# solve for x: sqrt(x^2 + 5x -3) = x+3

hala718 | Certified Educator

sqrt(x^2 + 5x -3) = x+3

To solve, let us square both sides:

==> [sqrt(x^2+5x-3)]^2 = (x+3)^2

Open btackets:

==> x^2 + 5x -3 = x^2 +6x +9

Reduce and group similar:

==> -x = 12

==> x= -12

krishna-agrawala | Student

Given:

sqrt(x^2 + 5x - 3) = x + 3

Taking square of both sides of the equation:

x^2 + 5x - 3 = (x + 3)^2

==> x^2 + 5x - 3 = x^2 + 6x + 9

==> x^2 - x^2 + 5x - 6x = 9 + 3

==> -x = 12

Therefore:

x = -12

giorgiana1976 | Student

Before solving the equation, we'll set the constraint of existence of the sqrt.

x^2 + 5x -3 >= 0

We'll calculate the equation:

x^2 + 5x -3 = 0

x1 = [-5+sqrt(25+12)]/2

x1 = (-5+sqrt37)/2

x2 = (-5-sqrt37)/2

The expression x^2 + 5x -3 is positive in intervals:

(-inf. , (-5-sqrt37)/2) U ((-5+sqrt37)/2 , +inf.)

Now, we'll solve the equation and we'll raise to square both sides:

[sqrt(x^2 + 5x -3)]^2 = (x+3)^2

x^2 + 5x -3 = x^2 + 6x + 9

We'll eliminate like terms:

5x - 6x - 3 - 9 = 0

-x - 12 = 0

x = -12

Because the solution doesn't belong to the intervals of admissible values, the equation has no solutions!

thewriter | Student

We have to solve the equation sqrt(x^2+5x-3)=x+3 for x.

First square both the sides of the equation, we get:

x^2+5x-3=(x+3)^2=x^2+6x+9

=>x^2-x^2+5x-6x-3-9 =0

=>-x-12=0

=>x=-12

Therefore x=-12

neela | Student

To solve sqrt(x^2+5x-3) = x+3

Solution:

Square both sides:

x^2+5x-3 = (x+3)^2

x^2+5x-3 = x^2+6x+9

0 = (x^2+6x+9 - (x^2+5x-3)

0 =  6x-5x+9+3

0 = x+12.

x = -12.