Question

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

Answer 1

Hint: 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 \[{i^2} = - 1\].