Solving Quadratic Equations for University Admission

Solving Quadratic Equations by Factorising

Factorisation involves expressing a quadratic expression in the form $(ax + b)(cx + d) = 0$. Once in this form, the zero product property allows us to solve the equation by setting each linear factor to zero, specifically $ax + b = 0$ or $cx + d = 0$. This method is most effective when $a, b, c$ and $d$ are rational numbers, often integers.

Worked Example: Solving by Factorising

Solve the equation: $6x^2 - 7x - 3 = 0$

To factorise this, we can use the method of splitting the middle term. We look for two numbers that multiply to give $6 \times -3 = -18$ and add up to $-7$. These numbers are $-9$ and $+2$.

1. Split the middle term: $6x^2 - 9x + 2x - 3 = 0$
2. Factorise in pairs: $3x(2x - 3) + 1(2x - 3) = 0$
3. Take out the common bracket: $(2x - 3)(3x + 1) = 0$

Now, solve the resulting linear equations:

If $2x - 3 = 0$, then $x = \frac{3}{2}$. If $3x + 1 = 0$, then $x = -\frac{1}{3}$.

A quadratic equation may result in zero, one, or two real solutions.

Disguised Quadratics

In many exam problems, equations do not initially look like quadratics but can be transformed into them through rearrangement or substitution. These are often called disguised quadratics.

Worked Example: Rearrangement

Solve the equation: $\frac{3}{x^2} + \frac{7}{x} = 6$

First, multiply every term in the equation by $x^2$ to remove the denominators: $3 + 7x = 6x^2$

Now, rearrange the equation into the standard form $ax^2 + bx + c = 0$: $6x^2 - 7x - 3 = 0$

This is the same quadratic solved in the previous example, giving $x = \frac{3}{2}$ or $x = -\frac{1}{3}$.

Worked Example: Substitution

Solve the equation: $6p^6 = 7p^3 + 3$ to find expressions for $p$.

Observe that $p^6 = (p^3)^2$. We can substitute $x = p^3$ to create a quadratic equation: $6x^2 - 7x - 3 = 0$

Using our previous solutions for $x$, we have: $p^3 = \frac{3}{2}$ or $p^3 = -\frac{1}{3}$

Finally, solve for $p$ by taking the cube root: $p = \sqrt[3]{\frac{3}{2}}$ or $p = \sqrt[3]{-\frac{1}{3}}$

Completing the Square

Completing the square is the process of writing a quadratic expression as the difference of two squares. For an expression $x^2 + ax$, we can write:

$x^2 + ax = (x + \frac{a}{2})^2 - (\frac{a}{2})^2$

Worked Example: Completing the Square

Express $x^2 - 3x$ as a difference of two squares in the form $(x + a)^2 - a^2$.

Comparing $x^2 + 2ax$ with $x^2 - 3x$, we find $2a = -3$, so $a = -\frac{3}{2}$. $(x - \frac{3}{2})^2 = x^2 - 3x + (\frac{3}{2})^2$ $x^2 - 3x = (x - \frac{3}{2})^2 - (\frac{3}{2})^2$

Solving Equations by Completing the Square

Once a quadratic is expressed in the form $(ax + b)^2 = c$, we can solve for $x$ by taking the square root of both sides.

Worked Example: Solving by Completing the Square

Solve $x^2 - 4x - 5 = 0$ by completing the square.

First, complete the square for $x^2 - 4x$: $x^2 - 4x = (x - 2)^2 - 4$

Substitute this back into the original equation: $(x - 2)^2 - 4 - 5 = 0$ $(x - 2)^2 = 9$

Take the square root of both sides: $x - 2 = \pm 3$

If $x - 2 = +3$, then $x = 5$. If $x - 2 = -3$, then $x = -1$.

The Quadratic Formula

For any quadratic equation in the form $ax^2 + bx + c = 0$, the solutions can be found using the quadratic formula. You must be able to recall this formula:

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

Worked Example: Using the Quadratic Formula

Solve the equation $3x^2 - 4x = 5$. Leave your answer in surd form.

First, set the equation to zero: $3x^2 - 4x - 5 = 0$. Here, $a = 3, b = -4$, and $c = -5$.

Substitute these into the formula: $x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(-5)}}{2(3)}$ $x = \frac{4 \pm \sqrt{16 + 60}}{6}$ $x = \frac{4 \pm \sqrt{76}}{6}$

Simplify the surd $\sqrt{76} = \sqrt{4 \times 19} = 2\sqrt{19}$: $x = \frac{4 \pm 2\sqrt{19}}{6} = \frac{2 \pm \sqrt{19}}{3}$

Finding Approximate Solutions Graphically

Approximate solutions to $ax^2 + bx + c = 0$ can be identified by plotting the function $y = ax^2 + bx + c$. The roots are the $x$ values where the curve crosses the $x$ axis (where $y = 0$).

In more detail, you can use a grid to read these values to a specified degree of accuracy, such as 1 decimal place.

From a graph like the one above, we can identify the roots by looking at the points where the parabola intercepts the horizontal axis. In this specific case for $y = 2x^2 + 6x + 3$, the roots are approximately $-0.6$ and $-2.4$.