6r^2+r-12=0?

3 Answers
Feb 23, 2018

r=4/3,-3/2

Explanation:

We got: 6r^2+r-12=0

Using the quadratic formula, the roots of a quadratic equation (ax^2+bx+c=0) are given by:

x=(-b+-sqrt(b^2-4ac))/(2a)

So here, we got a=6,b=1,c=-12, and so

r=(-1+-sqrt(1-4xx6xx(-12)))/12

r=(-1+-sqrt(1+288))/12

r=(-1+-sqrt(289))/12

r=(-1+-17)/12

r=cancel16^4/cancel12^3,cancel(-18)^-3/cancel12^2

r=4/3,-3/2

So, the roots of the equation 6r^2+r-12=0 are: r=4/3 and r=-3/2.

Feb 23, 2018

r = 4/3 , -3/2

Explanation:

6r^2 + r - 12 =0

Use the splitting the middle term method.

6r^2 + 9r -8r-12 =0

3r(2r +3) -4(2r+3) = 0

(3r-4)(2r+3) =0

So , 3r -4=0

3r =4

r=4/3

Or

2r+3 =0

2r =-3

r=-3/2

Feb 23, 2018

See a solution process below: r = -3/2 and r = 4/3

Explanation:

If you are trying to solve for r you can use the quadratic equation to solve this problem:

The quadratic formula states:

For color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0, the values of x which are the solutions to the equation are given by:

x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))

Substituting:

color(red)(6) for color(red)(a)

color(blue)(1) for color(blue)(b)

color(green)(-12) for color(green)(c) gives:

r = (-color(blue)(1) +- sqrt(color(blue)(1)^2 - (4 * color(red)(6) * color(green)(-12))))/(2 * color(red)(6))

r = (-1 +- sqrt(1 - (-288)))/12

r = (-1 +- sqrt(1 + 288))/12

r = (-1 +- sqrt(289))/12

r = (-1 - 17)/12 and r = (-1 + 17)/12

r = -18/12 and r = 16/12

r = -3/2 and r = 4/3