Math, asked by samueldaniel2468910, 10 months ago

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

Answers

Answered by kiki9876
129

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}

Answered by erinna
35

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

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