Question

Alpha and beta zeroes of the quadratic
polynomial x²-6x +y. And the value
of 'y' of 3 Alpha +2beta =20​

Answer 1

Given :

• Alpha and beta zeroes of the quadratic polynomial x²-6x +y.
• The value of 'y' of $\sf 3\alpha+2\beta = 20$

Solution :

General Equation of Quadratic polynomial :-

$\begin{gathered} \qquad \sf \: a {x}^{2} + by + c = 0 \end{gathered}$

Given Equation of Quadratic polynomial :-

$\begin{gathered} \qquad \sf {x}^{2} - 6x + y = 0 \end{gathered}$

Hence,

$\begin{gathered} \qquad \: \star \:\sf a = 1\\ \qquad \: \:\:\:\star \:\sf b = - 6 \\ \qquad \sf \star\: c = y \end{gathered}$

Sum of the roots :-

$\qquad {\sf \: \alpha + \beta = \dfrac{ - b}{a} = - \bigg(\dfrac { - 6}{1} \bigg)= 6}$

Product Of the roots :-

$\qquad \: { \sf \alpha \beta = \dfrac {c}{a} = \dfrac{y}{1} = y}$

As by given,

$\qquad \sf 3 \alpha + 2 \beta = 20$

The equations are,

$\qquad : \longrightarrow \sf \alpha + \beta = 6$ ...(1)

$\qquad :\longrightarrow \sf \alpha\beta = y$ ...(2)

$\qquad :\longrightarrow \sf 3\alpha + 2 \beta = 20$ ...(3)

Now let's use elimination method,

Multiply equation (1) by 3

$\begin{array} {ccc} \sf \cancel{3\alpha} + & \sf 3 \beta& = 18 \\\sf \cancel{3\alpha} +&\sf 2 \beta & = \sf 20 \\\end{array}$

________________

$\qquad \boxed{\beta = -2}$

Substitute values of $\beta$ in (1),

$\qquad : \implies \sf \alpha + \beta = 6 \\\\:\implies \sf \alpha - 2 = 6\\\\ : \implies \sf \alpha = 6+2 \\\\ : \implies \boxed{\alpha = 8}$

Substitute $\alpha$ and $\beta$ in (2)

$\qquad : \implies \alpha\beta = y\\\\ : \implies (8)(-2)=y \\\\:\implies \boxed{\sf y = -16}$

Answer 2

Answer:

Given :

Alpha and beta zeroes of the quadratic polynomial x²-6x +y.

The value of 'y' of

$\sf 3\alpha+2\beta$

= 203α+2β=20

Solution :

General Equation of Quadratic polynomial :-

$\begin{gathered} \qquad \sf \: a {x}^{2} + by + c = 0 \end{gathered}$

Given Equation of Quadratic polynomial :-

$\begin{gathered} \qquad \sf {x}^{2} - 6x + y = 0 \end{gathered}$

Hence,

$\begin{gathered} \begin{gathered} \qquad \: \star \:\sf a = 1\\ \qquad \: \:\:\:\star \:\sf b = - 6 \\ \qquad \sf \star\: c = y \end{gathered}\end{gathered}$

Sum of the roots :

$\qquad {\sf \: \alpha + \beta = \dfrac{ - b}{a} = - \bigg(\dfrac { - 6}{1} \bigg)= 6}α+β=$

a

−b

=−(

1

−6

)=6

Product Of the roots :-

\qquad \: { \sf \alpha \beta = \dfrac {c}{a} = \dfrac{y}{1} = y}αβ

$\qquad \: { \sf \alpha \beta = \dfrac {c}{a} = \dfrac{y}{1} = y}αβ$

=

a

c

=

1

y

=y

As by given,

$\qquad \sf 3 \alpha + 2 \beta = 203α+2β=20$

The equations are,

$\qquad :\longrightarrow \sf \alpha\beta = y:⟶αβ=y ...(2)$

$qquad :\longrightarrow \sf 3\alpha + 2 \beta = 20:⟶3α+2β=20 ...(3)$

Now let's use elimination method,

Multiply equation (1) by 3

$\begin{gathered}\begin{array} {ccc} \sf \cancel{3\alpha} + & \sf 3 \beta& = 18 \\\sf \cancel{3\alpha} +&\sf 2 \beta & = \sf 20 \\\end{array}\end{gathered}$

3α

+

3α

+

3β

2β

=18

=20

________________

$\qquad \boxed{\beta = -2}$

β=−2

Substitute values of \betaβ in (1),

$\begin{gathered}\qquad : \implies \sf \alpha + \beta = 6 \\\\:\implies \sf \alpha - 2 = 6\\\\ : \implies \sf \alpha = 6+2 \\\\ : \implies \boxed{\alpha = 8} \end{gathered}$

Substitute \alphaα and \betaβ in (2)

Step-by-step explanation:

$\sf\pink{hope\:it\:helps}$