If the roots of the given equation (c^2 - ab)x^2 - 2(a^2 - bc)x + (b^2 - ac) = 0 are real and equal, then show/prove that either a = (a + b + c) = 0 or a = b = c
You should remember that the condition for the roots of a quadratic equation `mx^2 + px + q = 0`to be real and equal.
`Delta = p^2 - 4mq = 0`
You need to identify the coefficients of the given quadratic such that:
`m = c^2 - ab, p = - 2(a^2 - bc), q = (b^2 - ac)`
Substituting these coefficients in formula representing `Delta` yields:
`Delta = 4(a^2 - bc) - 4(c^2 - ab)(b^2 - ac)`
Factoring out 4 yields:
`Delta = 4(a^2 - bc - c^2b^2 + ac^3 + ab^3 - a^2bc)`
Since `Delta = 0 => 4(a^2 - bc - c^2b^2 + ac^3 + ab^3 - a^2bc) = 0 => (a^2 - bc - c^2b^2 + ac^3 + ab^3 - a^2bc) = 0.`
This equation that relates the coefficients expresses the condition for the two roots to be equal.
Considering `a=b=c` , you need to check if the condition holds substituting a for `b` and c, such that:
`a^2 - a*a - a^2*a^2 + a*a^3 + a*a^3 - a^2*a*a = 0`
`a^2 - a^2 - a^4 + a^4 + a^4 - a^4 = 0`
Reducing like terms yields:
Considering `a=b=c ` and substituting a for b and c in equation that expresses the condition for the roots to be equal, the condition holds, hence, the supposition that `a=b=c ` is valid.