Question

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

Answer 1

Answer:

Given:-

A quadratic polynomial 2x² - 5x + 7

The zeroes of the polynomial are α and β.

To Find:-

• The quadratic equation whose zeroes are 3α + 4β and 4α + 3β.

Concept used:-

• For a quadratic polynomial ax² + bx + c

$\sf Sum\: of \: zeroes = \sf \dfrac{-b}{a}$

$\sf Product\: of \: zeroes = \sf \dfrac{c}{a}$

Solution:-

• Comparing 2x² - 5x + 7 with ax² + bx + c

Here:-

$a = 2 \:\:\:\: , b = - 5 \:\:\:\: , c = 7$

$\implies \sf Sum \: of \: roots = \dfrac{-(-5)}{2} = \dfrac{5}{2}$

$\implies \bf \alpha + \beta = \dfrac{5}{2}$

$\sf Product \: of \: zeroes = \dfrac{7}{2}$

$\implies \bf \alpha \beta = \dfrac{7}{2}$

• Now, we need to find the quadratic equation whose zeroes are 3α + 4β and 4α + 3β.

$\sf Sum \: of \: zeroes = (3\alpha + 4\beta) + (4\alpha + 3\beta)$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 7\alpha + 7\beta = 7(\alpha + \beta)=7α+7β=7(α+β)$

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

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \dfrac{35}{2}$

$Product \: of \: zeroes=(3α+4β)(4α+5β)$

$\sf = 12\alpha^2 + 9\alpha \beta + 16\alpha \beta + 12\beta^2$

$\sf = 12\alpha^2 + 12\beta^2+ 25\alpha \beta$

$\sf = 12(\alpha^2 + \beta^2)+ 25\alpha \beta$

$\sf = 12(\alpha + \beta) ^{2} - 2 \alpha \beta + 25\alpha \beta$

$\sf = 12(\alpha + \beta) ^{2} + 23\alpha \beta$

$\sf = 12 \bigg( \dfrac{5}{2} \bigg) ^{2} + 23 \times \dfrac{7}{2}$

$\sf = \dfrac{150}{2} + \dfrac{161}{2}$

$\sf = \dfrac{311}{2}$

$Product \: of \: the \: roots \: =2311 \\ General \: form \: of \: a \: quadratic \: equation \: is:-$

General form of a quadratic equation is:-

$\sf x^2 - (sum \: of \: zeroes)x + product \: of \: zeroes = 0$

$\sf \implies x^2 - \dfrac{35}{2}x + \dfrac{311}{2} = 0$

$\sf \implies 2x^2 - 35x + 311 = 0$

Hence, Required quadratic polynomial is:-

$\large\dashrightarrow \underline{\boxed{\bf 2x^2 - 35x + 311}}$

Answer 2

Answer:

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