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Till now, we have come across and solved various types of linear equations. However, solving quadratic equations requires a different method. Before learning how to solve quadratic equations, we should be able to identify the quadratic equations. Let us learn that with the help of this video.

Consider the equations

In equation (1), variable occurs only in second degree. The quadratic equations involving a variable only in second degree is a Pure quadratic equation.

Thus, an equation that can be expressed in the form , where a and c are real numbers and a ≠ 0 is a pure quadratic equation.

In equation (2) , variable occurs both in second and first degree. The quadratic equation involving a variable in both second and first degree is an Adfected quadratic equation.

Thus, an equation that can be expressed in the form , where a, b and c are real numbers and a ≠ 0 is an Adfected quadratic equation.

Now that you have learnt to identify a quadratic equation, let us look at an example in the given video to get a better idea.

Roots of a given quadratic equation are the values of the variables which satisfy that quadratic equation.

For example, consider the quadratic equation x2 – 8x + 15 = 0.

Here, L.H.S. =  x2 – 8x + 15 and R.H.S. = 0

Let us substitute the different values of variable x in this equation and observe the results.

When x = 1 then x2 – 8x + 15 = 12 – 8(1) + 15 = 1 – 8 + 15 = 8    ∴ L.H.S. ≠ R.H.S.

When x = –1 then x2 – 8x + 15 = (–1)2 – 8(–1) + 15 = 1 + 8 + 15 = 24    ∴ L.H.S. ≠ R.H.S.

When x = 3 then x2 – 8x + 15 = 32 – 8(3) + 15 = 9 – 24 + 15 = 0    ∴ L.H.S. = R.H.S.

When x = 4 then x2 – 8x + 15 = 42 – 8(4) + 15 = 16 – 32 + 15 = –1    ∴ L.H.S. ≠ R.H.S.

When x = 5 then x2 – 8x + 15 = 52 – 8(5) + 15 = 25 – 40 + 15 = 0    ∴ L.H.S. = R.H.S.

When x = –5 then x2 – 8x + 15 = (–5)2 – 8(–5) + 15 = 25 + 40 + 15 = 80    ∴ L.H.S. ≠ R.H.S.

It can be observed that for x = 3 and x = 5, we have L.H.S. = R.H.S. which means that the equation x2 – 8x + 15 = 0 is satisfied. Thus, 3 and 5 are roots of this quadratic equation.

In other words, we can say that x = 3 and x = 5 are the solutions of quadratic equation x2 – 8x + 15 = 0.

Also, for x = 1, x = –1, x = 4 and x = –5, we have L.H.S. ≠ R.H.S. which means that the equation x2 – 8x + 15 = 0 is not satisfied. Thus, 1, –1, 4 and –5 are not the roots or solution of this quadratic equation.

Let us go through few examples to understand the concept better.

Example 1:

Is the equation  a quadratic equation?

Solution:

The given equation is

We can also write it as

Hence, we can say that th...

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