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Revision:OCR Core 1 - Quadratics

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > OCR Core 1 - Quadratics

1 4. Quadratics
2 Also See
3 Comments

4. Quadratics

Quadratic expressions

As seen before, we referred to quadratic expressions as expressions in the form of $ax^{2} + bx + c$ . The term "quadratic" means that the polynomial one is referring to is simply one in which the highest power (of an unknown) is 2. (Cubic expression have a term of an unknown to the power of three, quartic expressions have a term of an unknown to the power of four, etc.).

One refers to a, b, and c as coefficients (of $x^{2}$ , , and constant term respectively).

Throughout the work in Core 1, one can assume that $x \in \mathbb{R}$ .

Alternative forms for quadratics

One can express quadratics in terms of their factors; this is called factor form, and can be useful for finding the solutions to the equation $ax^{2} + bx + c = 0$ quickly.

The completed square form is also useful. In this form one can tell what the minimum, or maximum point of a quadratic is (when it is graphed $y = ax^{2} + bx + c$ ).

Factorising quadratics

The ability to factorise a quadratic expression is essential, offering the ability to solve some quadratic equations easily, and also draw sketch graphs of quadratic functions in an effective manner.

It is a process that one can do using a trial and error process, however, with some insight it can be a quick and powerful tool.

Example

1. Factorise the quadratic $x^{2} + 2x + 1$ .

This one is very simple.

When one wishes to factorise a quadratic, one is looking for two linear algebraic expressions that will multiply together, and sometimes with a constant term, to produce a quadratic function.

Evidently, this expression cannot be divided by an integer value (such that it is simplified), so one merely has to look for two linear algebraic expressions that will multiply together to form this function.

Thinking about it logically, one wishes to solve the following:

$(Ax + B)(Cx + D) = x^{2} + 2x + 1$

This means that:

$B \times D = 1$

$A \times C = 1$

$(B \times C) + (A \times D) = 2$

As we would prefer to have integer values for these variables (A, B, C, and D), one would deduce that A = B = C = D = 1.

Hence, one has deduced that:

$(x + 1)(x + 1) = x^{2} + 2x + 1$

Which is true. Hence one has factorised the quadratic.

The sum of two squares will never factorise.

This is a useful piece of information, and it is easily explained.

Take the quadratic:

$x^{2} + n$

$m \in \mathbb{Z}$

$n = m^{2}$

If the expression did factorise, it would imply that for a value of "x" (or two values), it would be possible to have the expression to be equal to zero.

One can see that this is incorrect from the explanations of the rudimentary features of the quadratic form ( $ax^{2} + bx + c$ ). The constant value dictates the position in the y-axis, and hence if it is a positive, the lowest value of the curve ( $y = ax^{2} + bx + c$ ) would be above zero, and hence there is a contradiction.

Example

2. Solve the quadratic equation:

$x^{2} - 6x + 8 = 0$

First one can factorise this.

$(Ax + B)(Cx + D) = x^{2} - 6x + 8$

$A \times C = 1$

Hence, A = C = 1

$B \times D = 8$

$(B \times C) + (A \times D) = -6$

Hence:

B + D = -6

These algebraic expressions suggest that B = -2, D = -4 (or the other way around; as A = C, it does not matter).

Hence:

$(x - 2)(x - 4) = x^{2} - 6x + 8 = 0$

Hence:

$(x - 2) = 0, \or\ (x - 4) = 0$

Hence:

$x = 2, \or\ x = 4$

These are the solutions.

Completing the square

Completed square form is a form of quadratic expression. It is expressing the quadratic expression in the form of a square (or multiple of a square), and constant remainder.

Consider the quadratic expression:

$x^{2} + bx + c$

If one wishes to obtain a square such that the terms of $x^{2}$ , and are made, one know the following:

$\left( x + \frac{b}{2} \right) ^{2} = x^{2} + bx + \left( \frac{b}{2} \right) ^{2}$

Hence, the remainder will be $c - \left( \frac{b}{2} \right) ^{2}$ .

Example

1. Express the following expression in completed square form:

$x^{2} + 4x + 5$

First calculate the following:

$\left( x + \frac{b}{2} \right) ^{2} = (x + 2)^{2} = x^{2} + 4x + 4$

Now, one can calculate the remainder ( $c - \left( \frac{b}{2} \right) ^{2}$ ).

$c - \left( \frac{b}{2} \right) ^{2} = 5 - 4 = 1$

Hence:

$(x + 2)^{2} + 1 = x^{2} + 4x + 5$

This can be helpful for finding the vertex of a curve. If one was to graph the function in this example, such that $y = x^{2} + 4x + 5$ , one can deduce that the vertex would be at (-2, 1).

