Find the zeroes of the polynomial …
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Khushboo Kumari 5 years, 6 months ago
- 2 answers
Test
Related Questions
Posted by Hari Anand 5 months, 2 weeks ago
- 0 answers
Posted by Parinith Gowda Ms 2 months, 3 weeks ago
- 1 answers
Posted by Kanika . 1 week ago
- 1 answers
Posted by Vanshika Bhatnagar 1 year, 3 months ago
- 2 answers
Posted by Sahil Sahil 1 year, 3 months ago
- 2 answers
Posted by Parinith Gowda Ms 2 months, 3 weeks ago
- 0 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Rajan Kumar Pasi 5 years, 6 months ago
{tex}\huge {=>\ x = {-b \pm \sqrt{b^2-4ac} \over 2a}}\\ \huge {=>\ x = {-(-2) \pm \sqrt{(-2)^2-4(1)(8)} \over 2\times 1}}\\ \huge {=>\ x = {2 \pm \sqrt{4-32} \over 2}}\\ \huge {=>\ x = {2 \pm \sqrt{-28} \over 2}}\\ \huge {=>\ x = {1 \pm \sqrt{-7} }}\\ \text{Since,} \huge{ {b^2-4ac}}\ is\ smaller\ than\ 0.\ Thus,\ no\ real\ roots\ or\ zeros\ exist.{/tex}
0Thank You