Question

Solve \((x+3)^{2}-4(x+3)+3=0\)

Short answer

The solutions are \( x = 0 \) and \( x = -2 \).

Step-by-step solution

Let a new variable

Set a new variable to simplify the equation. Let’s define a new variable: \( y = x + 3 \)

Substitute the new variable

Rewrite the equation using the new variable \(y\). Substitute \(y\) for \(x + 3\) in the original equation: \( y^{2} - 4y + 3 = 0 \)

Solve the quadratic equation

Solve the quadratic equation \( y^{2} - 4y + 3 = 0 \). Use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \(a = 1\), \(b = -4\), and \(c = 3\): \( y = \frac{4 \pm \sqrt{(-4)^{2} - 4(1)(3)}}{2(1)} = \frac{4 \pm \sqrt{16-12}}{2}= \frac{4 \pm 2}{2} \)

Find the solutions for y

Solve for \(y\): \( y = \frac{4 + 2}{2} = 3 \) and \( y = \frac{4 - 2}{2} = 1 \)

Substitute back to find x

Substitute back \( x + 3 = y \) into the solutions for \(y\): For \(y = 3\): \( x + 3 = 3 \implies x = 0 \) For \(y = 1\): \( x + 3 = 1 \implies x = -2 \)

Verify the solutions

Verify both solutions by substituting back into the original equation: For \( x = 0 \): \((0 + 3)^{2} - 4(0 + 3) + 3 = 9 - 12 + 3 = 0\) (True) For \( x = -2 \): \((-2 + 3)^{2} - 4(-2 + 3) + 3 = 1 - 4 + 3 = 0\) (True)

Key concepts

quadratic formula

When solving a quadratic equation, an invaluable tool at your disposal is the quadratic formula. A quadratic equation typically takes the form of \[ ax^2 + bx + c = 0 \] where \(a\), \(b\), and \(c\) are constants. The quadratic formula allows you to find the values of \(x\) that make the equation true. It is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This might look intimidating at first, but it's essentially a step-by-step method to solve any quadratic equation. Let's break it down:

• Inside the square root: Calculate the discriminant \(b^2 - 4ac\). This tells you how many real solutions the equation has.
• Numerator: \(-b\) plus or minus the square root of the discriminant gives two possible solutions.
• Denominator: Dividing everything by \(2a\) completes the solutions to the equation.

In our exercise, we simplified the original quadratic equation to \(y^2 - 4y + 3 = 0\), where \(a = 1\), \(b = -4\), and \(c = 3\). Using the quadratic formula, we found the solutions for \(y\) as \(y = 3\) and \(y = 1\).

substitution method

The substitution method simplifies complex equations by introducing a new variable. This technique is particularly helpful for equations that can be transformed into simpler forms. Let's use our exercise as an example:
We started with \((x+3)^2 - 4(x+3) + 3 = 0\). This can be hard to solve directly. So, we set \(y = x + 3\), creating a new, simpler equation in terms of \(y\). The original equation became \(y^2 - 4y + 3 = 0\).
Using substitution:

• Choose a substitution variable: Select a part of the original equation to set as a new variable. Here, \(y = x + 3\).
• Rewrite the original equation: Substitute the variable into the equation to simplify it. Our new equation became \(y^2 - 4y + 3 = 0\).

After solving for \(y\), don't forget to substitute back to find the values of the original variable. Thus, substituting \(x + 3 = y\) back into the solved values for \(y\) gives us the final solutions for \(x\).

verification of solutions

Once you've found potential solutions to an equation, it's crucial to verify them. This ensures your solutions are correct and applicable to the original problem. Here's how you can verify solutions effectively:

• Substitute back: Plug each solution back into the original equation to see if it holds true.
• Check both sides: Make sure both sides of the equation are equal after substitution.

In our exercise, we found the solutions \(x = 0\) and \(x = -2\). To verify:
For \(x = 0\): \[(0 + 3)^2 - 4(0 + 3) + 3 = 9 - 12 + 3 = 0\] This is true, so \(x = 0\) works.
For \(x = -2\): \[(-2 + 3)^2 - 4(-2 + 3) + 3 = 1 - 4 + 3 = 0\] This is also true, so \(x = -2\) works as well.
By verifying, you confirm that your solutions are indeed correct, making this step a critical part of solving equations.