Find the value. Of a and …
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Test
Related Questions
Posted by Lakshay Kumar 1 year, 1 month ago
- 0 answers
Posted by Parinith Gowda Ms 4 months, 1 week ago
- 1 answers
Posted by Vanshika Bhatnagar 1 year, 5 months ago
- 2 answers
Posted by Parinith Gowda Ms 4 months, 1 week ago
- 0 answers
Posted by Hari Anand 7 months ago
- 0 answers
Posted by Sahil Sahil 1 year, 5 months ago
- 2 answers
Posted by Kanika . 1 month, 4 weeks ago
- 1 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, 7 months ago
We have the equation,
{tex}ax^2+bx-6=0{/tex}
Putting x = {tex}\frac{3}{4}{/tex} , we get
{tex}\therefore{/tex} a({tex}\frac{3}{4}{/tex})2 + b({tex}\frac{3}{4}{/tex}) - 6 = 0
{tex}\Rightarrow{/tex} a{tex}\frac{9}{16}{/tex} +b({tex}\frac{3}{4}{/tex}) - 6 = 0
{tex}\Rightarrow{/tex} 9a + 12b - 96 = 0
{tex}\Rightarrow{/tex} 3a + 4b = 32 ..(i)
Putting x = -2, we get
a(-2)2 + b(-2) - 6 = 0
4a -2b = 6
2a - b = 3 ..... (ii)
Multiplying (ii) by 4 adding the result from (i), we get
11a = 44 {tex}\Rightarrow{/tex} a = 4
Putting a = 4 in (i) , we get
3(4) + 4b = 32
{tex}\Rightarrow{/tex} 4b = 32 - 12
{tex}\Rightarrow{/tex} b = {tex}\frac{20}{4}{/tex} = 5
Therefore, the required value of a = 4 and b = 5.
0Thank You