Question

If & and ß case the zeroes of quadratic polynomiai 2x^2+5x + 7
Find a polynomial whose Zeroes are 2a+ 3B and 3a + 2B​

Answer 1

Given that

$\rm :\longmapsto\:\rm \: \alpha \: and \: \beta \: are \: zeroes \: {2x}^{2} + 5x + 7$

We know,

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

$\rm :\implies\: \alpha + \beta = - \: \dfrac{5}{2} - - (1)$

Also,

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

$\rm :\implies\: \alpha \beta = \dfrac{7}{2} - - - (2)$

Now, we have to find a quadratic polynomial whose zeroes are

$\rm :\longmapsto\:3 \alpha + 2 \beta \: and \: 2 \alpha + 3 \beta$

Consider,

Sum of zeroes,

$\rm :\longmapsto\:3 \alpha + 2 \beta + 2 \alpha + 3 \beta$

$\rm \: = \:5 \alpha + 5 \beta$

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

$\rm \: = \:5 \times ( - \dfrac{5}{2})$

$\rm \: = \:- \dfrac{25}{2}$

Now,

Consider,

Product of zeroes

$\rm :\longmapsto\:(3 \alpha + 2 \beta)( 2 \alpha + 3 \beta)$

$\rm \: = \: {6 \alpha }^{2} + 9 \alpha \beta + 4 \alpha \beta + {6 \beta }^{2}$

$\rm \: = \: {6 \alpha }^{2} + 13 \alpha \beta + {6 \beta }^{2}$

$\rm \: = \: {6 \alpha }^{2} + 12 \alpha \beta + {6 \beta }^{2} + \alpha \beta$

$\rm \: = \: 6({\alpha }^{2} + 2 \alpha \beta + {\beta }^{2}) + \alpha \beta$

$\rm \: = \: 6({\alpha + \beta ) }^{2} + \alpha \beta$

$\rm \: = \: 6 \times \dfrac{25}{4} + \dfrac{7}{2}$

$\rm \: = \: \dfrac{75}{2} + \dfrac{7}{2}$

$\rm \: = \: \dfrac{75 + 7}{2}$

$\rm \: = \: \dfrac{82}{2}$

$\rm \: = \: 41$

Hence,

The required polynomial is given by,

$\rm \: f(x) = k( {x}^{2} - (Sum \: of \: zeroes)x + Product \: of \: zeroes)$

$\rm \: f(x)= \: k \bigg({x}^{2} + \dfrac{25}{2}x + 41 \bigg) \: where \: k \ne \: 0$