Question

What is the discriminant of $- 9{x^2} + 10x = - 2x + 4$ and what does that mean?

Answer 1

Let us consider the given quadratic equation as $P(x) = a{x^2} + bx + c = 0$, where $a,b,c$ are real numbers with leading coefficient $a \ne 0$.
Discriminant Rule: Discriminant is the nature of the roots by finding or knowing whether the given quadratic equation has real, equal, unequal or imaginary complex roots.
If ${b^2} - 4ac = 0$, then the quadratic equation has real and equal roots.
If ${b^2} - 4ac < 0$, then the quadratic equation has an imaginary complex root.
If ${b^2} - 4ac > 0$, then the quadratic equation has real and unequal (distinct) roots.

Complete step-by-step solution:
From the given equation, we need to find a discriminant of $- 9{x^2} + 10x = - 2x + 4$.
By simplifying the terms on both the sides, we get
$- 9{x^2} + 10x + 2x - 4 = 0$
Now add the like terms of $x$ we get,
$- 9{x^2} + 12x - 4 = 0$
From the above quadratic equation we find the discriminant, ${b^2} - 4ac$
And here we have from the given information $a = - 9\,,\,\,b = 12\,,\,\,c = - 4$
Now, substitute the value of $a,b,c$ in ${b^2} - 4ac$ we get,
${b^2} - 4ac = {\left( {12} \right)^2} - 4 \times \left( { - 9} \right) \times \left( { - 4} \right)$
We simplify the above equation as follows.
${b^2} - 4ac = 144 - 4 \times \left( {36} \right)$
We again, simplify the above equation as
${b^2} - 4ac = 144 - 144 = 0$.
Therefore, we can conclude that the discriminant of $- 9{x^2} + 10x = - 2x + 4$ is ${b^2} - 4ac = 0$, that is the roots are real and equal.

Note: While finding the nature of the roots by discriminant method, we should always remember that we should avoid the square root in the calculations. Also recall that $\sqrt { - 1}$ is not defined in the real numbers. And if we get the value $\sqrt { - 1}$,it is called as imaginary units in complex numbers.
And it is denoted as $\sqrt { - 1} = i$ , we define the complex number as \mathbb{C} = \left\\{ {z = x + iy:x,y \in \mathbb{R}} \right\\} in which the equation ${x^2} + 1 = 0$is satisfied when we substitute $x = \sqrt { - 1}$or ${i^2} = - 1$.