Question

Suppose A, B, C are defined as $A = a^2b+ ab^2 - a^2c - ac^2$, $B = b^2c + bc^2 - a^2b - ab^2$, and $C = a^2c + ac^2-b^2c - bc^2$, where a > b > c > 0 and the equation $Ax^2 + Bx + C = 0$ has equal roots, then a, b, c are in

• A. A.P.
• B. G.P.
• C. H.P.
• D. A.G.P.

Answer 1

Answer: (C)

H.P.

Answer 2

The problem states that A, B, C are defined as: $A = a^2b+ ab^2 - a^2c - ac^2$ $B = b^2c + bc^2 - a^2b - ab^2$ $C = a^2c + ac^2-b^2c - bc^2$

First, let's find the sum $A+B+C$: $A+B+C = (a^2b+ ab^2 - a^2c - ac^2) + (b^2c + bc^2 - a^2b - ab^2) + (a^2c + ac^2-b^2c - bc^2)$ By observing the terms, we can see that all terms cancel out: $a^2b$ cancels with $-a^2b$ $ab^2$ cancels with $-ab^2$ $-a^2c$ cancels with $a^2c$ $-ac^2$ cancels with $ac^2$ $b^2c$ cancels with $-b^2c$ $bc^2$ cancels with $-bc^2$ Therefore, $A+B+C = 0$.

The given equation is $Ax^2 + Bx + C = 0$. Since $A+B+C = 0$, we can substitute $C = -(A+B)$ into the equation: $Ax^2 + Bx - (A+B) = 0$ $Ax^2 + Bx - A - B = 0$ $A(x^2-1) + B(x-1) = 0$ $A(x-1)(x+1) + B(x-1) = 0$ $(x-1)[A(x+1) + B] = 0$ $(x-1)(Ax + A + B) = 0$

This equation implies that one root is $x=1$. The problem states that the quadratic equation has equal roots. Since one root is $x=1$, the other root must also be $x=1$. This means $x=1$ is a double root.

For a quadratic equation $Ax^2+Bx+C=0$ to have equal roots at $x=1$, it must be of the form $A(x-1)^2 = 0$. Expanding $A(x-1)^2$, we get $A(x^2 - 2x + 1) = Ax^2 - 2Ax + A = 0$. Comparing the coefficients with $Ax^2+Bx+C=0$, we must have: $B = -2A$ $C = A$

We can use the condition $C=A$ to find the relationship between a, b, c. Let's factorize A and C. $A = a^2b+ ab^2 - a^2c - ac^2$ $A = a^2(b-c) + a(b^2-c^2)$ $A = a^2(b-c) + a(b-c)(b+c)$ $A = a(b-c)[a + (b+c)]$ $A = a(b-c)(a+b+c)$

$C = a^2c + ac^2-b^2c - bc^2$ $C = c(a^2+ac-b^2-bc)$ $C = c[(a^2-b^2) + (ac-bc)]$ $C = c[(a-b)(a+b) + c(a-b)]$ $C = c(a-b)(a+b+c)$

Now, set $A=C$: $a(b-c)(a+b+c) = c(a-b)(a+b+c)$

Given $a > b > c > 0$, it implies $a+b+c \ne 0$. So we can divide both sides by $(a+b+c)$: $a(b-c) = c(a-b)$ $ab - ac = ac - bc$ $ab + bc = 2ac$

To find the relationship between a, b, c, divide the entire equation by $abc$ (which is non-zero since $a,b,c > 0$): $\frac{ab}{abc} + \frac{bc}{abc} = \frac{2ac}{abc}$ $\frac{1}{c} + \frac{1}{a} = \frac{2}{b}$

This equation shows that the reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in an Arithmetic Progression (A.P.). If the reciprocals of three numbers are in A.P., then the numbers themselves are in Harmonic Progression (H.P.). Therefore, a, b, c are in H.P.