Quadratic Equations
1. Introduction
We have already studied linear equations which involve variables whose power is one. Some times we come across the equations like \({x^2} - 14{\rm{ }} = {\rm{ }}0\) or \({x^2} - 5x + {\rm{ }}60{\rm{ }} = {\rm{ }}0,\) these types of equations have two powers of \(x\) and have two solutions which means that these equations can be balanced by two values of \(x\) which are known roots of the equations. For example: \({x^2} - 5x{\rm{ }} + 6{\rm{ }} = {\rm{ }}0\) can be written as \((x - 2)(x - 3) = 0\) which means \(x = 2\) or \(x = 3\). Hence equation has two roots or solutions. These type of equations are known as quadratic equation, in general any equation of the type \(a{x^2} + bx + c = {\rm{ }}0\) is known as quadratic equation, where \(a, b, c\) are real numbers. In other words \(A\) quadratic equation is an equation where the highest power of \(x\) is \({x^2}\). There are various methods of solving quadratic equations, as shown below.
(a) Completing the Square:
Notice that the Principle of Square Roots requires an algebraic expression to be squared and set equal to a constant. Not all quadratic equations come in this form. However, we can always algebraically create this form through a process called completing the square.
Given a quadratic expression of the form\({x^2} + bx\), if we add the constant term \({\left( {\frac{b}{2}} \right)^2}\) to the expression \({x^2} + bx\) we obtain
\({x^2} + bx + {\left( {\frac{b}{2}} \right)^2} = {\left( {x + \frac{b}{2}} \right)^2}\).
Example 1: Solve \({x^2} - 6x + 2 = 0\) by completing the square.
Solution: The given equation is \({x^2} - 6x = - 2\)
(To complete the square on the LHS (left hand side), we must add \({6^2}/4 = 9\). We must, of course, do this to the RHS also).
\({x^2} - 6x + 9 = 7 \Rightarrow {(x - 3)^2} = 7\)
\((x - 3) = \pm 2.646 \Rightarrow x = \pm 5.646\)
(b) Using Factors
Many times the expression \(a{x^2} + bx + c\) can be factorised and by equating each factor to zero, we get two solutions for example\({x^2}-{\rm{ }}10x + {\rm{ }}21{\rm{ }} = {\rm{ }}0\) can be written
\(\left( {x-{\rm{ }}3} \right)\left( {x-{\rm{ }}7} \right){\rm{ }} = {\rm{ }}0 \Rightarrow \;x = {\rm{ }}3\) or 7
Example 2: In a college there are some students, seven times the square root of the total number of students are playing, one fourth students are studying and remaining 49 students are sleeping. Find the total number of students.
Solution: Suppose the total number of student is \({x^2}\) then \(7x\) are playing \(\frac{{{x^2}}}{4}\) are studying and 49 are sleeping. Hence \({x^2}\) = \(7x + \frac{{{x^2}}}{4}+ 49\)
\(3{x^{2\;}} = {\rm{ }}28x\; + {\rm{ }}196\) or \(3{x^2}-{\rm{ }}28x\;-{\rm{ }}196{\rm{ }} = {\rm{ }}0\)
\(\left( {3x + {\rm{ }}14} \right){\rm{ }}\left( {x-{\rm{ }}14} \right) = 0 \Rightarrow x\; = {\rm{ }}14\) or \( - \frac{{14}}{3}\)
Total number of students = 14.
(c) Using the formula \(\frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\)
Some times the quadratic expression can not be factorised. In such cases we use formula to find roots if the equation\[a{x^2} + bx{\rm{ }} + {\rm{ }}c = {\rm{ }}0\] Where \(a \ne {\rm{ }}0\) and \(a, b\) and \(c\) belong to the set of real numbers, is called a quadratic equation. Simplifying the equation, we have \[a\left( {{x^2} + \frac{b}{a}x + \frac{c}{a}} \right) = 0\] \[{\left( {x + \frac{b}{{2a}}} \right)^2} - \frac{{{b^2}}}{{4{a^2}}} + \frac{c}{a} = 0\] \[{\left( {x + \frac{b}{{2a}}} \right)^2} = \frac{{{b^2} - 4ac}}{{4{a^2}}} \]\[\color{blue}{\Rightarrow x=\frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}}\]