +3\beta )$ and $(3\alpha +2\beta )$.

Answer 1

Step 1: Find the sum and product of the polynomial:

Here, the quadratic polynomial is:

$p\left(x\right)=2x^2-5x+7$

Given that, $\alpha$ and $\beta$ are the zeroes of the given quadratic polynomial.

Then we get,

$$\begin{gathered} \alpha +\beta =-\left(\frac{-5}{2}\right)=\frac{5}{2} \\ \alpha \beta =\frac{7}{2} \end{gathered}$$

Step 2: Find a polynomial whose zeroes are of $(2\alpha +3\beta )\text{and}(3\alpha +2\beta ).$

Sum of the zeroes $=(2\alpha +3\beta )+(3\alpha +2\beta )$

$$\begin{gathered} =5\alpha +5\beta \\ =5(\alpha +\beta ) \\ =5\times \frac{5}{2} \\ =\frac{25}{2} \end{gathered}$$

product of the zeroes $=(2\alpha +3\beta )\times (3\alpha +2\beta )$

$$\begin{gathered} =6{\alpha }^2+13\alpha \beta +6{\beta }^2 \\ =6({\alpha }^2+{\beta }^2)+13\alpha \beta \\ =6[{(\alpha +\beta )}^2-2\alpha \beta ]+13\alpha \beta \\ =6{(\alpha +\beta )}^2-12\alpha \beta +13\alpha \beta \\ =6{(\alpha +\beta )}^2+\alpha \beta \\ =6{\left(\frac{5}{2}\right)}^2+\frac{7}{2} \\ =6\times \frac{25}{4}+\frac{7}{2} \\ =\frac{75}{2}+\frac{7}{2} \\ =\frac{82}{2} \\ =41 \end{gathered}$$

Thus, the required polynomial becomes:

$$\begin{gathered} x^2-\text{(sum of zeroes})x+(\text{product of zeroes)} \\ {x}^2-\frac{25}{2}+41 \end{gathered}$$

Hence, the required quadratic polynomial is $(x^2-12.5x+41)$.