Question

1.Equation of (x+1)^2-x^2=0 has 1 point
number of real roots equal to:
O (a)1
O (b)2
O (c)3
O (d)4​

Answer 1

Answer:

${(x + 1)}^{2} - {x}^{2} = 0 \\ \\ {x}^{2} + 1 + 2x - {x }^{2} = 0 \\ \\ 2x + 1 = 0$

Since this is not a quadratic equation and the degree of equation is 1.

Therefore number of real roots also equals to 1.

Which is :-

$2x + 1 = 0 \\ \\ 2x = - 1 \\ \\ x = \frac{ - 1}{2}$

Hope it helps.