Question

Write the sum and product of the zeroes of the polynomial R(x) = x2 – 5.

Answer 1

Answer:

0 and -5

Step-by-step explanation:

as sum of root = -(coeff. of x)/coeff. of x^2

but we are not given the x term so it is 0

product of root= constant/ coeff. of x^2

so it is -5(given in question)

The equation given is x^2 + 0x -5

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

If $\alpha \: \: and \: \: \beta \:$are the zeroes of the quadratic polynomial $a {x}^{2} + bx + c$

Then

$\displaystyle \: \alpha + \beta \: = - \frac{b}{a} \: \: and \: \: \: \alpha \beta \: = \frac{c}{a}$

TO DETERMINE

The sum of the Zeroes and the product of the zeroes of the quadratic polynomial

$\sf{ \: {x}^{2} - 5 \: }$

CALCULATION

The given Quadratic polynomial is

$\sf{ {x}^{2} - 5 \: }$

Comparing with

$a {x}^{2} + bx + c$

We get

$\sf{a = 1 \: , \: b = 0 \: , \: c = - 5}$

So

$\sf{Sum \: of \: the \: Zeroes \: } = \displaystyle \: - \frac{b}{a} = 0$

$\sf{Product \: of \: the \: Zeroes \: } = \displaystyle \: \frac{c}{a} = \frac{ - 5}{1} = - 5$

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ADDITIONAL INFORMATION

A general equation of quadratic equation is

$a {x}^{2} + bx + c = 0$

Now one of the way to solve this equation is by SRIDHAR ACHARYYA formula

For any quadratic equation

$a {x}^{2} + bx + c = 0$

The roots are given by

$\displaystyle \: x = \frac{ - b \pm \: \sqrt{ {b}^{2} - 4ac } }{2a}$