28 Apr Quadratic Equations Formula
At Miracle Learning Centre, we teach you interesting stuff in maths tuition. We offer secondary maths tuition, primary mathematics tuition, JC maths tuition in Singapore. Today, we are going to take a look at quadratic equations. Quadratic equations are a bit different from linear equations and they have some unique properties. In this maths tuition, we are going to learn about quadratic equations and see how to solve them. In order to understand the topics discussed in this maths tuition, you must already be familiar with basic algebra as well as linear equations. Without any further delay, let us get started straight away.
Quadratic equations take the following form:
ax2+ bx + c = 0
The term ax2 which involves the 2nd power of x makes all the difference and distinguishes these types of equations from other equations.
Some examples are quadratic equations are as follows:
2×2 + 3x + 6 = 0 for which a=2, b=3 and c=6
5×2-3x+8=0 for which a=5,b=-3 and c=8
7×2 + 3x -19 =0 for which a=7, b=3 and c =-19
When a=0, the equation will become a linear equation because the term ax2 will no longer be present. As a result, a cannot be 0 in any quadratic equation.
Solving quadratic equations
In this part of this math tuition, we are going to see how we can solve quadratic equations for the unknown value “x”. The technique was first shown by Sridhara Acharya – an Indian mathematician in the 9th century.
So, consider the standard form of the quadratic equation which goes as follows:
ax2+ bx + c = 0
Now, we multiply both sides by 4a:
4a2x2+ 4abx + 4ac = 0
4a2x2+ 4abx = – 4ac
Now, we add b2 to both sides:
4a2x2+ 4abx +b2 = – 4ac +b2
(2ax )2 + 2.(2ax).(b) + b2 = – 4ac +b2
(2ax+b)2 = b2 – 4ac
The term on the right side of the equation b2-4ac is known as the discriminant.
By continuing with the above equation, we get:
2ax+b = ±√(b^2-4ac)
2ax = -b ± √(b^2-4ac)
x = (-b± √(b^2-4ac))/2a
This formula is also known as the sridharacharya formula.
The ± sign means that there can be two solutions:
x = (-b+ √(b^2-4ac))/2a or x = (-b- √(b^2-4ac))/2a
Let us see with an example. Consider the following equation:
x2 – 5x – 6 =0
Those of you who are familiar with factoring quadratic equations will be able to use the technique of middle-term factorization to solve for x. However, in this math tuition, we are going to see the alternative way of solving.
So, for the above equation, a = 1, b= -5 and c =-6
By plugging these values in to our above formula:
x = (-(-5)± √(〖(-5)〗^2-4.(1).(-6)))/(2.(1)) = (5± √(25+24))/2 =(5± 7)/2
As mentioned above, the ± means there can be two values.
So from the above, we get :
x = (5+ 7)/2 or x =(5- 7)/2
x = 6 or x = -1
The solutions 6 and -1 are also known as the roots of the quadratic equation. As we can in this case, there are two roots. The discriminant plays an extremely important role and from the discriminant of a quadratic equation, we could predict some properties of the roots although we might not know the exact roots. However, we are going to discuss that in another article. In case you find any difficulty in understanding any of the topics discussed above, do not hesitate to take help from your teachers at math tuition or your peers. We hope you enjoyed this math tuition from Miracle Learning Centre. Look out for the next math tuition from Miracle Learning Centre.