Question

2) it l and m are zeroes of the polynomial p(x)=2x2-5x+7.find a polynomial whose zeroes are 2l+3and 2m+3.
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Answer 1

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

Given polynomial is

$\rm :\longmapsto\:p(x) = {2x}^{2} - 5x + 7$

and

$\rm :\longmapsto\:l \: and \: m \: are \: zeroes \: of \: p(x).$

We know, that

In a quadratic polynomial,

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

$\bf\implies \:l + m = - \dfrac{( - 5)}{2} = \dfrac{5}{2}$

and

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

$\bf\implies \:lm = \dfrac{7}{2}$

Now,

We have to find the quadratic polynomial whose zeroes are 2l + 3 and 2m + 3.

So,

Sum of zeroes, is

$\rm \: = \: \:2l + 3 + 2m + 3$

$\rm \: = \: \:2(l +m) + 6$

$\rm \: = \: \:2 \times \dfrac{5}{2} + 6$

$\rm \: = \: \:5 + 6$

$\rm \: = \: \:11$

Now,

Product of zeroes, is

$\rm \: = \: \:(2l + 3)(2m + 3)$

$\rm \: = \: \:4lm + 6l + 6m + 9$

$\rm \: = \: \:4lm + 6(l + m) + 9$

$\rm \: = \: \:4 \times \dfrac{7}{2} + 6 \times \dfrac{5}{2} + 9$

$\rm \: = \: \:14 + 15 + 9$

$\rm \: = \: \:38$

So, the required polynomial be f(x) having zeroes (2l + 3) and (2m + 3), is given by

$\boxed{\rm\:f(x) = k\bigg( {x}^{2} - (Sum \: of \: zeroes)x + Product \: of \: zeroes \bigg) \: where \: k \: \ne \: 0}$

So On substituting all values evaluated above, we get

$\rm :\longmapsto\:f(x) = k\bigg( {x}^{2} - 11x + 38 \bigg) \: where \: k \: \ne \: 0$

Additional Information :-

$\rm :\longmapsto\: \alpha \: and \: \beta \: are \: zeroes \: of \: {ax}^{2} + bx + c \: then$

$\red{\boxed{ \bf{ \: \alpha + \beta = - \frac{b}{a}}}}$

and

$\red{\boxed{ \bf{ \: \alpha\beta = \frac{c}{a}}}}$

$\rm :\longmapsto\: \alpha, \: \beta \: and \gamma \: are \: zeroes \: of \: {ax}^{3} + b {x}^{2} + cx + d \: then$

$\red{\boxed{ \bf{ \: \alpha + \beta + \gamma = - \frac{b}{a}}}}$

$\red{\boxed{ \bf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}}}}$

$\red{\boxed{ \bf{ \: \alpha\beta \gamma = - \: \frac{d}{a}}}}$

Answer 2

Answer:

Step-by-step explanation:

Given polynomial is

and

We know, that

In a quadratic polynomial,

and

Now,

We have to find the quadratic polynomial whose zeroes are 2l + 3 and 2m + 3.

So,

Sum of zeroes, is

Now,

Product of zeroes, is

So, the required polynomial be f(x) having zeroes (2l + 3) and (2m + 3), is given by

So On substituting all values evaluated above, we get

Additional Information :-

and