23 Questions on Quadratic Equations - Factoring, Completing Square, Quadratic Formula + Explanations

Step 1: Move the constant term to the other side of the equation.

\(x^2 - 6x = -5\)

Step 2: Complete the square. Take half the coefficient of \(x\), square it, and add it to both sides.

\(\text{Coefficient of } x = -6, \text{Half} = -3, \text{Square} = 9\)

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

\((x - 3)^2 = 4\)

Step 3: Take the square root of both sides.

\(x - 3 = \pm 2\)

Step 4: Solve for \(x\).

\(x = 3 + 2 \quad \Rightarrow \quad x = 5\)

\(x = 3 - 2 \quad \Rightarrow \quad x = 1\)

The solutions are \(x = 5\) and \(x = 1\).

Step 1: Write the equation in standard form.

\(x^2 - 9x + 20 = 0\)

Step 2: Factorize the quadratic expression.

Find two numbers that multiply to \(20\) and add to \(-9\). These numbers are \(-4\) and \(-5\).

\((x - 4)(x - 5) = 0\)

Step 3: Solve for \(x\) by setting each factor equal to 0.

\(x - 4 = 0 \quad \Rightarrow \quad x = 4\)

\(x - 5 = 0 \quad \Rightarrow \quad x = 5\)

The solutions are \(x = 4\) and \(x = 5\).

Step 1: Identify the coefficients \(a\), \(b\), and \(c\).

\(a = 2, b = -4, c = -6\)

Step 2: Write the quadratic formula:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Step 3: Substitute the values of \(a\), \(b\), and \(c\) into the formula.

\(x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(-6)}}{2(2)}\)

\(x = \frac{4 \pm \sqrt{16 + 48}}{4}\)

\(x = \frac{4 \pm \sqrt{64}}{4}\)

Step 4: Simplify the square root and solve for \(x\).

\(x = \frac{4 \pm 8}{4}\)

Case 1: \(x = \frac{4 + 8}{4} = 3\)

Case 2: \(x = \frac{4 - 8}{4} = -1\)

The solutions are \(x = 3\) and \(x = -1\).

Step 1: Write the quadratic equation in standard form:

\(x^2 - 2x - 8 = 0\)

Step 2: Factorize the quadratic expression. Find two numbers that multiply to \(-8\) and add to \(-2\). These numbers are \(-4\) and \(2\).

\((x - 4)(x + 2) = 0\)

Step 3: Solve for \(x\) by setting each factor equal to 0:

\(x - 4 = 0 \quad \Rightarrow \quad x = 4\)

\(x + 2 = 0 \quad \Rightarrow \quad x = -2\)

The solutions are \(x = 4\) and \(x = -2\).

Step 1: Move the constant term to the other side:

\(x^2 - 4x = -4\)

Step 2: Complete the square. Take half the coefficient of \(x\), square it, and add it to both sides:

\(\text{Half of } -4 = -2, \text{Square} = 4\)

\(x^2 - 4x + 4 = -4 + 4\)

\((x - 2)^2 = 0\)

Step 3: Take the square root of both sides:

\(x - 2 = 0 \quad \Rightarrow \quad x = 2\)

The solution is \(x = 2\).

Step 1: Identify the coefficients \(a\), \(b\), and \(c\):

\(a = 3, b = -12, c = 12\)

Step 2: Write the quadratic formula:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Step 3: Substitute the values of \(a\), \(b\), and \(c\) into the formula:

\(x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(3)(12)}}{2(3)}\)

\(x = \frac{12 \pm \sqrt{144 - 144}}{6}\)

\(x = \frac{12 \pm \sqrt{0}}{6}\)

Step 4: Solve for \(x\):

\(x = \frac{12}{6} = 2\)

The solution is \(x = 2\) (a repeated root).

Step 1: Write the quadratic equation in standard form:

\(x^2 + 5x - 14 = 0\)

Step 2: Factorize the quadratic expression. Find two numbers that multiply to \(-14\) and add to \(5\). These numbers are \(7\) and \(-2\).

\((x + 7)(x - 2) = 0\)

Step 3: Solve for \(x\) by setting each factor equal to 0:

\(x + 7 = 0 \quad \Rightarrow \quad x = -7\)

\(x - 2 = 0 \quad \Rightarrow \quad x = 2\)

The solutions are \(x = -7\) and \(x = 2\).

Step 1: Identify the coefficients \(a\), \(b\), and \(c\):

\(a = 2, b = -3, c = -5\)

Step 2: Write the quadratic formula:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Step 3: Substitute the values of \(a\), \(b\), and \(c\) into the formula:

\(x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-5)}}{2(2)}\)

\(x = \frac{3 \pm \sqrt{9 + 40}}{4}\)

\(x = \frac{3 \pm \sqrt{49}}{4}\)

Step 4: Solve for \(x\):

Case 1: \(x = \frac{3 + 7}{4} = \frac{10}{4} = 2.5\)

Case 2: \(x = \frac{3 - 7}{4} = \frac{-4}{4} = -1\)

The solutions are \(x = 2.5\) and \(x = -1\).

