Question

p(x)=x square+x-1 find the zeroes of the polynomial​

Answer 1

Solution:

Given,

→ x² + x - 1 = 0

We have to find out the zeros of the given equation.

Comparing with ax² + bx + c = 0, we get,

→ a = 1, b = 1, c = -1

So,

→ Discriminant (D) = b² - 4ac

→ D = 1² - 4 × 1 × (-1)

→ D = 1 + 4

→ D = 5

So,

→ x = (-b ± √D)/2a

→ x = (-1 ± √5)/2

Therefore,

→ x = (-1 + √5)/2 and (-1 -√5)/2

Answer:

• x = (-1 + √5)/2 and (-1 -√5)/2

•••♪l

Answer 2

Given=

${x}^{2} + x - 1 = 0$

we have find out the zeero of the given equetion

$comparing \: with \: {ax}^{2} + bx + c = 0 \: we \: get$

= a =1 ,b=1,c=1

so =

$= discriment (d) = {b}^{2} - 4 \: ac$

$= d = {1}^{2} - 4 \times 1 \times ( - 1) \\ = d = 1 + 4 \\ d = 5$

SO :

x = ( -b +√D)/2a

x = (-1 +√5)/2

therefore :

x = (-b + √ D)/a and x = (-1 +√5)/2

answer : x =(-b+√D)/2a and x=(-1+√5)/2