Question

The difference between the two roots of the equation 3x^2+10x+c=0 is 4 2/3 . Find the solutions for the equation.

Answer 1

Given the equation:
`3x^2+10x+c =0`

A quadratic equation is in the form:
`ax^2+bx+c = 0` where a, b ,c are the coefficient and a≠0 then the solution is given by :

`x_(1,2) = (-b\pm √(b^2-4ac))/(2a)` ......[1]

On comparing with given equation we get;

a =3 , b = 10

then, substitute these in equation [1] to solve for c;

`x_(1,2) = (-10\pm √(10^2-4\cdot 3 \cdot c))/(2 \cdot 3)`

Simplify:

`x_(1,2) = (-10\pm √(100- 12c))/(6)`

Also, it is given that the difference of two roots of the given equation is
`4(2)/(3) = (14)/(3)`

i.e,

`x_1 -x_2 = (14)/(3)`

Here,

`x_1 = (-10 + √(100- 12c))/(6)` , ......[2]

`x_2= (-10 - √(100- 12c))/(6)` .....[3]

then;

`(-10 + √(100- 12c))/(6) - ((-10 + √(100- 12c))/(6)) = (14)/(3)`

simplify:

`(2 √(100- 12c) )/(6) = (14)/(3)`

or

`√(100- 12c) = 14`

Squaring both sides we get;

`100-12c = 196`

Subtract 100 from both sides, we get

`100-12c -100= 196-100`

Simplify:

-12c = -96

Divide both sides by -12 we get;

c = 8

Substitute the value of c in equation [2] and [3]; to solve
`x_1 , x_2`

`x_1 = (-10 + √(100- 12\cdot 8))/(6)`

or

`x_1 = (-10 + √(100- 96))/(6)` or

`x_1 = (-10 + √(4))/(6)`

Simplify:

`x_1 = (-4)/(3)`

Now, to solve for
`x_2` ;

`x_2 = (-10 - √(100- 12\cdot 8))/(6)`

or

`x_2 = (-10 - √(100- 96))/(6)` or

`x_2 = (-10 - √(4))/(6)`

Simplify:

`x_2 = -2`

therefore, the solution for the given equation is:
`-(4)/(3)` and -2.