Does $(x-1)^{2}+2(x+1)=0$ have a real root? Justify your answer.
(NO) To determine if the equation has real roots,we first simplify it:
$(x-1)^{2}+2(x+1)=0$
$(x^{2}-2x+1)+2x+2=0$
$x^{2}+3=0$
Now,we compare this with the standard quadratic form $ax^{2}+bx+c=0$,where $a=1, b=0, c=3$.
The discriminant $D$ is given by $D = b^{2}-4ac$.
$D = (0)^{2}-4(1)(3) = -12$.
Since the discriminant $D < 0$,the equation has no real roots.
$(x-1)^{2}+2(x+1)=0$
$(x^{2}-2x+1)+2x+2=0$
$x^{2}+3=0$
Now,we compare this with the standard quadratic form $ax^{2}+bx+c=0$,where $a=1, b=0, c=3$.
The discriminant $D$ is given by $D = b^{2}-4ac$.
$D = (0)^{2}-4(1)(3) = -12$.
Since the discriminant $D < 0$,the equation has no real roots.
Explore More
Similar Questions
If the following quadratic equation has two equal and real roots,then find the value of $k$: $x^{2} - kx + 25 = 0$.
Medium
View SolutionState whether the quadratic equation $(x+1)(x-2)+x=0$ has two distinct real roots. Justify your answer.
Medium
View SolutionSolve the equation $x^{2}-8x-21=0$ using the quadratic formula.
Difficult
View SolutionFind the discriminant of the following quadratic equation: $4x^{2} - 12x + 9 = 0$.
Easy
View SolutionState whether the quadratic equation $\sqrt{2} x^{2}-\frac{3}{\sqrt{2}} x + \frac{1}{\sqrt{2}} = 0$ has two distinct real roots. Justify your answer.
Difficult
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