That's one way of solving it. Here's another.
Either way you'll have to check for extraneous solutions.
Let the two roots be x=a and x=b,
Then we have a system of three equations and 3 unknowns:
Substitute for in
Solve for P
Substitute in
It's easy to see that 1 is a solution to
1 | 1 0 -13 12
| 1 1 -12
1 1 -12 0
So we have factored the polynomial equation as
We further factor it as
So we have four solutions for a
a=0, a=1, a=-4, a=3
although one or more may be extraneous.
Substituting a=0 in
Substituting a=1 in
Substituting a=-4 in
Substituting a=3 in
------------------
Checking:
P=0
,
So P=0 is extraneous
P=-11
,
So P=-11 is extraneous
P=64
,
P=64 is a solution, since (-4)2=16
P=-27
,
P=-27 is a solution, since 32=9
There are 2 actual solutions, P=64 and P=-27
Edwin
If one root of the equation x^2-12x-P=0 is the square of the other, then find the value of P?
The roots’ sum: ______, or 12
Let smaller root be S
Then larger root =
Since the 2 roots’ sum is 12, then we can say that:
(S + 4)(S - 3) = 0
S, or smaller root = - 4, or 3
When smaller root = - 4, then larger root = , or 16
When smaller root = 3, then larger root = , or 9
The roots’ product: ______, or – P
Roots’ product, when roots are – 4 and 16: - 4(16), or – 64
Thus, - P = - 64
, or
Roots’ product, when roots are 3 and 9 = 3(9), or 27
Thus, - P = 27
, or