Let’s learn quadratic equation definition

The generalized form of quadratic equation is ay2 + by + c = 0, where a,b and c are constants with a being non-zero. This gives us only one unknown variable y and hence quadratic equations are also called 'uni-variate' as per the quadratic equation definition. Since they are of second order, they have two real or complex solutions. There are many ways of solving quadratic equations. Some popular methods are solving by quadratic formula, factorizing, graphing and completion by square. We will see the quadratic equation graph method in detail.

Method of plotting quadratic equation graphs

The graph of quadratic equations are always parabolic in shape. Hence even though they can be plotted like linear graphs, we need to take some more points into consideration. Let us consider an example quadratic equation and plot it on the graph thereby finding the roots. Consider the equation y = x2 – 3x + 2. Let us consider values of x from -3 to +3 and see whether the graph touches x-axis or not. The graph has to touch the x-axis at minimum 1 point and maximum 2 points. This point becomes the roots of quadratic equation. The values of equation for different values of x are given in the table below.

 x -3 -2 -1 0 1 2 3 y 20 12 6 2 0 0 2

Here two points x=1 and x=2 have corresponding y-axis values zero. Hence they are the roots of the equation. Let us plot a graph for the same. Thus the graph comes out to be parabolic in shape as shown. Now that we have solved it by graph, let us see related quadratic equation word problems.

Check out quadratic equation word problems

The word problem helps us to understand practical application of quadratic equations. An example is given below to have the practical idea.

(i) An object is falling from 27.3 m tall building with a speed of 18.2 m/s. The height after the launch follows an equation h(t) = -9.1t2 + 18.2t + 27.3. Here h is in meters and t in seconds. In how many seconds, the object will fall on the ground?

When object touches the ground, height becomes 0, i.e.

0 = -9.1t2 + 18.2t + 27.3

0 = t2 – 2t – 3 i.e. (t + 1)(t – 3) = 0. This gives two different values for t, t=-1 and t=3. The time can't be negative. Hence t = 3 or the object will touch the ground in 3 seconds.

 More topics in Quadratic Equations Quadratic Formula Quadratic Formula Proof Multiplying Negative Exponents
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