Question

Explain real roots with examples?

Answer 1

Hint: We have to explain real roots. We know that whenever we solve a linear or a quadratic equation, we get the value variable of the equation or in other words, we find the solution of the equation. It is this ‘solution’ that we call real roots. For example – for the equation \[{{x}^{2}}-7x+12=0\], on solving it, we have the real roots as 3 and 4.

Complete step-by-step solution:
According to the given question, we have to explain about real roots.
Let us suppose we are given an equation with one variable (either linear or quadratic) and we are asked to solve it. When we solve that equation, we get the value of that variable which we call the solution of that equation. It is about this solution that we talk about as real roots.
For example – we have a quadratic equation \[{{x}^{2}}-7x+12=0\] (let’s say)
We will now find the discriminant, D which is,
\[D={{b}^{2}}-4ac\]
\[\Rightarrow D={{(-7)}^{2}}-4(1)(12)\]
\[\Rightarrow D=49-48\]
\[\Rightarrow D=1>0\]
Since, the discriminant gives a value greater than 0, so there exists real roots.
\[{{x}^{2}}-7x+12=0\]
Sum, ‘b’ = -7, product of the terms, ‘ac’ = 12
We need factors which when added and multiplied give the above values.
\[12=2\times 6=3\times 4\]
We will now take 3 and 4, we have
\[\Rightarrow {{x}^{2}}-(3+4)x+12=0\]
Now, opening the bracket, we have,
\[\Rightarrow {{x}^{2}}-3x-4x+12=0\]
Taking the common terms out from each pair, we get,
\[\Rightarrow x(x-3)-4(x-3)=0\]
Now, we will take \[(x-3)\] as common and we get,
\[\Rightarrow (x-4)(x-3)=0\]
\[\Rightarrow x=4,3\]
We get the real roots as \[x=4,3\].

Note: The discriminant comes in handy when we have to check if real roots exist in an equation or not. We have,
\[D={{b}^{2}}-4ac\]
When \[D>0\], then two real roots exist
When \[D=0\], then one real root exist
And when \[D<0\], then we get two imaginary roots.
Imaginary roots are represented using the iota (\[i\]) symbol.