Solving Equations By Factorization
To understand minimum and maximum of quadratic functions let us first understand how to solve a quadratic equation
. A quadratic equation is an equation of the form ax2 + bx + c = 0.
When the product of two or more numbers is equal to zero, then one of the numbers must be zero. This idea can be used to solve quadratic equations.
Example
Solve the quadratic equation (x +1)(x - 2) = 0 .
Answer
(x +1)(x-2) = 0
Either, x +1 = 0 = x = - 1
Or, x-2 = 0 = x = 2
The solutions are x = 1 and 2.
To solve an equation like x2 + 5x + 4 = 0, we first need to factorise x2 + 5x + 4 into a product of two factors (x +)(x +), where and are some numbers. To understand how to do this, we expand the product (x +)(x +), giving:
(x +)(x +) = x2 +x +x + = x2 + ( +)x + .
You can see that we can write x2 + 5x + 4 in the form (x + )(x + ) if we can find two numbers and with + = 5 and = 4 . We now try to guess the numbers and .
The product of and is 4, so the numbers could be 1 & 4 or 2 & 2.
The sum of and is 5, so and must be 1 & 4
This shows that x2 + 5x + 4 = (x +1)(x + 4).
If we couldnt guess these numbers, then we would need to use another method!
Example
Write x2 + x - 6 as a product of two factors.
Answer
The numbers and have product 6, and sum 1.
Pairs of numbers with product 6 are 1 & 6; 2 & 3; 3 & 2; and 6 & 1.
Their sum is 1, so and must be 3 & 2, and x2 + x - 6 = (x + 3)(x - 2).
A common factor is a factor which is a factor of every term. It is a good idea to factorise out the common factors first of all.
Example
Factorise 10x2 - 10x - 60.
Answer
10x2 - 10x 60
= 10(x2 - x - 6)
= 10(x - 3)(x + 2)
Example
Solve
3x2 + 3x - 36 = 0, by factorising.
Answer
3x2 + 3x 36 = 0
= 3(x2 + x - 12) = 0
= 3(x + 4)(x - 3) = 0
= 3(x + 4)(x - 3) = 0
Either, x + 4 = 0 = x = - 4
Or, x - 3 = 0 = x = 3
The solutions are x = 3 and 4.
Example
Solve x2 = 7x.
= x2 = 7x
= x2 - 7x = 0
= x(x - 7) = 0.
So,
Either, x = 0;
Or, x - 7 = 0 = x = 7;
The answers are x = 0 and 7.
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If we tried to solve x2 = 7x by dividing both sides of the equation by x, then we end up with only one solution: x = 7. The reason is that when we divide both sides by x we need to assume that x 0, so we end up with only the non-zero solution.
Solving Equations by Using the Quadratic Formula and Finding Minimum and Maximum
The quadratic formula is useful for solving quadratic equations when the equation has complicated terms. If you only want an approximate answer, then the quadratic formula is best. If you want an exact answer, then you may prefer the methods of factorisation or completing the square. However, all three methods will lead to the same results.
The Quadratic Formula:
(1) If b2 - 4ac 0, then the equation ax2 + bx + c = 0 has no solutions.
(2) If b2 - 4ac 0, then the equation ax2 + bx + c = 0 has two solutions, given by x = '(-b +-sqrt(b^2-4ac))/(2a)'
(3) If b2 - 4ac = 0 the two solutions are the same.
Example
Solve the equation
x2 + x - 2.01= 0.
Answer
a =1, b =1, c = -2.01
b2 - 4ac =12 - 4 x 1 x (- 2.01) = 9.04 0 = there are two solutions
x = '(-b +-sqrt(b^2-4ac))/(2a)' = - 1 9.04 2
=1.0033 and - 2.0033 (4 d.p.)
Examples of Quadratic Functions and Finding Minimum and Maximum of Parabolas
We often see parabolas in the world around us, in equipment and in visual design.
Example
The mirrors in torches and car headlights are shaped like parabolas; microwave receivers on the roofs of buildings and satellite TV receivers also have parabolic shapes. Parabolas have the special property that radiation generated at a point, called the focus, is reflected in parallel rays off the parabola. Torch bulbs are placed at this point. Also, parallel rays of incoming radiation are concentrated at the focus. Receptor devices are placed at this point of high intensity. Every parabola has this property, and no other type of curve does.
parabola_0
We use x and y as the independent and dependent variables when we study the general properties of quadratic functions, and place no restriction on the domain of the functions. The important characteristics of quadratic functions are found from their graphs. These are the x- and y-intercepts, the vertex (or turning point), the line of symmetry.
Example
The graph of the quadratic function y = x2 + 4x + 3 is shown below. The x-intercepts of the parabola are (1, 0) and (3, 0), the y-intercept is (0, 3) and the vertex or turning point is (2, 1).
You can see that the parabola is symmetric about the line x = 2, in the sense that this line divides the parabola into two parts, each of which is a mirror image of the other.
parabola1
The Shape of a Parabola and Minimum and Maximum of Quadratic Equations
Parabolas have two orientations: concave up and concave down. The orientation of a parabola can be found from its equation.
parabola2
Example
The parabola y = x2 is concave up. To see this, imagine how the value of y will change when we substitute very large x-values into the equation, such as x = 1 000, x = 1 000 000, x = 1000 000 000, etc. As x is given larger and larger values, the value of y becomes very large and positive. So the parabola must be concave up.
In comparison, the parabola y = -x2 is concave down. As x is given larger and larger values, the value of y becomes very large and negative. So the parabola must be concave down.
Example
The parabola y = 50x2 - 230x + 107 is concave up. To see this, imagine how y changes when very large values of x are substituted into the equation. When x is very large, the value of 50x2 will be much larger than the value of 230x and also much larger than 107, so the value of y would be very large and positive (check this). This would be true for all very large values of x, so the parabola must be concave up.
In general:
The parabola y = ax2 + bx + c is concave up if a 0 and concave down if a 0.
Minimum and Maximum of Quadratic Functions
To find the minimum or maximum of a quadratic function we complete the square expressing the function in the form:
f(x) = a(x-p)2 + q
But in this method, there are restrictions as to whether a maxima or a minima exists or not. Since a quadratic equation actually transforms into a parabola, we need to know whether it's an upward open parabola(minima) or a downward open parabola(maxima). It is determined by the sign of the coefficient of the square term of the variable('x').
by: mathqa
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