# Solve for x the equation square root(x^2+x+1)=3-x

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### 3 Answers

We have to solve for x given that sqrt(x^2+x+1)=3-x

sqrt(x^2+x+1)=3-x

square both the sides

=> x^2 + x + 1 = (3 - x)^2

=> x^2 + x + 1 = 9 - 6x + x^2

=> x^2 - x^2 + x + 6x + 1 - 9 = 0

=> 7x - 8 = 0

=> x = 8/7

**The solution of the equation is x = 8/7**

square root(x^2+x+1)=3-x

(x^2+x+1)=(3-x)^2 [squaring both side]

x^2 +x +1 = 3^2 -2.3.x+x^2 [(a-b)^2 = a^2-2ab+b^2]

x^2 +x +1 =9-6x+x^2 [x^2 get cancelled]

x+1=9-6x

x+6x = 9-1[adding 6x-1 on both side]

7x=8

x=8/7

We notice that we don't have to determine the values of x for the square root to exist, because the expression x^2+x+1 is positive for any value of x.

We'll raise to square both sides:

x^2 + x + 1 = (3-x)^2

We'll expand the square:

x^2 + x + 1 = 9 - 6x + x^2

We'll shift all terms to one side:

x^2 + x + 1 - 9 + 6x - x^2 = 0

We'll eliminate like terms:

7x - 8 = 0

We'll isoalte 7x to the left:

7x = 8

x = 8/7

**The solution of the equation is x = 8/7.**