# What is the value of a and b if the roots of ax^2 + bx + 2 = 0 is 8 and 6.

*print*Print*list*Cite

You need to remember that substituting the value of root for x in equation yields 0 such that:

`f(8)=0 =gt a*8^2 + b*8 + 2 = 0 =gt 64a + 8b + 2 = 0`

You need to divide by 2 both sides such that:

`32a + 4b = -1`

`f(6) = a*6^2 + b*6 + 2 = 0 =gt 36a + 6b + 2 = 0`

You need to divide by 2 both sides such that:

`18a + 3b = -1`

You need to solve simultaneous equations using elimination, hence you should multiply the first equation by -3 and the second equation by 4 such that:

`-3*(32a + 4b) = 3 =gt -96a - 12b =3`

`4(18a + 3b ) = -4 =gt 72a + 12b = -4`

Adding these equations yields:

`-96a - 12b + 72a + 12b= 3 - 4 =gt -24a = -1 =gt a = 1/24`

Substituting `1/24` for a in equation `18a + 3b = -1` yields:

`18/24 + 3b = -1 =gt 3b = -1 - 18/24 =gt b = -42/(3*24)`

`b = -21/(3*12) =gt b = -7/12`

**Hence, evaluating the values of coefficients a and b under given conditions yields: `x^2/24 - (7x)/12 + 2 = 0` **

The roots of the quadratic equation ax^2 + bx + 2 = 0 are 8 and 6

(x - 8)(x - 6) = ax^2 + bx + 2

=> x^2 - 14x + 48 = ax^2 + bx + 2

=> (1/24)x^2 - (14/24)x + 2 = ax^2 + bx + 2

=> a = 1/24 and b = 14/24

**The value of a and b for which the roots of ax^2 + bx + 2 = 0 are 8 and 6 is a = 1/24 and b = -14/24**