Question 925613
If one root of the equation x^2-12x-P=0 is the square of the other, then find the value of P?
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{{{x^2 - 12x - P = 0}}} 
The roots’ sum: {{{s = - b/a}}}______{{{s = -(- 12)/1}}}, or 12

Let smaller root be S
Then larger root = {{{S^2}}}
Since the 2 roots’ sum is 12, then we can say that: {{{S + S^2 = 12}}}
{{{S^2 + S - 12 = 0}}}
(S + 4)(S - 3) = 0
S, or smaller root = - 4, or 3 

When smaller root = - 4, then larger root = {{{(- 4)^2}}}, or 16
When smaller root = 3, then larger root = {{{3^2}}}, or 9

The roots’ product: {{{p = c/a}}}______{{{p = (- P)/1}}}, or – P
Roots’ product, when roots are – 4 and 16: - 4(16), or – 64
Thus, - P = - 64
{{{P = (- 64)/(- 1)}}}, or {{{highlight_green(64)}}}

Roots’ product, when roots are 3 and 9 = 3(9), or 27
Thus, - P = 27
{{{P = 27/(- 1)}}}, or {{{highlight_green(- 27)}}}