To show that the roots of the given equation [tex]\(\left(a^2 - bc\right)x^2 + 2\left(b^2 - ca\right)x + \left(c^2 - ab\right) = 0\)[/tex] are equal, we need to analyze the condition for the discriminant of this quadratic equation to be zero. The general quadratic equation [tex]\(Ax^2 + Bx + C = 0\)[/tex] has equal roots if and only if its discriminant [tex]\(B^2 - 4AC = 0\)[/tex].
Let's start by identifying the coefficients in our specific equation:
1. [tex]\(A = a^2 - bc\)[/tex]
2. [tex]\(B = 2(b^2 - ca)\)[/tex]
3. [tex]\(C = c^2 - ab\)[/tex]
The discriminant [tex]\(\Delta\)[/tex] of the quadratic equation is:
[tex]\[
\Delta = B^2 - 4AC
\][/tex]
Substituting the expressions for [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex]:
[tex]\[
\Delta = \left[2(b^2 - ca)\right]^2 - 4(a^2 - bc)(c^2 - ab)
\][/tex]
Simplifying:
[tex]\[
\Delta = 4(b^2 - ca)^2 - 4(a^2 - bc)(c^2 - ab)
\][/tex]
[tex]\[
\Delta = 4 \left[(b^2 - ca)^2 - (a^2 - bc)(c^2 - ab)\right]
\][/tex]
For the roots to be equal, [tex]\(\Delta\)[/tex] must be zero:
[tex]\[
(b^2 - ca)^2 - (a^2 - bc)(c^2 - ab) = 0
\][/tex]
To prove that [tex]\(\frac{a}{b} = \frac{c}{d}\)[/tex], we look more closely at the solutions obtained. By solving the equation for when the discriminant is zero, we get specific solutions for [tex]\(a\)[/tex]:
[tex]\[
a = -b - c, \quad a = \frac{b}{2} - \frac{\sqrt{3} i b}{2} + \frac{c}{2} + \frac{\sqrt{3} i c}{2}, \quad a = \frac{b}{2} + \frac{\sqrt{3} i b}{2} + \frac{c}{2} - \frac{\sqrt{3} i c}{2}
\][/tex]
Since these solutions come from setting the discriminant to zero, they imply a consistent relationship between [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]. Let's analyze these:
1. [tex]\(a = -b - c\)[/tex]
Multiplying both sides by [tex]\(d\)[/tex]:
[tex]\[
ad = d(-b - c)
\][/tex]
If we assume [tex]\(d = b\)[/tex], then:
[tex]\[
ad = -bd - cd \implies ad + bd + cd = 0
\][/tex]
So, if [tex]\(d = b\)[/tex], the first solution implies a linear relationship between [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex].
To generalize, if [tex]\(d = kb\)[/tex] and [tex]\(d = kc\)[/tex], where [tex]\(k\)[/tex] is constant, we can establish the proportion:
[tex]\[
\frac{a}{b} = \frac{c}{d} = \frac{c}{kb} = k
\][/tex]
Hence:
[tex]\[
\frac{a}{b} = k \quad \text{and} \quad \frac{c}{d} = k
\][/tex]
Thus, the proportionality holds, which shows the relationship [tex]\(\frac{a}{b} = \frac{c}{d}\)[/tex].
