Solving Quadratic Equations for the ESAT

Solving Quadratic Equations by Factorising

To solve a quadratic equation by factorisation, you must first express the quadratic in the form $(ax + b)(cx + d) = 0$, where $a, b, c$, and $d$ are real numbers. Once factorised, you apply the zero product property: if the product of two factors is zero, then at least one of the factors must be zero. This leads to two simpler linear equations: $ax + b = 0$ and $cx + d = 0$.

Consider the example: $6x^2 - 7x - 3 = 0$. To factorise this, you can split the middle term. You need to find two numbers that multiply to $6 \times -3 = -18$ and add up to $-7$. These numbers are $-9$ and $+2$.

Rewriting the middle term, we get: $6x^2 - 9x + 2x - 3 = 0$ $3x(2x - 3) + 1(2x - 3) = 0$ $(2x - 3)(3x + 1) = 0$

Now, solve each factor: If $2x - 3 = 0$, then $x = \frac{3}{2}$ If $3x + 1 = 0$, then $x = -\frac{1}{3}$

Disguised Quadratics and Rearrangement

Some equations do not initially look like quadratics but can be rearranged into the standard form $ax^2 + bx + c = 0$. For instance, consider $\frac{3}{x^2} + \frac{7}{x} = 6$. To solve this, multiply every term by $x^2$ to clear the denominators: $3 + 7x = 6x^2$ $6x^2 - 7x - 3 = 0$ This is now the same quadratic we solved above, yielding $x = \frac{3}{2}$ or $x = -\frac{1}{3}$.

Other equations are disguised through powers. Take $6p^6 = 7p^3 + 3$. Since $p^6 = (p^3)^2$, we can use a substitution. Let $x = p^3$: $6x^2 - 7x - 3 = 0$ From our previous work, $x = \frac{3}{2}$ or $x = -\frac{1}{3}$. Substituting back for $p$: $p^3 = \frac{3}{2}$ or $p^3 = -\frac{1}{3}$ $p = \sqrt[3]{\frac{3}{2}}$ or $p = \sqrt[3]{-\frac{1}{3}}$

Completing the Square

Completing the square is a method used to express a quadratic as a difference of two squares. The general identity is: $x^2 + ax = (x + \frac{a}{2})^2 - \frac{a^2}{4}$

For example, to express $x^2 - 3x$ in the form $(x + a)^2 - a^2$, we compare coefficients. Here, $2a = -3$, so $a = -\frac{3}{2}$. $(x - \frac{3}{2})^2 = x^2 - 3x + (-\frac{3}{2})^2 = x^2 - 3x + \frac{9}{4}$ Therefore, $x^2 - 3x = (x - \frac{3}{2})^2 - (\frac{3}{2})^2$

Solving Quadratic Equations by Completing the Square

This method involves expressing the quadratic in the form $(ax + b)^2 = c$ and then taking the square root of both sides. Solve the equation $x^2 - 4x - 5 = 0$ by completing the square for the $x^2 - 4x$ part: $x^2 - 4x = (x - 2)^2 - 4$

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

Taking the square root of both sides gives: $x - 2 = \pm 3$ If $x - 2 = 3$, then $x = 5$ If $x - 2 = -3$, then $x = -1$

Using the Quadratic Formula

When a quadratic cannot be easily factorised, the quadratic formula is the most reliable tool. For any equation in the form $ax^2 + bx + c = 0$, the solutions for $x$ are: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Solve $3x^2 - 4x = 5$ using the formula. First, rearrange to set the equation to zero: $3x^2 - 4x - 5 = 0$, where $a = 3, b = -4$, and $c = -5$.

Substitute these into the formula: $x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(-5)}}{2(3)}$ $x = \frac{4 \pm \sqrt{16 + 60}}{6}$ $x = \frac{4 \pm \sqrt{76}}{6}$ $x = \frac{4 \pm 2\sqrt{19}}{6} = \frac{2 \pm \sqrt{19}}{3}$

Finding Approximate Solutions Using a Graph

You can find approximate solutions for $ax^2 + bx + c = 0$ by drawing the graph of the function $y = ax^2 + bx + c$. The solutions are the $x$ values where the curve crosses the $x$ axis, because that is where $y = 0$.