2. Express the following expression in completed square form:

$6 - 8x - 4x^{2}$

This is slightly different from the ones that have been encountered so far.

$6 - 8x - 4x^{2} = -4x^{2} - 8x + 6 = -2(2x^{2} + 4x - 3)$

Now one can use the method on the bracket with the positive term of $x^{2}$ .

$-2(2x^{2} + 4x - 3) = -2(2(x^{2} + 2x) - 3) = -2(2(x + 1)^{2} - 5) = 10 - 4(x + 1)^{2}$

In a situation such as this one, one must be careful to ensure that all the signs are correct. A good habit is to check the answer by multiplying out the brackets, and checking that it is identical to the original expression.

Solving quadratic equations

Some quadratic equations can be very hard to factorise, and some cannot be factorised, it is in these circumstances that a more versatile solution is necessary. There is a formula for finding the roots (or solutions) to a quadratic equation of the form $ax^{2} + bx + c = 0$ . The quadratic equation formula is not a method that one would want to use for every quadratic equation as it is slower (in many cases) than factorising, however it is very useful in those cases where one is unable to factorise a quadratic equation.

The quadratic equation formula is derived from the completion of the square on the generic quadratic equation:

$ax^{2} + bx + c = 0$

Hence:

$x^{2} + \frac{b}{a}x + \frac{c}{a} = 0$

$x^{2} + \frac{b}{a}x + \frac{c}{a} = \left( x+ \frac{b}{2a} \right) ^{2} - \left( \frac{b^{2}}{4a^{2}} + \frac{c}{a} \right)$

$\left( x+ \frac{b}{2a} \right) ^{2} - \left( \frac{b^{2}}{4a^{2}} + \frac{c}{a} \right) = 0$

$\left( x+ \frac{b}{2a} \right) ^{2} = \left( \frac{b^{2}}{4a^{2}} - \frac{c}{a} \right)$

$\left( x+ \frac{b}{2a} \right) ^{2} = \left( \frac{b^{2}}{4a^{2}} - \frac{4ac}{4a^{2}} \right) = \frac{b^{2} - 4ac}{4a^{2}}$

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

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

This formula gives the solution to any quadratic equation of the form $ax^{2} + bx + c = 0$

Example

1. Solve the equation $x^{2} - 6x + 8 = 0$ using the quadratic equation formula.

Establish the coefficients:

a = 1, b = -6, c = 8

Hence:

$x = \frac{ -b \pm \sqrt{b^{2} - 4ac}}{2a} = \frac{ 6 \pm \sqrt{36 -32}}{2} = \frac{ 6 \pm 2}{2} = 4, \ or\ 2$

The discriminant

It might have occurred to some students that as there is a square root involved, it could be possible for this to have no real value.

This is precisely true. For those quadratic functions who have no x-axis intercepts, and thus no solution such that $x \in \mathbb{R}$ , this square root has not real value.

This can only be true if the value of $b^{2} - 4ac < 0$ .

This value ( $b^{2} - 4ac$ ) is called the discriminant, and the value of it has an effect on the number of real roots of the equation.

One can see (from the above) that if the discriminant is less than zero, there are no real roots.

Consider the situation when the discriminant is equal to zero. One would be doing the following:

$x = \frac{ -b \pm \sqrt{0}}{2a} = x = \frac{-b}{2a}$

This is only one value, and hence it is said that the equation has a repeated root. This occurs (graphically) when the x-axis is a tangent to the curve (if the quadratic expression is graphed, such that $y = ax^{2} + bx + c$ ).

The other situation is when the discriminant is greater than zero. Quite obviously this will cause there to be two distinct roots, as the square root will have a value (that is not zero).

(It is noteworthy that if the discriminant is, itself, a perfect square, the solutions to the equation will be integers, or fractions, unless one of the coefficients was an irrational number, and another was such that the discriminant could be a perfect square).

Simultaneous equations

As seen before, one can form a quadratic equation through the existence of two simultaneous equations, one, or both of whom are quadratic in order.

Example

1. Find the solutions to the simultaneous equations:

$x^{2} + 4x + 4 = 0$

2x + 3 = 0

Hence one should form a single quadratic equation to solve:

$x^{2} + 2x + 1 = 0 = (x + 1)(x + 1)$

Hence, x = -1, or x = -1.

Equations which reduce to quadratic equations

Some equations may be quartic, or some other order, however they can be reduced to be solvable using the methods one has to solve quadratic equations.

Example

1. Solve the equation:

$x^{4} - 6x^{2} + 8 = 0$