i need a lil help with this
If x=2 is a root of the equation
6x^3 - px^2 - 14x + 24=0, find p,hence, solve the equation
i got p to be 11
how do i solve the equation?
You did it.
http://www.jiskha.com/display.cgi?id=1165710846.1165713920
To solve the equation, we can use the fact that x=2 is a root. This means that when we substitute x=2 into the equation, it should equal zero.
Substituting x=2 into the equation:
6(2)^3 - p(2)^2 - 14(2) + 24 = 0
Simplifying this expression:
48 - 4p - 28 + 24 = 0
-4p + 44 = 0
-4p = -44
p = -44/-4
p = 11
So, you correctly found that p = 11.
To solve the equation, we need to factorize it or use the quadratic formula. However, this equation is a cubic equation, so we need to use a different method.
One possible method is to use synthetic division to find the remaining roots. The root x=2 means that (x-2) is a factor of the equation. We can use synthetic division to divide the equation by (x-2) and find the remaining quadratic equation.
Performing synthetic division:
2 | 6 -p -14 24
| 12 24 20
-----------------
6 12 10 44
The result of synthetic division is 6x^2 + 12x + 10, with a remainder of 44.
Now, we solve this quadratic equation: 6x^2 + 12x + 10 = 0
We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. However, upon inspection, we can see that the quadratic equation has no real roots because its discriminant (b^2 - 4ac) is negative.
Therefore, the original equation 6x^3 - px^2 - 14x + 24 = 0 has one real root x = 2 and no additional real roots.