Solving radical equation

Examples

Solve the equation $\sqrt{2x + 1} + 1 = x$

Explanation:

Given equation is : $\sqrt{2x + 1} + 1 = x$

Step 1: Isolate the radical term and subtract 1 from both sides of the equation to get $\sqrt{2x + 1} = x - 1$.

Step 2: Square both sides $(\sqrt{2x + 1})^2 = (x - 1)^2$ Square both sides of the equation to eliminate the radical term. This gives $2x + 1 = x^2 - 2x + 1$.

Step 3: Simplify Simplify the equation by combining like terms $x^2 - 2x -2x + 1-1=0$. This gives $x^2 - 4x = 0$.

Step 4: Factor Factor out the common factor of $x$ to get $x(x - 4) = 0$.

Step 5: Solve for $x$ Use the zero product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. So we have $x = 0$ or $x - 4 = 0$, which gives $x = 0$ or $x = 4$.

Step 6: Check the solutions:

It is possible to get extraneous solutions when solving radical equations, which means that a solution may satisfy the equation after simplification, but may not be a valid solution for the original equation. By checking the solution in step 6, we can make sure that we do not include any extraneous solutions in our final answer.

For $x=0$:

$\sqrt{2(0)+1}+1=0$.

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

$1 ≠ -1$ This solution does not work.

For $x=4$:

$\sqrt{2(4)+1}+1=4$
$\sqrt{9}+1=4$
$3+1=4$ $4=4$ This solution works

It is important to check that the value of $x$ obtained in step $5$ actually satisfies the original equation. If it does not satisfy the equation, then it is not a valid solution. In this case, the solution $x=4$ was obtained in step $5$, and step $6$ verifies that it satisfies the original equation $\sqrt{2x+1}+1=x$.

Solution:

Given equation is : $\sqrt{2x + 1} + 1 = x$

$\sqrt{2x + 1} = x - 1$
Square both sides $(\sqrt{2x + 1})^2 = (x - 1)^2$ $2x + 1 = x^2 - 2x + 1$
or $x^2 - 2x -2x + 1-1=0$
or $x^2 - 4x = 0$

$x(x - 4) = 0$

$x=0$ ; $x=4$:

Verification

$x=0$:

$\sqrt{2(0)+1}+1=0$.

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

$1 ≠ -1$
Now, $x=4$:

$\sqrt{2(4)+1}+1=4$
$\sqrt{9}+1=4$
$3+1=4$ $4=4$

Thus the solution set $= \{4\}$