SOLUTION: For what value of {{{ k }}} does {{{ P(x) }}} have exactly one root where that root is real?
{{{ P(x) = x^3 - 6x^2 + kx -8 }}}
Please show working. Thanks.
Algebra ->
Polynomials-and-rational-expressions
-> SOLUTION: For what value of {{{ k }}} does {{{ P(x) }}} have exactly one root where that root is real?
{{{ P(x) = x^3 - 6x^2 + kx -8 }}}
Please show working. Thanks.
Log On
You can put this solution on YOUR website! set k = 13 and try x = 1
Solve for x over the real numbers:
x^3 -6x^2 +13x -8 = 0
1 -6 +13 -8 = 0
now use synthetic division to find the other factor, (x-1) being one factor
1 | 1 -6 13 -8 |
| 1 -5 8 |
| 1 -5 8 0 |
Our two factors are:
(x-1) (x^2 -5x +8)
Split into two equations:
x-1 = 0 or x^2 -5x +8 = 0
Add 1 to both sides:
x = 1 or
x^2 -5x +8 = 0
Subtract 8 from both sides:
x = 1 or x^2 -5x = -8
Add 25/4 to both sides:
x = 1 or x^2 -5x +25/4 = -7/4
Write the left hand side as a square:
(x -5/2)^2 = -7/4
(x-5/2)^2 = -7/4 has no solution since for all x on the real line, (x-5/2)^2 >=0 and -7/4<0:
Answer: x = 1 and k = 13
