Solving Quadratic Equations

\( 3x^2 = 48 \).

Step 1: Isolate the squared variable.

\( 3x^2 = 48 \)

\(x^2 = 16\)

Step 2: Solve your quadratic equation by calculating the square root of both sides of the equation. Remember that you will have two solutions because the square root of a number can either be positive or negative.

\( \sqrt{x^2} = ±\sqrt{16} x = ±4 \)

\( 14x^2 - 35x = 0\)

Step 1: Find the greatest common factor by identifying the numbers and variables that each term has in common.

\( 14x^2 = 2 \cdot 7\cdot x \cdot x \)

\( 35x = 5 \cdot 7 \cdot x\)

As we can see the factors that are common to both terms are 7 and x, therefore the GCF = 7x.

Step 2: Write out each term as a product of the greatest common factor and another factor, i.e. the two parts of the term. The other factor can be determined by dividing your term by your GCF.

\(\frac{14x^2}{7x} = 2x \therefore 14x^2 = (2x) \cdot 7x\)

\(\frac{35x}{7x} = 5 \therefore 35x = 7x \cdot 5\)

Step 3: Having rewritten each term, rewrite the quadratic equation in the following form: ab+ac=0

\( 14x^2 - 35x = 7x \cdot (2x) - 7x(5)\)

Step 4: Apply the law of distributive property and factor out the greatest common factor.

\( 7x(2x) -7x(5) = 7x(2x-5)\)

Step 5: Equate the factored expression to 0 and find the x-intercepts.

\( x_1: \begin{split} 7x = 0 \\ x = 0 \end {split} \)

\( x_2: \begin{split}2x - 5= 0 \\ x = \frac{5}{2}\end {split}\)

Step 1: Transform your equation from standard form, \( ax^2 + bx + c = 0\), into a perfect square trinomial, \( a^2 + 2ab + b^2\).

\( 9x^2 - 12x +4 = (3x)^2 -2(3x)(2) - 2^2\)

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

\( 3x - 2 = 0\)

\( x = \frac {2}{3} \)

\( 2x^2 - 7x - 15\)

Step 1: List out the values of a, b and c.

\( a = 2, b = -7, c = -15\)

Step 2: Find the factors that when multiplied equal \(a \cdot c\) , and when added equal b. T where numbers that product ac and also add to b.

\( ac = -30, b = -7\)

\(1 \cdot 30 = 30\)

\(2 \cdot 15 = 30\)

\(3 \cdot 10 = 30 \qquad 3-10 = -7\)

\(5 \cdot 6 = 30\)

The two numbers are therefore 3 and -10, as they add to -7. The other factors of 30 cannot be arranged in any way that would make them equal to -7.

Step 3: Use these factors to rewrite the x-term (bx) in the original expression/equation.

\( 2x^2 - 7x - 15 = 2x^2 + 3x - 10x - 15\)

Step 4: Use grouping to factor the expression. Group the first two terms and the last two terms together, then pull out common factors from both groups and combine like terms.

\( \begin{split} (2x^2 + 3x) - (10x - 15) = x (2x + 3) - 5(2x + 3) \\ = (x-5)(2x-3)\end{split}\)

Step 5: Equate the factored expression to 0 and solve for the x-intercepts.

\( (x - 5) (2x - 3)\)

\( x_1: \begin{split} x - 5 = 0 \\ x = 5 \end {split} \)

\( x_2: \begin{split}2x + 3= 0 \\ x = \frac{-3}{2}\end {split}\)

\( 3x^2 - 5x - 7 = 0 \)Step 1: List out the values of a, b and c.

\(a = 3, b = -5, c = -7 \)

Step 2: Calculate the value of m by using the following equation: \( m = \frac{b}{2a}\)

\( \begin{split}m = \frac{-5}{2(3)} \\ = -\frac{5}{6}\end{split} \)

Step 3: Calculate the value of n by using the following equation: \( n = c - \frac{b^2}{4a}\)

\(n = -7 - \frac {(-5)^2}{4(3)}\)

\(\begin{split} n = -7 - \frac {25}{12} \\ = - \frac{109}{12} \end{split}\)

\(3(x - \frac{5}{6})^2 - \frac{109}{12} = 0\)

\(3(x - \frac{5}{6})^2 = \frac{109}{12}\)

\(x - \frac{5}{6}^2 = \frac{109}{36}\)

\(\sqrt{(x - \frac{5}{6}^2} = ±\sqrt{\frac{109}{36}}\)

\(x - \frac {5}{6} = \pm \frac {\sqrt{109}}{6}\)

\(x_{1,2} = \pm \frac {\sqrt{109}}{6} + \frac{5}{6}\)

\(x_{1} : x = \frac{5 + \sqrt{109}}{6} = 2.57\)

\(x_2: x = \frac {5-\sqrt{109}}{6} = -0.91 \)

\( x^2 - 7x + 12 = 0 \)

Step 1: List out the values of a, b and c.

\(a = 1, b = -7, c = 12 \)

Step 2: Calculate the value of the discriminant.

\(\begin{split} \Delta =(-7)^2 -4(1)(12) \\ = 1 \end{split}\)

Step 3: Substitute the values of a,b and c into the Quadratic Formula and solve for both roots/solutions.

\(\begin{align}x_{1} : x&= \frac{-b + \sqrt{\Delta}}{2a} \\ &=\frac{-(-7)+\sqrt{1}}{2(1)} \\ &= 4 \end{align}\)

\(\begin{align}x_{2} : x&= \frac{-b - \sqrt{\Delta}}{2a} \\ &=\frac{-(-7)-\sqrt{1}}{2(1)} \\ &= 3 \end{align}\)