Question

Q15:-If alpha and bita are zeroes of the quadratic polynomial x^2-6x + a, find the value of a if 3 alpha + 2 bita = 20

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Answer 1

Solution:

Given:

⇒ Quadratic polynomial = x² - 6x + a      ........(1)

⇒ 3α + 2β = 20      

To find:

⇒ value of a

Now, α and β are zeroes of the quadratic polynomial x² - 6x + a.

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

$\sf{\implies \alpha+\beta =-\bigg(\dfrac{-6}{1}\bigg)}$

$\sf{\implies \alpha +\beta =6\;\;\;\;........(2)}$

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

$\sf{\implies \alpha \beta =\dfrac{a}{1}}$

$\sf{\implies \alpha \beta =a\;\;\;\;........(3)}$

From above we get 3 equations,

⇒ 3α + 2β = 20      ........(1)

⇒ α + β = 6        ........(2)

⇒ αβ = a            ........(3)

Now, By substitution method we will find α and β.

⇒ α + β = 6      .........(2)

⇒ α = 6 - β  

Put the value of α in Equation (1), we get

⇒ 3α + 2β = 20      ........(1)

⇒ 3(6 - β) + 2β = 20

⇒ 18 - 3β + 2β = 20

⇒ -β = 20 - 18

⇒ -β = 2

⇒ β = -2

Now, put the value of β in equation (2), we get

⇒ α + β = 6

⇒ α - 2 = 6

⇒ α = 6 + 2

⇒ α = 8

Now, put the value of α and β in equation (3), we get

⇒ αβ = a            ........(3)

⇒ 8 × (-2) = a

⇒ a = -16

Hence, the value of a is -16.

Answer 2

Given:

α and β are the roots of the quadratic equation

x²-6x+a

To find:

finding the value of a

Explanation:

from the given information

$3 \alpha + 2 \beta = 20........(1)$

Now let us find sum of roots

we know that

$\boxed{\sf sum \: of \: roots =\dfrac{-b}{a}}$

here

f(x)=x²-6x+a

a=1. b=-6. c=a

$= > sum \: of \: roots \: = \dfrac{ - ( -6 )}{1} \\ \\ = > \alpha + \beta = 6..........(2)$

multiply (2) with 2

it becomes

$= > 2 \alpha + 2 \beta = 12$

equating (1)-equation(2)

$= > 3 \alpha + 2 \beta - 2 \alpha - 2 \beta = 20 - 12 \\ \\ \boxed{ \alpha = 8}$

then

$= > 8 + \beta = 6 \\ \\ \boxed{ \beta = - 2}$

as α and β are the roots

f (α)=f (β)=0

substituting f(8)=0

$= > 8 ^{2} - 6(8) + a = 0 \\ \\ = > 64 - 48 + a = 0 \\ \\ = > 16 + a = 0 \\ \\ \boxed{a = - 16}$

f(-2)=0

$= > {( - 2)}^{2} - 6( - 2) + a = 0 \\ \\ = > 4 + 12 + a = 0 \\ \\ \boxed{a = - 16}$