If p, q are zeroes of …
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Neha Neha 6 years, 5 months ago
- 1 answers
Test
Related Questions
Posted by Hari Anand 6 months, 1 week ago
- 0 answers
Posted by Parinith Gowda Ms 3 months, 2 weeks ago
- 1 answers
Posted by Vanshika Bhatnagar 1 year, 4 months ago
- 2 answers
Posted by Kanika . 1 month ago
- 1 answers
Posted by Sahil Sahil 1 year, 4 months ago
- 2 answers
Posted by Parinith Gowda Ms 3 months, 2 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
Sia ? 6 years, 5 months ago
According to the question, p and q are the zeroes of polynomial f(x) = 2x2 - 7x + 3,
f(x) = 2x2 -7x + 3
Sum of roots = p + q = {tex}- \frac { \text { Coefficient of } x } { \text { Coefficient of } x ^ { 2 } }{/tex}
{tex}= - \left( \frac { - 7 } { 2 } \right) = \frac { 7 } { 2 }{/tex}
Product of roots= pq {tex}= \frac { \text { Constant term } } { \text { Coefficient of } x ^ { 2 } } = \frac { 3 } { 2 }{/tex}
Since, (p + q)2 =p2 + q2 + 2pq
So, p2 + q2 = (p + q)2 - 2pq
{tex}= \left( \frac { 7 } { 2 } \right) ^ { 2 } - 3 = \frac { 49 } { 4 } - \frac { 3 } { 1 } = \frac { 37 } { 4 }{/tex}
Hence, the value of p2+ q2={tex}\frac{37}{4}{/tex}
0Thank You