Question

Find the value of a and b for which `x=3/4`and `x =-2` are the roots of the equation `ax^2+bx-6=0` 

 

Answer 1

It is given that `3/4` is a root of ax^2+bx-6=0; therefore, we have: 

`axx(3/4)^2+bxx3/4-6=0` 

`⇒(9a)/16+(3b)/4=6` 

`⇒ (9a+12b)/16=6` 

`⇒ 9a+12b-96=0` 

`⇒3a+4b=32`                         ........(i)  

Again, (-2) is a root of ax^2+bx-6=0; therefore, we have: 

`axx(-2)^2+bxx(-2)-6=0`

`⇒ 4a-2b=6` 

`⇒2a-b=3`                                  ..................(2) 

On multiplying (ii) by 4 and adding the result with (i), we get: 

`⇒3a+4b+8a-4b=32+12` 

`⇒11a=44` 

`⇒a=4` 

Putting the value of a in (ii), we get: 

`2xx4-b=3` 

`⇒8-b=3 `

`⇒ b=5` 

Hence, the required values of a and b are 4 and 5, respectively.