Question

if alpha and beta are the zeros of the quadratic polynomial f(x)=x2-x-4 find the value of (1) 1/alpha +1/beta -alfabeta

Answer 1

Answer:

$\frac{15}{4}$

Step-by-step explanation:

$f(x) = {x}^{2} - x - 4 \\ \alpha + \beta = \frac{ - b}{a} = 1 \\ \alpha \beta = \frac{c}{a} = - 4 (1) \: \: \frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{ \alpha + \beta }{ \alpha \beta } - \alpha \beta = \frac{ - 1}{ 4} - ( - 4) \\ = \frac{ - 1}{4} + 4 \\ = \frac{ - 1 + 16}{4} \\ = \frac{15}{4}$

Answer 2

The value of $\dfrac{1}{\alpha}+\dfrac{1}{\beta}-\alpha\beta$ is 3.75.

Step-by-step explanation:

The given polynomial is

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

If a polynomial is defined as $p(x)=ax^2+bx+c$, then

Sum of zeroes = -b/a

Product of zeroes = c/a

It is given that α and β are the zeros of the quadratic polynomial f(x).

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

$\alpha+\beta=1$

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

$\alpha\beta=-4$

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

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

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

$\dfrac{1}{\alpha}+\dfrac{1}{\beta}-\alpha\beta=-0.25+4$

$\dfrac{1}{\alpha}+\dfrac{1}{\beta}-\alpha\beta=3.75$

Therefore, the value of $\dfrac{1}{\alpha}+\dfrac{1}{\beta}-\alpha\beta$ is 3.75.

#Learn more

If alpha and beta are the zeros of the polynomial P of x is equal to X square + 12 x + 35 and find the value of one by Alpha Plus One by beta​.

https://brainly.in/question/9051524