Question

The number of real zeroes of x^4−1=0 is​

Answer 1

Answer:

${x}^{4} - 1 = 0 \\ ( {x}^{2} + 1)( {x}^{2} - 1) = 0 \\ ( {x}^{2} + 1)(x + 1)(x - 1) = 0 \\ x = \sqrt{ + 1 \: } and \: \sqrt{ - 1} \: and \: - 1 \: and \: + 1$

Which are the required real zeroes

Answer 2

Answer:

x^2+1 is no real roots

therefore,The number of zeros of the given polynomial =2

Step-by-step explanation:

The given polynomial:

x^4-2

To find, the number of zeros of the given polynomial=?

•°• x^4 -1

(x^2)^2 -(1^2)^2

Using the algebraic identity

a^2-b^2 = (a+b) (a-b)

={( x^{2}+1 ^{2})(x^{2}-1^{2})}

= (x^2+1) (x+1) ( x-1)

x=1 and -1

x^2+1 is no real roots

Therefore,The no. of zeros of the given polynomial=2

Hence,the no. of zeros of the polynomial is 2.

Thank you hope it can help you