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QUADRATIC EQUATIONS
1. Introduction
A quadratic equation in the variable x is an equation that can be written in the form:
ax2 + bx + c = 0
(1)
where a, b and c represent real number coefficients.
This form is sometimes called the standard form. The term quadratic is used for any equation
where the highest power of the variable x is 2. This means an equation that is a constant times
the variable squared plus a constant times the variable plus a constant equals zero, where the
coefficient a of the variable squared cannot be zero, since otherwise it would be a linear
equation.
Example 1
Consider the following quadratic equations:
(a) 2x2 +3x +1 = 0
(b) x2 + x = 3
(c) (x+2)(x+3) = 5
(d) 5x2 – 7x = 0
(e) 3x2 + 8 = 0
Write these equations in standard form.
Solution
Equation (a) is already in standard form.
Equation (b) requires the term 3 to be transferred to the left-hand-side.
Equation (c) requires an expansion of the brackets, the term 5 to be transferred to the lefthand-side and like terms collected
Equation (d) is already in standard form (there is no constant term).
Equation (e) is already in standard form (there is no term in x).
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Example 2
Write the following equation in standard form and identify a, b, and c.
(x + 4)(x – 3) = 10
Solution
Expand the brackets on the left-hand-side and then subtract the 10 from both sides.
x2 – x – 12 = 10
x2 – x – 22 = 0
Therefore
a = 1 b = – 1 c = – 22
2. SOLVING QUADRATIC EQUATIONS BY FACTORING
Quadratic equations are more difficult to solve than linear equations, since once you have
them in standard form it is hard to simplify them any further. Also, in this form there are still
generally two values of the variable (called roots) that satisfy the equation, so it's hard to see
what can be done to get the variable alone.
There are several techniques available to overcome the problem.
One way is to solve the equation by factoring. This method works because of the following
principle of zero products. This states:
If A and B are numbers or expressions and A x B = 0, then either A = 0 or B = 0.
That is, the only way to multiply two numbers or expressions and obtain zero is if one of
these numbers or expressions itself is zero.
The principle of zero products allows us to reduce a complicated equation to a simpler
equation provided the right side of the equation is zero and the left-hand-side of the equation
can be factored, because then each of the factors can be set equal to zero.

To solve a quadratic equation by factoring, it must first be written in standard form. It is
very important that the right-hand-side is set equal to zero since the principle of zero
products only works for zero on the right-hand-side.

Factor the left-hand-side of the equation.

Set each of the factors equal to zero and solve the two resulting linear equations to find
the solutions (roots) to the original quadratic equation.
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Example 3
Solve the following quadratic equation using factoring.
2x2 + 7x – 15 = 0
Solution
Factor the left-hand-side:
(x + 5)(2x – 3) = 0
Then using the principle of zero products:
x+5=0
or
2x – 3 = 0
Therefore the two solutions (or roots) are:
x = – 5 and
x = 1.5
3. SOLVING QUADRATIC EQUATIONS USING THE QUADRATIC
FORMULA
Some quadratics are difficult (or even impossible) to factor and so for these equations other
methods are required. Perhaps the simplest method in this case is to use the quadratic formula
which provides the two roots of any quadratic equation, whether it can be factored or not.
The quadratic formula for an equation written in standard form is shown in Equation (1).
(2)
The expression b2 – 4ac shown under the square root sign is called the discriminant.
If the value of the discriminant is positive then the equation has two roots which are different
in value.
If the value of the discriminant is zero then the equation has two roots but they are both the
same.
If the value of the discriminant is negative then the equation has no roots that are real
numbers. (In this case complex numbers may be used but this is beyond the scope of this
discussion.).
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Example 4
Solve the following quadratic equation using the quadratic formula.
2x2 + 7x – 15 = 0
Solution
In this case a = 2 b = 7 c = – 15
The value of the discriminant is b2 – 4ac = 72 – 4(2)(–15) = 169
From Equation (2):
x = (– 7  169)/4
= – 5 or 1.5
Example 5
Solve the following equation using the quadratic formula.
4x2 – 20x + 25 = 0
Solution
In this case a = 4 b = – 20 c = 25
The value of the discriminant is b2 – 4ac = 202 – 4(4)(25) = 0
From Equation (2):
x = (20  0)/8
= 2.5
That is, in this case since the value of the discriminant is zero, the two roots of the equation
have the same of 2.5.
Example 6
Solve the following quadratic equation using the quadratic formula.
5x2 + 2x + 3 = 0
5
Solution
In this case a = 5 b = 2 c = 3
The value of the discriminant is b2 – 4ac = 42 – 4(5)(3) = – 44
Since the value of the discriminant is negative, this equation has no roots that are real
numbers.
4. REMARKS
This document has dealt with quadratic equations that are also known as second degree
equations since the highest exponent of the x variable is 2. In the case where the highest
exponent of x is 3, these are called cubic equations (where there is generally three roots) and
where it is 4 these are called quartic equations (where there is generally four roots) .
Formulas exist for finding the roots of cubic and quartic equations but they are quite complex
and beyond the scope of this discussion.
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EXERCISES
1. Solve the following quadratic equations by factoring.
(a) 2x2 – 7x = 0
(b) x2 + x – 12 = 0
(c) 2x2 – x – 0 = 0
(d) 12x2 – 13x + 3 = 0
(e) 4x2 + 27x + 35 = 0
2. Solve the following quadratic equations using the quadratic formula.
(a) x2 – 4x – 12 = 0
(b) 6x2 + 31x + 35 = 0
(c) 3x2 – 5x – 5 = 0
(d) 9x2 – 30x + 25 = 0
(e) 3x2 + 4x + 2 = 0
3. Construct an equation where b2 = 4ac and the value of b is negative. Show that both roots
to your equation are identical.
4. The square of a number is equal to two more than the value of the number itself. Construct
an appropriate quadratic equation to find all the possible values of the number.
5. The area of a rectangular paddock is 525 square metres. The length of the paddock is 5
metres more than twice the breadth.
(a) Construct an appropriate quadratic equation to find the length and breadth of the paddock.
(b) Solve the equation in (a) and hence find both the length and breadth of the paddock.
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SOLUTIONS TO EXERCISES
1. (a) x = 0, x = 3.5
(b) x = 3, x = – 4
(c) x = 2.5, x = – 2
(d) x = 1/3, x = ¾
(e) x = – 5, x = – 7/4
2. (a) x = 6, x = – 2
(b) x = – 3.5, x = – 5/3
(c) x = 1.82, x = – 0.15
(d) x = 5/3, x = 5/3. (Both roots are identical since the value of the discriminant is zero.)
(e) There is no real number solution since the value of the discriminant is negative.
3. An example of such an equation is:
x2 – 4x + 4 = 0
4. Let the number = x. Then:
x2 = x + 2
or
x2 – x – 2 = 0
The two roots are x = –1 and x = 2.
5. Let x = the breadth of the paddock (in metres)
Then the length = 2x + 5 metres
Since the area is 525 square metres, then:
x(2x + 5) = 525
2x2 + 5x – 525 = 0
(b) The only positive solution is x = 15.
The breadth of the paddock is x = 15 metres
The length of the paddock = 2x + 5 = 35 metres