Question

if alpha and beta are the zeros of the quadratic polynomial x square - 5 x + 4 find the value of one by Alpha Beta + 1 by beta minus 2 alpha beta

Answer 1

Hey

$f(x) = x {}^{2} - 5x + 4 \\ \\ \mathbf{here} \\ \alpha + \beta = - ( - 5) = 5 \\ \alpha \beta = 4 \\ \\ now \\ \\ \frac{1}{ \alpha } + \frac{1}{ \beta } - 2 \alpha \beta \\ \\ = \frac{ \alpha + \beta }{ \alpha \beta } - 2 \alpha \beta \\ \\ = \frac{5}{4} - 2(4) \\ \\ = \frac{5}{4} - 8 = \frac{5 - 32}{4} = \frac{ - 27}{4}$

I hope it helps you!!!

Answer 2

The value of $\dfrac{1}{\alpha}+\dfrac{1}{\beta}-2\alpha\beta$ is -6.75.

Step-by-step explanation:

If a equation is defined as $ax^2+bx+c=0$ , then  

Sum of roots = -b/a

Product of roots =c/a

The given quadratic polynomial is

$f(x)=x^2-5x+4$

It is given that α and β are the zeros of the quadratic polynomial.

Here, a=1, b=-5 and c=4.

$\alpha+\beta=-\dfrac{-5}{1}=5$

$\alpha\beta=\dfrac{4}{1}=4$

We need to find the value of $\dfrac{1}{\alpha}+\dfrac{1}{\beta}-2\alpha\beta$.

$\dfrac{1}{\alpha}+\dfrac{1}{\beta}-2\alpha\beta=\dfrac{\beta+\alpha}{\alpha\beta}-2\alpha\beta$

Substitute α+β=5 and αβ=4.

$\dfrac{1}{\alpha}+\dfrac{1}{\beta}-2\alpha\beta=\dfrac{5}{4}-2(4)=-6.75$

Therefore, the value of $\dfrac{1}{\alpha}+\dfrac{1}{\beta}-2\alpha\beta$ is -6.75.

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