Step 1: Write the quadratic equation in standard form:

\(x^2 - 7x + 10 = 0\)

Step 2: Factorize the quadratic expression. Find two numbers that multiply to \(10\) and add to \(-7\). These numbers are \(-5\) and \(-2\).

\((x - 5)(x - 2) = 0\)

Step 3: Solve for \(x\) by setting each factor equal to 0:

\(x - 5 = 0 \quad \Rightarrow \quad x = 5\)

\(x - 2 = 0 \quad \Rightarrow \quad x = 2\)

The solutions are \(x = 5\) and \(x = 2\).

Step 1: Move the constant term to the other side of the equation:

\(x^2 - 4x = 21\)

Step 2: Complete the square. Take half the coefficient of \(x\), square it, and add it to both sides:

\(\text{Half of } -4 = -2, \text{Square} = 4\)

\(x^2 - 4x + 4 = 21 + 4\)

\((x - 2)^2 = 25\)

Step 3: Take the square root of both sides:

\(x - 2 = \pm 5\)

Step 4: Solve for \(x\):

\(x = 2 + 5 \quad \Rightarrow \quad x = 7\)

\(x = 2 - 5 \quad \Rightarrow \quad x = -3\)

The solutions are \(x = 7\) and \(x = -3\).

Step 1: Identify the coefficients \(a\), \(b\), and \(c\):

\(a = 2, b = 5, c = -3\)

Step 2: Write the quadratic formula:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Step 3: Substitute the values of \(a\), \(b\), and \(c\) into the formula:

\(x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-3)}}{2(2)}\)

\(x = \frac{-5 \pm \sqrt{25 + 24}}{4}\)

\(x = \frac{-5 \pm \sqrt{49}}{4}\)

Step 4: Simplify the square root and solve for \(x\):

Case 1: \(x = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2}\)

Case 2: \(x = \frac{-5 - 7}{4} = \frac{-12}{4} = -3\)

The solutions are \(x = \frac{1}{2}\) and \(x = -3\).

Step 1: Write the quadratic equation in standard form:

\(x^2 - 8x + 16 = 0\)

Step 2: Factorize the quadratic expression. Recognize that this is a perfect square trinomial:

\((x - 4)^2 = 0\)

Step 3: Solve for \(x\) by setting the factor equal to 0:

\(x - 4 = 0 \quad \Rightarrow \quad x = 4\)

The solution is \(x = 4\) (a repeated root).

Step 1: Divide through by \(4\) to simplify:

\(x^2 - 3x + \frac{9}{4} = 0\)

Step 2: Move the constant to the other side:

\(x^2 - 3x = -\frac{9}{4}\)

Step 3: Complete the square. Take half the coefficient of \(x\), square it, and add it to both sides:

\(\text{Half of } -3 = -\frac{3}{2}, \text{Square} = \frac{9}{4}\)

\(x^2 - 3x + \frac{9}{4} = -\frac{9}{4} + \frac{9}{4}\)

\((x - \frac{3}{2})^2 = 0\)

Step 4: Solve for \(x\):

\(x - \frac{3}{2} = 0 \quad \Rightarrow \quad x = \frac{3}{2}\)

The solution is \(x = \frac{3}{2}\) (a repeated root).

Step 1: Write the quadratic equation in standard form:

\(x^2 - 10x + 25 = 0\)

Step 2: Recognize that this is a perfect square trinomial:

\((x - 5)^2 = 0\)

Step 3: Solve for \(x\) by setting the factor equal to 0:

\(x - 5 = 0 \quad \Rightarrow \quad x = 5\)

The solution is \(x = 5\) (a repeated root).

Step 1: Write the quadratic equation in standard form:

\(x^2 + 6x + 8 = 0\)

Step 2: Factorize the quadratic expression. Find two numbers that multiply to \(8\) and add to \(6\). These numbers are \(4\) and \(2\).

\((x + 4)(x + 2) = 0\)

Step 3: Solve for \(x\) by setting each factor equal to 0:

\(x + 4 = 0 \quad \Rightarrow \quad x = -4\)

\(x + 2 = 0 \quad \Rightarrow \quad x = -2\)

The solutions are \(x = -4\) and \(x = -2\).

