Without solving examine the nature of roots of equation ×2-6x+9=0
Answers
Answered by
16
Heya mate!!
Here's your answer!!
For examining the nature of the roots of an equation, we should find it's discriminant(D).
If D = 0, roots are real and equal
If D > 0, roots are real
If D < 0, roots are imaginary
In this equation,
a = 1, b = -6 and c = 9
D = b² - 4ac
= (-6)² - 4 × 1 × 9
= 36 - 36
= 0
Therefore we can conclude that the roots of the quadratic equation x² - 6x + 9 are real and equal.
Hope it helps you!!
Cheers ☺☺
Mark as Brainliest if it helps!!
#foreverJungkook
Here's your answer!!
For examining the nature of the roots of an equation, we should find it's discriminant(D).
If D = 0, roots are real and equal
If D > 0, roots are real
If D < 0, roots are imaginary
In this equation,
a = 1, b = -6 and c = 9
D = b² - 4ac
= (-6)² - 4 × 1 × 9
= 36 - 36
= 0
Therefore we can conclude that the roots of the quadratic equation x² - 6x + 9 are real and equal.
Hope it helps you!!
Cheers ☺☺
Mark as Brainliest if it helps!!
#foreverJungkook
Answered by
1
Answer:
x² - 6x+ 9 =0
Step-by-step explanation:
nature of the roots of the quadratic equation
Similar questions