Question

If the equation \( a(b - c)x^2 + b(c - a)x + c(a - b) = 0 \) has equal roots, where \( a + c = 15 \) and \( b = \frac{36}{5} \), then \( a^2 + c^2 \) is equal to               .

Hint:

When solving for the sum or product of squares in equations, use the identity \( (a + c)^2 = a^2 + 2ac + c^2 \) to simplify calculations.

Answer 1

The given equation is: \[ a(b - c)x^2 + b(c - a)x + c(a - b) = 0. \] We are told that the equation has equal roots, and that \( x = 1 \) is one root. The other root is also 1. Thus, the sum of the roots \( \alpha + \beta \) is: \[ \alpha + \beta = \frac{b(c - a)}{a(b - c)} = 2. \] Now, from the condition of equal roots, we have: \[ \alpha + \beta = 2 \quad \Rightarrow \quad bc + ab + ac = 2ab - 2ac. \] Simplifying further: \[ 2ac = ab + bc \quad \Rightarrow \quad 2ac = b(a + c). \] Substituting \( a + c = 15 \) and \( b = \frac{36}{5} \): \[ 2ac = 15 \times \frac{36}{5} = 108. \] Thus, \( ac = 54 \). Next, we are given \( a + c = 15 \), so: \[ a^2 + c^2 = (a + c)^2 - 2ac = 15^2 - 2 \times 54 = 225 - 108 = 117. \] Thus, the value of \( a^2 + c^2 \) is: \[ \boxed{117}. \]

Answer 2

Given quadratic \[ a(b-c)x^2 + b(c-a)x + c(a-b)=0, \] and it has equal roots with one root \(x=1\). So both roots are \(1\). 

Sum of roots \(=1+1=2\). By comparing with \(\displaystyle -\frac{\text{coefficient of }x}{\text{coefficient of }x^2}\), \[ \frac{b(c-a)}{a(b-c)}=2. \] Rearranging gives \[ b(c-a)=2a(b-c)\;\Rightarrow\; bc-ba = 2ab-2ac. \] Bringing terms together, \[ 2ac = ab + bc = b(a+c). \]

Using \(a+c=15\) and \(b=\dfrac{36}{5}\), \[ 2ac = b(a+c)=\frac{36}{5}\cdot 15 =108, \] so \(ac=54\).

Finally, \[ a^2+c^2=(a+c)^2-2ac=15^2-2\cdot54=225-108=117. \]

Answer:

\(\boxed{117}\)