Step 1: Identify the coefficients \(a\), \(b\), and \(c\):

\(a = 1, b = -2, c = -15\)

Step 2: Write the quadratic formula:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Step 3: Substitute the values of \(a\), \(b\), and \(c\) into the formula:

\(x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-15)}}{2(1)}\)

\(x = \frac{2 \pm \sqrt{4 + 60}}{2}\)

\(x = \frac{2 \pm \sqrt{64}}{2}\)

Step 4: Simplify the square root and solve for \(x\):

Case 1: \(x = \frac{2 + 8}{2} = 5\)

Case 2: \(x = \frac{2 - 8}{2} = -3\)

The solutions are \(x = 5\) and \(x = -3\).

Step 1: Write the quadratic equation in standard form:

(-3) = -6[/latex] and whose sum is \(-5\). These numbers are \(-6\) and \(1\).

Rewrite the middle term:

\(2x^2 - 6x + x - 3 = 0\)

Group terms:

\((2x^2 - 6x) + (x - 3) = 0\)

Factor each group:

\(2x(x - 3) + 1(x - 3) = 0\)

Factor out \((x - 3):\)

\((2x + 1)(x - 3) = 0\)

Step 3: Solve for \(x\) by setting each factor equal to 0:

\(2x + 1 = 0 \quad \Rightarrow \quad x = -\frac{1}{2}\)

\(x - 3 = 0 \quad \Rightarrow \quad x = 3\)

The solutions are \(x = -\frac{1}{2}\) and \(x = 3\).

Step 1: Recognize that this equation is a perfect square trinomial:

\(x^2 - 12x + 36 = 0\)

Step 2: Write it as a square:

\((x - 6)^2 = 0\)

Step 3: Solve for \(x\) by setting the square equal to 0:

\(x - 6 = 0 \quad \Rightarrow \quad x = 6\)

The solution is \(x = 6\) (a repeated root).

Step 1: Recognize that this is a difference of squares:

\(x^2 - 16 = (x - 4)(x + 4)\)

Step 2: Factorize the expression:

\((x - 4)(x + 4) = 0\)

Step 3: Solve for \(x\) by setting each factor equal to 0:

\(x - 4 = 0 \quad \Rightarrow \quad x = 4\)

\(x + 4 = 0 \quad \Rightarrow \quad x = -4\)

The solutions are \(x = 4\) and \(x = -4\).

Step 1: Factor out the greatest common factor (GCF):

\(3(x^2 - 6x + 9) = 0\)

Step 2: Recognize the perfect square trinomial:

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

\(3(x - 3)^2 = 0\)

Step 3: Solve for \(x\) by setting the factor equal to 0:

\(x - 3 = 0 \quad \Rightarrow \quad x = 3\)

The solution is \(x = 3\) (a repeated root).

Step 1: Identify the coefficients \(a\), \(b\), and \(c\):

\(a = 1, b = 3, c = -10\)

Step 2: Write the quadratic formula:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Step 3: Substitute the values of \(a\), \(b\), and \(c\) into the formula:

\(x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-10)}}{2(1)}\)

\(x = \frac{-3 \pm \sqrt{9 + 40}}{2}\)

\(x = \frac{-3 \pm \sqrt{49}}{2}\)

Step 4: Solve for \(x\):

Case 1: \(x = \frac{-3 + 7}{2} = \frac{4}{2} = 2\)

Case 2: \(x = \frac{-3 - 7}{2} = \frac{-10}{2} = -5\)

The solutions are \(x = 2\) and \(x = -5\).

Step 1: Divide through by \(4\) to simplify:

\(x^2 - x - 2 = 0\)

Step 2: Move the constant to the other side:

\(x^2 - x = 2\)

Step 3: Complete the square. Take half the coefficient of \(x\), square it, and add it to both sides:

\(\text{Half of } -1 = -\frac{1}{2}, \text{Square} = \frac{1}{4}\)

\(x^2 - x + \frac{1}{4} = 2 + \frac{1}{4}\)

\((x - \frac{1}{2})^2 = \frac{9}{4}\)

Step 4: Solve for \(x\):

\(x - \frac{1}{2} = \pm \frac{3}{2}\)

Case 1: \(x = \frac{1}{2} + \frac{3}{2} = 2\)

Case 2: \(x = \frac{1}{2} - \frac{3}{2} = -1\)

The solutions are \(x = 2\) and \(x = -1\).

Step 1: Move the constant to the other side:

\(x^2 + 2x = 8\)

Step 2: Complete the square. Take half the coefficient of \(x\), square it, and add it to both sides:

\(\text{Half of } 2 = 1, \text{Square} = 1\)

\(x^2 + 2x + 1 = 8 + 1\)

\((x + 1)^2 = 9\)

Step 3: Take the square root of both sides:

\(x + 1 = \pm 3\)

Step 4: Solve for \(x\):

Case 1: \(x = -1 + 3 = 2\)

Case 2: \(x = -1 - 3 = -4\)

The solutions are \(x = 2\) and \(x = -4\).