Question

If alpha and beta are the zeroes of the polynomial 2x^2-5x +7 find the quadratic polynomial whose zeroes are 2alpha+3 beta and 3 alpha +2 beta
​

Answer 1

When roots are given, equation in quadratic form

x

2

−(α+β)x+αβ=0

Here roots are (2α+3β) and (3α+2β)

∴x

2

−(2α+3β+3α+2β)x+(2α+3β)(3α+2β)=0

x

2

−5(α+β)x+6α

2

+4αβ+4αβ+6β

2

=0

f(x)=2x

2

−5x+7

α+β=

2

5

, αβ=

2

7

∴α

2

+β

2

=(α+β)

2

−2αβ

=

4

25

−7=

4

−3

x

2

−5(

2

5

)x+6(−

4

3

)+13×

2

7

=0

∴2x

2

−50x−18+182=0

2x

2

−50x+164=0

∴x

2

−25x+82=0.

Answer 2

Given -

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

To find -

$the \: quadratic \: polynomial \: whose \: \\ zeroes \: are \: 2 \alpha + 3 \beta and \: 3 \alpha + 2 \beta$

Solution -

$p(x) = 2{x}^{2} - 5x + 7$

$\alpha \: and \: \beta \: are \: zeroes \: of \: given \: quadratic \: polynomial$

$\alpha + \beta = \frac{ - b}{a}$

$\alpha + \beta = - \frac{( - 5)}{2} = \frac{5}{2} .....1)$

$\alpha \times \beta = \frac{c}{a}$

$\alpha \beta = \frac{7}{2} ....2)$

$zeroes \: of \: polynomial \: are \\ (2 \alpha + 3 \beta ) \: and \: (3 \alpha + 2 \beta )$

$then \: \alpha + \beta = (2 \alpha + 3 \beta ) + (3 \alpha + 2 \beta )$

$\alpha + \beta = 5 \alpha + 5 \beta \\ \: \: \: \: \: \: \: \: \: \: \: \: \: = 5( \alpha + \beta )$

From 1) we will get-

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

Now,

$\alpha \beta = (2 \alpha + 3 \beta )(3 \alpha + 2 \beta ) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \alpha \beta = 6 { \alpha }^{2} + 4 \alpha \beta + 9 \alpha \beta + {6 \beta }^{2} \\ \alpha \beta = 6( { \alpha }^{2} + { \beta }^{2} ) + 13 \alpha \beta$

We know that,

(a²+b²) =(a+b) ²-2ab

$\alpha \beta = 6[ {( \alpha + \beta )}^{2} - 2 \alpha \beta] + 13 \alpha \beta$

$\alpha \beta =6[ ( { \frac{5}{2} )}^{2} - 2 \times \frac{7}{2}] + 13 \times \frac{7}{2} (by \: 1 \: and \: 2)$

$\alpha \beta = 6( \frac{25}{4} - 7) + \frac{91}{2}$

$\alpha \beta = 6 \times \frac{ - 3}{4} + \frac{91}{2}$

$\alpha \beta = \frac{ - 9}{2} + \frac{91}{2}$

$\alpha \beta = \frac{ - 9 + 91}{2}$

$\alpha \beta = \frac{82}{2}$

$\alpha \beta = 41$

$\therefore \: the \: polynomial \: obtained \: is \:$

$g(x) = {x}^{2} - ( \alpha + \beta )x + \alpha \beta$

$g(x) = {x}^{2} - \frac{25}{2} x + 41$