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

Step-by-step explanation:

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.

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

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

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

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

$\sf Product \: of \: zeroes = (3\alpha + 4\beta) (4\alpha + 5\beta)$

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

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

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

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

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

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

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

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

$\sf Product\: of \: the \: roots = \dfrac{311}{2}$

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

Given-

• if $\alpha$ and $\beta$ are the zeroes of the polynomial 2x^2 -5x+7

To Find-

• Quadratic polynomial whose zeroes are 3$\alpha$ +4$\beta$ and 4$\alpha+3\beta.$

Concept used

• $\rm \: \alpha + \beta = \dfrac{ - b}{a}$
• $\rm \alpha \beta = \dfrac{c}{a}$

$\bf \: k \{ {x}^{2} - (s) x + p \}$

Solution-

For Given Quadratic equation

$\bf \: 2{x}^{2} - 5x + 7 = 0$

Here,

• a=2

• b=-5

• c=7

We know,

$\rm \alpha + \beta = \dfrac{ - b}{a} \\ \\ \implies \boxed{ \mathbf{ \alpha + \beta = \frac{5}{2} }}$

We also know,

$\rm \: \alpha \beta = \dfrac{c}{a} \\ \\ \implies \boxed{ \bf \alpha \beta = \frac{7}{2} }$

Required equation

$\bf k \{ {x}^{2} - (s)x + p \}$

Here

$\rm \: sum = 3 \alpha + 4 \beta + 4 \alpha + 3 \beta \\ \\ \implies \rm sum = 7 \alpha + 7 \beta \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \rm \: sum = 7( \frac{5}{2} ) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \boxed{ \mathbf{ \: sum = \frac{35}{2} }} \: \: \: \: \: \: \: \: \: \: \: \:$

$\rm \: product \: = (3 \alpha + 4 \beta )(4 \alpha + 3 \beta ) \\ \\ \rm \implies product = 12 { \alpha }^{2} + 9 \alpha \beta + 16 \alpha \beta + 12 { \beta }^{2} \\ \\ \implies \rm product = 12( { \alpha }^{2} + { \beta }^{2} ) + 25 \beta \alpha \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \rm \: product \ = 12( \alpha + \beta )^2 - 2 \alpha \beta + 25 \alpha \beta \\ \\ \rm \implies \: product = 12( \frac{5}{2} )^2 + 23 \times \frac{7}{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \rm \implies product = \frac{300}{4} + \frac{161}{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \boxed{\bf product = \frac{311}{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Putting values

$\bf k \{ {x}^{2} - ( \frac{35}{2} x) + ( \frac{311}{2} ) \} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \bf k\{ \frac{2 {x}^{2} - 35x + 311}{2} \} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \underline { \boxed{ \huge \bf \: 2 {x}^{2} - 35x + 311}} \\ \bf is \: the \: required \: equation \: where \: k = 2$