Introduction
Quadratic Equation
The word “quadratic” came from “quadrature”, the Latin word for square. Hence, we can define a quadratic equation as an equation where the variable is of the second degree. Therefore, a quadratic equation is also called an “Equation of degree 2”.
Consider this example. Suppose you want a plot of land and want to build a house there.
Assume the area of the plot is 300 m^{2}. You want to make your house have a breadth equal to twice its length. Let \(‘x’\) be the length then breadth \(= 2x.\) Also, we might have Length×Breadth = Area of the House.
How will you optimize the area of land you have? We can write the area of the house as \((x) × (2x) = 2x^2.\) If \(2x^2 = 300,\) we will have optimized the area available to us.
This is called a quadratic equation. Solving for \(x\) gives us the value of \(x = 12.247 \text{ m}. \)So to make optimal use of the land, we just have to make the length of our house \(= 12.247 \text{ m}\) and the breadth \(= 24.495 \text{ m.}\)
General form of Quadratic Equation
The Standard Form
We must know that a quadratic polynomial can be written as \(ax^2 + bx + c.\) This is known as the general form or quadratic form.
If the quadratic polynomial = 0, it forms a quadratic equation. Therefore, the standard quadratic form of a quadratic equation can be written as: \(ax^2 + bx + c = 0;\) where x is an unknown variable, and a, b, c are constants with ‘a’ ≠ 0 (if a = 0, then it becomes a linear equation).
Source: giphy.com
The constants ‘a’, ‘b’, and ‘c’ are called coefficients. Let us look at some examples of a quadratic form:
 \(2x^2+5x+3=0;\) In this, a=2, b=3 and c=5
 \(x^23x=0;\) Here, a=1 since it is 1 times \(x^2,\) \(b=3\) and \(c=0,\) not shown as it is zero.
But sometimes, the quadratic equation does not come in the standard quadratic form. These are the hidden quadratic equations that we might have to reduce to the quadratic form. Here are some examples:
Equation

Standard Form

Coefficients

Explanation

x^{2} – 3x = 1

x^{2} – 3x – 1 = 0

a = 1, b = 3, c = 1

Compare it to the general form of the quadratic equation and subtract 1 from both sides.

2(z^{2} – 2z)=5

2z^{2} – 4z – 5 = 0

a = 2, b = 4, c = 5

We need to expand (open the brackets)
by multiplying 2 with z^{2} and 2z and also we need to bring 5 to the left side to equate the equation with 0.

y(y2)=0

y^{2} – 2y = 0

a = 1, b = 2, c = 0

We need to expand, multiply y with both y and 2 and the output you get is in the desired general form.

