Quadratic Equations – Sum of roots – 23 January 2025 (Shift 1)

Solution:

Observe the coefficients: $A = a(b-c)$, $B = b(c-a)$, $C = c(a-b)$.
Sum of coefficients: $A+B+C = ab-ac + bc-ab + ac-bc = 0$.
Since the sum of coefficients is 0, one root is $x=1$.
Given that roots are equal, the other root is also 1.
Product of roots $\alpha \cdot \beta = 1 \cdot 1 = 1$.
Using formula $\text{Product} = \frac{C}{A}$:
$$\frac{c(a-b)}{a(b-c)} = 1 \Rightarrow ac – bc = ab – ac \Rightarrow 2ac = ab + bc$$
$$2ac = b(a+c)$$
Substitute given values $a+c=15$ and $b=\frac{36}{5}$:
$$2ac = \frac{36}{5}(15) \Rightarrow 2ac = 36(3) = 108$$
$$ac = 54$$
Now find $a^2 + c^2$:
$$a^2 + c^2 = (a+c)^2 – 2ac$$
$$a^2 + c^2 = (15)^2 – 108 = 225 – 108 = 117$$
Ans. 117