Solve the Quadratic: Expanding (x+3)^2 to Find x in 2x+5

Question

Solve the following equation:

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

00:00 Find X

00:03 Use the abbreviated multiplication formulas

00:13 Substitute appropriate values according to the data and open the parentheses

00:30 Substitute in our equation

00:50 Arrange the equation so that one side equals 0

00:58 Collect terms

01:09 Identify coefficients

01:24 Use the roots formula

01:48 Substitute appropriate values and solve

02:07 Calculate the square and products

02:14 The square root of 0 is always equal to 0

02:21 When the root equals 0, the equation will have only one solution

02:48 And this is the solution to the question

To solve the equation $(x+3)^2 = 2x + 5$ , we proceed as follows:

Step 1: Expand the left side. Using the identity $(a+b)^2 = a^2 + 2ab + b^2$ , we find:
$(x+3)^2 = x^2 + 6x + 9$ .
Step 2: Set the equation to zero by moving all terms to one side:
$x^2 + 6x + 9 = 2x + 5$
Subtract $2x + 5$ from both sides:
$x^2 + 6x + 9 - 2x - 5 = 0$
This simplifies to:
$x^2 + 4x + 4 = 0$ .
Step 3: Solve the quadratic equation $x^2 + 4x + 4 = 0$ . Notice this can be factored as:
$(x+2)^2 = 0$ .
Step 4: Solve for $x$ by setting the factor equal to zero:
$x+2 = 0$ .
Thus, $x = -2$ .

Therefore, the solution to the equation $(x+3)^2 = 2x + 5$ is $x = -2$ .

$x=-2$