X2-(2b-1)x+(b2-b-20)=0 find the value of x
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Sakshi Bachles 6 years, 5 months ago
- 1 answers
Test
Related Questions
Posted by Hari Anand 6 months, 1 week ago
- 0 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
- 1 answers
Posted by Vanshika Bhatnagar 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
In the given quadratic equation, x2 - (2b - 1 ) x + (b2- b - 20) = 0
A= +1, B= - (2b - 1 ), C= (b2- b - 20)
{tex}x = {-B \pm \sqrt{B^2-4AC} \over 2A}{/tex}
{tex}x = \frac { ( 2 b - 1 ) \pm \sqrt { ( 2 b - 1 ) ^ { 2 } - 4 \left( b ^ { 2 } - b - 20 \right) } } { 2 }{/tex}
{tex}x = \frac { ( 2 b - 1 ) \pm 9 } { 2 }{/tex}
{tex}x = \frac { 2 b + 8 } { 2 } \quad and \quad x = \frac { 2 b - 10 } { 2 }{/tex}
{tex}\therefore x= b+4 \quad and \quad x=b-5{/tex}
0Thank You