Question

If α and β are the zeros of the quadratic polynomial f(x) = 2x² -5x + 7, find a polynomial whose zeros are 2α+ 3β and 3α+ 2β? [Class 10]

Answer 1

Answer:

K[$x^{2}$  -  25x +25]

Step-by-step explanation:

given , polynomial   f(x) =   $2x^{2}$ - 5x +7

zeroes  = α  and β

since,  sum of zeroes  = - b/a

           α+ β=   - (-5)/1

                    = 5

 product of zeroes  =   c/a  

              αβ  =   7/1 = 7

Now,    sum of zeroes                        ;     product of zeroes

 (2α +  3 β )  +   (3β  +2α)                               (2α +  3 β ) (3β  +2α)  

           =    2α +  3 β + 3β  +2α               =           (2α +  3 β )^2      

            =     5α + 5β                                =           (5)^2

            =   5 (   α+ β  )                             =        25

         =    5 ( 5)

          = 25

 hence,   required polynomial  

=   k [ x^2 -   (   (2α +  3 β )  +   (3β  +2α)  ) x + (2α +  3 β ) (3β  +2α)  ]

=    k [ x^2 - 25x +25]

Answer 2

Answer:

$\tt \purple{FORMULA \: USED : - }$

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

$\mathtt\purple{GIVEN:-}$

If α and β are the zeros of the quadratic polynomial f(x) = 2x² -5x + 7

$\mathtt\purple{SOLUTION:-}$

So,

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

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

And

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

$\alpha \beta = \purple{ \frac{7}{2} }$

Now,

$\tt \purple{SUM \: OF \: THE \: ZEROS (s)\:} \tt{= (2 \alpha + 3 \beta ) + (3 \alpha + 2 \beta )}$

$\implies \mathtt{5 \alpha + 5 \beta }$

$\implies \tt \purple{5(\alpha + \beta )}$

Now Substitute the value of α+β that we get above

$\implies \tt \purple{5 \times \frac{5}{2} }$

$\implies \tt \purple{ \frac{25}{2} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\tt\purple{PRODUCT \: OF \: THE \: ZEROS (p)\:} = \tt{\: (2 \alpha + 3 \beta )(3 \alpha + 2 \beta)}$

$\implies \tt{{6 \alpha }^{2} + 9 \alpha \beta + 3 \alpha \beta + {6 \beta }^{2} }$

$\implies \tt{6({ \alpha }^{2} + { \beta }^{2} )+ 13 \alpha \beta }$

$\implies \tt{6{ \alpha }^{2} + 6 { \beta }^{2} + 12 \alpha \beta + \alpha \beta }$

$\implies \tt{6( { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta ) - \alpha \beta }$

$\implies \tt{6( { \alpha } + { \beta })^{2} - \alpha \beta }$

Now Substitute the value of αβ that we get above

$\implies \tt{6 \times ( \frac {5}{2} )^{2} + \frac{7}{2} }$

$\implies \tt{6 \times ( \frac {25}{4} ) + \frac{7}{2} }$

$\implies \tt{ \frac {75}{2} + \frac{7}{2} }$

$\implies \tt{ \frac{82}{2} }$

$\implies \tt{ 41 }$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\tt\purple{PRODUCT \: OF \: THE \: ZEROS (p)\:} = \tt{ 41}$

$\tt \purple{SUM \: OF \: THE \: ZEROS (s)\:} \tt{= \frac{25}{2} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\mathtt{GENERAL \: FORM \: OF \: EQUATION}$

$\tt\purple{ {x}^{2} \: -SUM \: OF \: ZEROS(x) + PRODUCT \: OF \: ZEROS}$

$\tt{ {x}^{2} - \frac{25x}{2} + 41 }$