Math, asked by Victormarshall95, 1 year ago

If α, β are the zeroes of the polynomials f(x)= x²-3x+6, then find the value of 1/α + 1/β + α² +β² -2αβ.

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Answers

Answered by Swaroop11
12
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Answered by tardymanchester
14

Answer:

The value of \frac{1}{\alpha} +\frac{1}{\beta} + \alpha^2 +\beta^2 -2\alpha\beta=-14.5

Step-by-step explanation:

Given : If \alpha,\beta are the zeroes of the polynomials f(x)=x^{2} -3x+6

To find : The value of \frac{1}{\alpha} +\frac{1}{\beta} + \alpha^2 +\beta^2 -2\alpha\beta

Solution :

The zeros of the quadratic equation y=ax^2+bx+c is defined as:

Product of zeros \alpha\beta=\frac{c}{a}

Sum of zeros \alpha+\beta=-\frac{b}{a}

Comparing with given function the zeros are,

\alpha\beta=\frac{6}{1}=6  .....[1]

\alpha+\beta=-\frac{-3}{1}=3 ......[2]

Now, we solve the given value,

=\frac{\beta+\alpha}{\alpha\beta}+\alpha^2 +\beta^2 -2\alpha\beta+2\alpha\beta-2\alpha\beta

=\frac{\beta+\alpha}{\alpha\beta}+(\alpha+\beta)^2-4\alpha\beta

Substitute the value from [1] and [2],

=\frac{3}{6}+(3)^2-4(6)

=\frac{1}{2}+9-24

=\frac{1}{2}-15

=-14.5

Therefore, The value of \frac{1}{\alpha} +\frac{1}{\beta} + \alpha^2 +\beta^2 -2\alpha\beta=-14.5

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