« Solving Quadratic Equation 

A "radical" equation is an equation in which there is a variable inside the radical sign
Step 1: Isolate the radicals to left side of the equal sign.
Step 2: Square each side of the equation
Step 3: Solve the resulting equation
Step 4: Check all solutions
Example 1: Solve $\sqrt {2x + 3}  x = 0$
Solution
Step 1: Isolate the radicals to left side of the equal sign.
$$\sqrt {2x + 3}  x = 0$$ $$\sqrt {2x + 3} = x$$
Step 2: Square each side of the equation
$${\left( {\sqrt {2x + 3} } \right)^2} = {(x)^2}$$ $$2x + 2 = {x^2}$$ $${x^2}  2x  3 = 0$$
Step 3: Solve the resulting equation
$${x^2}  2x  3 = 0$$ $$a = 1, b =  2, c =  3$$ $${x_{1,2}} = \frac{{  b \pm \sqrt {{b^2}  4ac} }}{{2a}}$$ $${x_{1,2}} = \frac{{  (  2) \pm \sqrt {{{(  2)}^2}  4 \cdot 1 \cdot (  3)} }}{{2 \cdot 1}}$$ $${x_{1,2}} = \frac{{2 \pm \sqrt {4 + 12} }}{2}$$ $${x_{1,2}} = \frac{{2 \pm 4}}{2}$$ $${x_1} = \frac{{2 + 4}}{2},{x_2} = \frac{{2  4}}{2}$$ $${x_1} = 3,{x_2} =  1$$
Step 4: Check all solutions
Let's check to see if x_{1} = 3 is solution:
$$\sqrt {2x + 3}  x = 0$$ $$\sqrt {2 \cdot 3 + 3}  3 = 0$$ $$\sqrt 9  3 = 0$$ $$3  3 = 0$$ $$0 = 0, OK$$
Let's check to see if x_{2} = 1 is solution:
$$\sqrt {2x + 3}  x = 0$$ $$\sqrt {2 \cdot (  1) + 3}  (  1) = 0$$ $$\sqrt {  2 + 3} + 1 = 0$$ $$\sqrt 1 + 1 = 0$$ $$2 = 0,NOTOK$$
So, the original equation had a single solution x = 3 .
Level 1
Level 2
Example 2: Solve $\sqrt {4x + 3} + 2x  1 = 0$
Solution
$$\sqrt {4x + 3} + 2x  1 = 0$$ $${(\sqrt {4x  3} )^2} = {(1  2x)^2}$$ $$4x  3 = 1  4x + 4{x^2}$$ $$4x  3  1 + 4x  4{x^2} = 0$$ $$  4{x^2} + 8x  4 = 0/:(  4)$$ $${x^2}  2x + 1 = 0$$ $${(x  1)^2} = 0$$ $$x = 1$$
Let's check it to see if x = 1 is a solution to the original equation.
$$\sqrt {4x  3} + 2x  1 = 0$$ $$\sqrt {4 \cdot 1  3} + 2 \cdot 1  1 = 0$$ $$\sqrt 1 + 2  1 = 0$$ $$2 = 0$$
So, the original equation had no solutions.
Level 1
Level 2
Example 3: Solve $\sqrt {3x + 4}  \sqrt {2x + 1} = 1$
Solution
First thing to do is get one of the square roots by itself.
$$\sqrt {3x + 4}  \sqrt {2x + 1} = 1$$ $$\sqrt {3x + 4} = 1 + \sqrt {2x + 1} $$ $${(\sqrt {3x + 4} )^2} = {(1 + \sqrt {2x + 1} )^2}$$ $$3x + 4 = {1^2} + 2 \cdot 1 \cdot \sqrt {2x + 1} + {(\sqrt {2x + 1} )^2}$$ $$3x + 4 = 1 + 2\sqrt {2x + 1} + 2x + 1$$
We have managed to eliminate one of square roots!! We will continue to work this problem as we did in the previous examples.
$$3x + 4 = 1 + 2\sqrt {2x + 1} + 2x + 1$$ $$3x + 4 = 2 + 2x + 2\sqrt {2x + 1} $$ $$3x + 4  2  2x = 2\sqrt {2x + 1} $$ $${(x + 2)^2} = {(2\sqrt {2x + 1} )^2}$$ $${x^2} + 2 \cdot 2 \cdot x + {2^2} = {2^2}(2x + 1)$$ $${x^2} + 4x + 4 = 8x + 4$$ $${x^2}  4x = 0$$ $$x(x  4) = 0$$ $${x_1} = 0$$ $${x_2} = 4$$
Let's check both possible solutions. We will start with x = 0.
$$\sqrt {3x + 4}  \sqrt {2x + 1} = 1$$ $$\sqrt {3 \cdot 0 + 4}  \sqrt {2 \cdot 0 + 1} = 1$$ $$\sqrt 4  \sqrt 1 = 1$$ $$1 = 1 \Rightarrow OK$$
Now let's check x = 4.
$$\sqrt {3x + 4}  \sqrt {2x + 1} = 1$$ $$\sqrt {3 \cdot 4 + 4}  \sqrt {2 \cdot 4 + 1} = 1$$ $$\sqrt {16}  \sqrt 9 = 1$$ $$1 = 1 \Rightarrow OK$$
Level 1
Level 2