Question

if alpha and beta are the zeroes of the quadratic polynomial 2x²-5x +7, then find the quadratic polynomial whose zeros are 3alpha +4 beta 4 alpha +3 beta.​

Answer 1

Answer:

$\boxed{\sf \: g(x) = \frac{k}{2} \bigg(2 {x}^{2} - 35 x + 67 \bigg), \: \: where \:k \: \ne \: 0 \: }$

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

Given that,

$\sf \: \alpha \: and \: \beta \: are \: zeroes \: of \: polynomial \: {2x}^{2} - 5x + 7$

We know,

$\boxed{{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\sf \:  \implies \: \alpha + \beta = - \dfrac{( - 5)}{2} = \dfrac{5}{2}$

Also,

$\boxed{{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\sf \:  \implies \: \alpha \beta = \dfrac{7}{2}$

Now,

We have to find quadratic polynomial whom zeroes are

$3\alpha + 4\beta$ and $4\alpha + 3\beta$

Now, Consider

$\sf \: S = (3 \alpha + 4\beta) + (4 \alpha + 3 \beta )$

$\sf \: S = 7 \alpha + 7\beta$

$\sf \: S = 7 (\alpha + \beta)$

$\sf \: S = 7 \times \dfrac{5}{2}$

$\sf \: \bf\implies \:S = \dfrac{35}{2}$

Now, Consider

$\sf \: P = (3 \alpha + 4\beta) \times (4 \alpha + 3 \beta )$

$\sf \: P = {12 \alpha }^{2} + 9 \alpha \beta + 16 \alpha \beta + {12 \beta }^{2}$

$\sf \: P = {12 \alpha }^{2} + {12 \beta }^{2} + 25 \alpha \beta$

$\sf \: P = {12 \alpha }^{2} + {12 \beta }^{2} + 24 \alpha \beta + \alpha \beta$

$\sf \: P = 12({ \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta) + \alpha \beta$

$\sf \: P = 12 \times \dfrac{5}{2} + \dfrac{7}{2}$

$\sf \: P = \dfrac{60}{2} + \dfrac{7}{2}$

$\sf \: \bf\implies \:P = \dfrac{67}{2}$

So, Required quadratic polynomial having sum of zeroes as S and product of zeroes as P, is given by

$\sf \: g(x) = k( {x}^{2} - Sx + P), \: \: where \:k \: \ne \: 0$

$\sf \: g(x) = k\bigg( {x}^{2} - \dfrac{35}{2} x + \dfrac{67}{2} \bigg), \: \: where \:k \: \ne \: 0$

$\sf \: \bf\implies \:g(x) = \frac{k}{2} \bigg(2 {x}^{2} - 35 x + 67 \bigg), \: \: where \:k \: \ne \: 0$