Question

if alpha and beta are the zeros of the polynomial 2x^2-5x+7 then find a quadratic polynomial whose zeros are (3alpha +4beta) and (4alpha +3beta)

Answer 1

Last you just solve it by yourself I have done the main part

Answer 2

Answer:

16y²+56-40y

Step-by-step explanation:

α and β are zeroes of the equation p(x)=2x²-5x+7

Solving for α and β,

$2x^2-5x+7=0\\2x^2-5x=-7\\x^2-5x/2=-7/2\\\\x^2-5x/2+(5/4)^2=-7/2+25/16\\\\(x-5/4)^2=\frac{25-56}{16} \\\\x-5/4=\sqrt{\frac{-31}{16}}=\pm \frac{i\sqrt 31}{4} \\x=\frac{5\pm i\sqrt 31}{4}$

So,

$\alpha=\frac{5+i\sqrt{31}}{4}\\ \beta=\frac{5-i\sqrt{31}}{4}$

Hence,

$3\alpha+4\beta=\frac{15+3i\sqrt{31}+20-4i\sqrt{31}}{4}=\frac{35-i\sqrt{31}}{4} \\ \\4\alpha+3\beta=\frac{20+4i\sqrt{31}+15-3i\sqrt{31}}{4} =\frac{35+i\sqrt{31}}{4}$

So root of the polynomial is $\frac{5\pm i\sqrt{31}}{4}$

Then we will have

$y=\frac{5\pm i \sqrt{31}}{4} \\\\4y=5\pm i\sqrt{31}\\4y-5=\pm i\sqrt{31}\\(4y-5)^2=(\pm i\sqrt{31})^2\\16y^2+25-40y=-31\\16y^2+56-40y=0$

Hence, the polynomial is 16y²+56-40y.