Question

If alpha and bitta are the zeros of the quadratic polynomial f(x) =x^2+x-2,find the value of 1/alpha-1/bitta

Answer 1

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

If α and β are the zeroes of the quadratic polynomial f(x) = x² + x - 2.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The value of 1/α - 1/β.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

We have f(x) = x² + x - 2

Zero of the polynomial f(x) = 0

So;

$\longrightarrow\sf{x^{2} +x-2=0}\\\\\longrightarrow\sf{x^{2} +2x-x-2=0}\\\\\longrightarrow\sf{x(x+2)-1(x+2)=0}\\\\\longrightarrow\sf{(x+2)(x-1)=0}\\\\\longrightarrow\sf{x+2=0\:\:\:Or\:\:\:x-1=0}\\\\\longrightarrow\sf{\blue{x=-2\:\:Or\:\:x=1}}$

∴ The α = -2 and β = 1 are the zeroes of the polynomial.

Now;

As the given quadratic polynomial as we compared with ax² + bx + c

• a = 1
• b = 1
• c = -2

$\underline{\pink{\mathcal{SUM\:OF\:THE\:ZEROES\::}}}$

$\mapsto\tt{\alpha +\beta =\dfrac{-b}{a} }=\dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2} } }\\\\\\\mapsto\tt{\alpha +\beta =\dfrac{-1}{1} }\\\\\\\mapsto\tt{\alpha +\beta =-1}$

$\underline{\pink{\mathcal{PRODUCT\:OF\:THE\:ZEROES\::}}}$

$\mapsto\tt{\alpha \times\beta =\dfrac{c}{a} }=\dfrac{Constant\:term}{Coefficient\:of\:x^{2} } }\\\\\\\mapsto\tt{\alpha \times\beta =\dfrac{-2}{1} }\\\\\\\mapsto\tt{\alpha \times\beta =-2}$

Now;

$\longrightarrow\sf{\dfrac{1}{\alpha } -\dfrac{1}{\beta } }\\\\\\\longrightarrow\sf{\dfrac{\alpha -\beta }{\alpha\beta }} \\\\\\\longrightarrow\sf{\dfrac{-3}{-2} }\\\\\\\longrightarrow\sf{\blue{\dfrac{3}{2} }}$

Thus;

The value is 3/2.

Answer 2

Answer:

Step-by-step explanation:

Concept:

If the first term of this polynomial has a power of 2 and the second term has a power of 1, the polynomial's degree is the biggest exponent, which is 2. As a result, a polynomial of degree two is referred to as a quadratic polynomial.

Given:

The quadratic polynomial $f(x) =x^2+x-2$

Find:

The value of $\frac{1}{\alpha } -\frac{1}{\beta}$.

Solution:

Given, The quadratic polynomial $f(x) =x^2+x-2$

sum of roots $=\alpha+\beta=-1$

Product of roots$=\alpha-\beta=-2 \ldots .(1)$

Therefore, value of $\frac{1}{\alpha}-\frac{1}{\beta}=\frac{\beta-\alpha}{\alpha \beta}=\frac{-(\alpha-\beta)}{\alpha \beta} \ldots . .(2)$

$\Rightarrow \alpha-\beta=\frac{\sqrt{\mathrm{D}}}{\mathrm{a}}=\frac{\sqrt{\mathrm{b}^{2}-4 \mathrm{ac}}}{\mathrm{a}}=\frac{\sqrt{1+4 \times 2 \times 1}}{1}=3$

$\alpha-\beta=3 \ldots .(3)$

putting the value of$\alpha \beta$ and $\alpha-\beta$ in equation (2)

$\Rightarrow \frac{1}{\alpha}-\frac{1}{\beta}=\frac{-(3)}{-2}=\frac{3}{2}$

#SPJ2