Question

if α and β are the zeroes of the quadratic polynomial x2-3x+a find the value of a if 3α+2β=20

Answer 1

A = -154

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Given :

• p(x) = $x^{2} -3x + A$ *
• a = 1; b = -3; c = A
• $3\alpha + 2\beta = 20$

To Find :

• Value of a

Solution :

We know that,

$Sum \: of \: Zeroes( \alpha + \beta) = \frac{-b}{a}$

$=> \alpha + \beta = \frac{-(-3)}{1}\\=> \alpha + \beta = 3 ----(1)$

$Also, 3\alpha + 2\beta =20 ----(2)$ [Given]

Equating the 2 equations by Elimination method**......

Multiplying equation(1) by 2 and then subtract from equation(1) we get,

$\boxed{\alpha = 14}$

Substituting value of $\alpha$ in equation(1)*** we get,

$\boxed{\beta = -11}$

We know that,

$Product \: of\: Zeroes(\alpha \beta) = \frac{c}{a}$

$=> \frac{c}{1} = -154\\=> c = A = -154$

Hence, $\boxed{\textbf{A = -154}}$

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NOTE -

* - To avoid any confusions, I have taken 'a'(unknown value)  as 'A'

** - You can use any method for equating the equations as per your wish.

*** - You can take any equation as per your wish.

Answer 2

Step-by-step explanation:

x² - 6x + a

a = 1    b = -6   c = a

α +β = -b/a = 6

αβ = c/a = a

3α + 2β = 20

3α = 20 - 2β  

α = (20 - 2β) / 3

α + β = 6

(20 - 2β)/3 + β = 6

multiplying by 3 on both sides

20 - 2β + 3β = 18

20 + β = 18

β = 18 - 20  

β = -2

α = (20 -2β)/3

α = (20 - 2 × -2)/3

α = (20 + 4)/3

α = 8

αβ =  a  

8 × -2 = a

a = -16

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