Quadratic Equations

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.

Now, let us learn about the roots of a quadratic equation.

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

For example, consider the quadratic equation *x*^{2} – 8*x* + 15 = 0.

Here, L.H.S. = *x*^{2} – 8*x* + 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 *x*^{2} – 8*x* + 15 = 1^{2} – 8(1) + 15 = 1 – 8 + 15 = 8 ∴ L.H.S. ≠ R.H.S.

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

When *x* = 3 then *x*^{2} – 8*x* + 15 = 3^{2} – 8(3) + 15 = 9 – 24 + 15 = 0 ∴ L.H.S. = R.H.S.

When *x* = 4 then *x*^{2} – 8*x* + 15 = 4^{2} – 8(4) + 15 = 16 – 32 + 15 = –1 ∴ L.H.S. ≠ R.H.S.

When *x* = 5 then *x*^{2} – 8*x* + 15 = 5^{2} – 8(5) + 15 = 25 – 40 + 15 = 0 ∴ L.H.S. = R.H.S.

When *x* = –5 then *x*^{2} – 8*x* + 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 *x*^{2} – 8*x* + 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 *x*^{2} – 8*x* + 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 *x*^{2} – 8*x* + 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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