How to write a quadratic function in standard form?
The functions above are examples of quadratic functions in standard quadratic form. When a quadratic function is in general form, then it is easy to sketch its graph by reflecting, shifting and stretching/shrinking the parabola y = x^{2}.
The quadratic function \(f(x) = a(x  h)^2 + k,\) not equal to zero, is said to be in standard quadratic form. If a is positive, the graph opens upward, and if a is negative, then it opens downward.
The line of symmetry is that the vertical line \(x = h,\) and therefore the vertex is that the point (h,k).
So, how to write a quadratic function in standard form?
Any quadratic function is often rewritten in standard quadratic form or general form by completing the square. Note that when a quadratic function is in its general form it's also easy to seek out its zeros by the root principle.
Let us now look at some examples to fully understand how to write a quadratic function in standard form.
Consider this in its general form.
\(f(x) = 2x^2 + 2x + 3\)
Find the vertex of the graph of f.
Solution
\[\begin{align}f(x) &= 2x^2 + 2x + 3\\&= (2x^2 + 2x) + 3\\&= 2(x^2 x) + 3\\&= 2(x^2  x + \frac{1}{4} \frac {1}{4}) + 3\end{align}\]
We add and subtract \(\frac{1}{4},\) because \((\frac{1}{2})2 = \frac{1}{4},\) and 1 is that the coefficient of x.
\(= 2(x^2  x + \frac{1}{4}) 2(\frac{1}{4}) + 3.\)
Note that everything within the parentheses is multiplied by 2, so once we remove \(\frac{1}{4 }\)from the parentheses, we must multiply it by 2.
\(\begin{align}f(x) &= 2(x \frac{1} {2})2 + \frac{1}{2} + 3\\[5pt]&= 2(x  \frac{1}{2})2 + \frac{7}{2}\end{align}\)
.
The vertex is the point \((\frac{1}{2}, \frac{7}{2})\). So the graph opens downward \((2 < 0),\) the vertex is the highest point on the graph.
How to find the zeros of a quadratic function?
In precalculus, you might have used the zeroproduct property to find the roots of a factored equation. After you factor a polynomial into its different sets, you can identify each set equal to zero to solve for the roots with the zeroproduct property.
The zeroproduct property predicts that if several factors are multiplying to give you zero, at least one of them has to be zero.
Your work is to find all the values of x that make the polynomial equal to zero. If the polynomial is factored as you can set each factor equal to zero and solve for x.
So, how to find the zeros of a quadratic function?
Factoring \(x ^2+ 3x – 10 = 0\) gives you \((x + 5)(x – 2)\).
Moving ahead is easy because each factor is linear (first degree).
The equation \(x + 5 = 0\) gives you a solution, \(x = –5,\) and
\( x – 2 = 0\) gives you the other solution,
\(x = 2.\)
These answers each become an xintercept on the graph of the polynomial.
Sometimes after you’ve factored, one or both of the two factors can be factored again, in this case, you should continue factoring.
In other cases, they may be non factorable. If one of these factors is quadratic, you could find the roots only by using the quadratic form.
For example, \(6x^4 – 12x^3 + 4x^2= 0\) factors to \(2x^2(3x^2– 6x + 2) = 0.\)
The first term, \(2x^2 = 0,\) is solvable using algebra, but the second factor, \(3x^2– 6x + 2 = 0,\) is non factorable and requires the quadratic form.
In other cases, they may be non factorable, in which case you can solve them only by using the quadratic form.
Consider \(f(x) = 3x^2 + 12x + 8\) in general form. Sketch the graph of f ,find its vertex, and find the zeros of f.
Answer
An alternate method of finding the vertex
In some cases completing the square isn't the simplest thanks to finding the vertex of a parabola. If the graph of a quadratic function has two xintercepts, then the road of symmetry is the vertical line through the midpoint of the xintercepts.
The xintercepts of the graph above are at 5 and three.
The line of symmetry goes through 1, which is the average of 5 and three.
\(\frac{(5 + 3)}{2} = \frac{2}{2} = 1\)
Once we all know that the road of symmetry is \(x = 1,\) then we all know the primary coordinate of the vertex is \(1.\)
The second coordinate of the vertex is often found by evaluating the function at \(x = 1\)
Consider the function \(f(x) = x^2  6x + 7\) in general form. Sketch the graph of f and find its zeros and vertex.
Solution
\(f(x) = x^2  6x + 7\)
\(= (x2  6x )+ 7.\) Group the \(x^2\) and x terms then complete the square on these terms.
\(= (x^2  6x + 9  9) + 7\)
We need to feature 9 because it's the square of 1 half the coefficient of x, \((\frac{6}{2})2 = 9.\)
Hence we were solving an equation. We simply added 9 to each side of the equation. In this setting, we add and subtract 9 in order that we don't change the function.
\(= (x^2  6x + 9)  9 + 7\)
We see that \(x^2  6x + 9\) is a perfect square, namely \((x  3)^2.\)
\(f(x) = (x  3)^2  2\)
This is the standard form.
From this result, one easily finds the vertex of the graph of \(f\) is \((3, 2).\)
To find the zeros of \(f,\) we set f adequate to 0 and solve for \(x.\)
\[\begin{align}(x  3)^2  2 &= 0\\[5pt](x  3)^2 &= 2\\[5pt](x  3) &= ± \sqrt{2}\\[5pt]x &= 3 ± \sqrt{2}\end{align}\]
To sketch the graph of f we shift the graph of \(y = x^2\) three units to the proper and two units down.
If the coefficient of \(x^2\) isn't 1, then we must factor this coefficient from the \(x^2\) and \(x\) terms before proceeding.