Math, asked by mryoyo1034, 3 months ago

If the zeroes of polynomial x^3 -6x^2 + 3x + 10 are α , α + β, α +2β. Find α and β.23.If the zeroes of polynomial
Maths Class 10

Answers

Answered by kundankumarkk8440749
0

Step-by-step explanation:

The value of \alpha^2+\beta^2=\frac{106}{9}α

2

2

=

9

106

Step-by-step explanation:

Polynomial : p(x) = 3x^2 - 14x +15p(x)=3x

2

−14x+15 whose \alphaα and \betaβ are zeroes .

General equation : ax^2+bx+c=0ax

2

+bx+c=0

Where,

Sum of zeroes = -\frac{b}{a}−

a

b

Product of zeroes = \frac{c}{a}

a

c

Sum of zeroes of given equation : \alpha+\beta=\frac{14}{3}α+β=

3

14

Product of zeroes of given equation : \alpha\beta =\frac{15}{3}=5αβ=

3

15

=5

Formula used,

\begin{gathered}(a+b)^2=a^2+b^2+2ab\\(a+b)^2-2ab=a^2+b^2\end{gathered}

(a+b)

2

=a

2

+b

2

+2ab

(a+b)

2

−2ab=a

2

+b

2

The identity form is

(\alpha+\beta)^2-2 \alpha \beta=\alpha^2+\beta^2(α+β)

2

−2αβ=α

2

2

Substitute the values,

\begin{gathered}(\frac{14}{3})^2-2(5)=\alpha^2+\beta^2\\\\\frac{106}{9}=\alpha^2+\beta^2\end{gathered}

(

3

14

)

2

−2(5)=α

2

2

9

106

2

2

Therefore, the value of \alpha^2+\beta^2=\frac{106}{9}α

2

2

=

9

106

#Learn more:

If alpha and beta are the zeroes of the polynomial p(x) = 3x2 - 14x +15 ,find the value of alpha square + beta square

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