Find the discriminant of the following quadratic equation and hence determine the nature of the roots of the equation: $x(x - 5) = 36$.
(D) The given equation is $x(x - 5) = 36$.
Expanding the equation: $x^2 - 5x = 36$.
Rearranging into the standard form $ax^2 + bx + c = 0$: $x^2 - 5x - 36 = 0$.
Here,$a = 1$,$b = -5$,and $c = -36$.
The discriminant $D$ is given by the formula $D = b^2 - 4ac$.
Substituting the values: $D = (-5)^2 - 4(1)(-36)$.
$D = 25 + 144 = 169$.
Since $D > 0$ and $169$ is a perfect square,the roots are real,rational,and distinct.
Expanding the equation: $x^2 - 5x = 36$.
Rearranging into the standard form $ax^2 + bx + c = 0$: $x^2 - 5x - 36 = 0$.
Here,$a = 1$,$b = -5$,and $c = -36$.
The discriminant $D$ is given by the formula $D = b^2 - 4ac$.
Substituting the values: $D = (-5)^2 - 4(1)(-36)$.
$D = 25 + 144 = 169$.
Since $D > 0$ and $169$ is a perfect square,the roots are real,rational,and distinct.
Explore More
Similar Questions
Which of the following is a quadratic equation?
Medium
View SolutionThe roots of $4x^{2} + 2x + \frac{1}{4} = 0$ are $\ldots \ldots \ldots \ldots.$
Easy
View SolutionSolve the following equation using the method of factorization: $(x+2)^{2}-14(x+2)+45=0$
Easy
View SolutionIf one root of $x^{2}-5x+6=0$ is also the root of $x^{2}+3x+c=0$,then $c = \ldots$
Medium
View SolutionObtain the roots of the following quadratic equation by using the general formula for the solution: $2x^{2} - 2\sqrt{2}x + 1 = 0$.
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo