# For what real values of x the identity is true : sqrt(x^2 + x + 1)=2-x

hala718 | Certified Educator

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

Let us square both sides:

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

Now expand the square on the right side:

==> x^2 + x +1 = 4 -4x + x^2

Eliminate x^2

==> x+1 = 4-4x

Group similar:

==> 5x = 3

==> x= 3/5

giorgiana1976 | Student

First, we'll have to impose the constraint of sqrt existance:

x^2 + x + 1>0

It is obvious that it is true the inequality above, because if we calculate delta = 1-4 = -3<0, so the expression x^2 + x + 1 is positive for any value of x!

Now, we can solve the equation. For this reason, we'll raise to square both sides, to eliminate the square root:

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

We'll expand the square from the right side and we'll get:

(x^2 + x + 1) = 4 - 4x + x^2

We'll subtract x^2 both sides:

x+1 = 4-4x

5x + 1 = 4

We'll subtract 1 both sides:

5x = 3

We'll divide by 5 both sides:

x = 3/5

The identity is true for x = 3/5.

neela | Student

sqrt(x^2+x+1) = 2-x. To find the roots which are real.

Solution:

squaring both sides,

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

x2+x+1 = 4-4x+x^2.

x+4x = 4-1=3

5x/3 =3/5 =0.6

x = 0.6

Check:

LHS: sqrt(x^2+x+1) = (0.6)^2+0.6+1) = sqrt(1.96) = 1.4

RH: 2-0.6 = 1.4

x+1+2x =4