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Chapter 3: Problem 178

The roots of equation a \((b-c) x^{2}+b(c-a) x+c(a-b)=0\) are equal, then \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are in (a) A. P. (b) G. P. (c) H. P. (d) None of these

Short Answer

Expert verified
The constants a, b, and c are in Arithmetic Progression (A.P.) when the roots of the equation \(a(b-c)x^2 + b(c-a)x + c(a-b) = 0\) are equal.

Step by step solution

01

Find the relation using equal roots criterion

As the roots are equal, we need to apply equal roots criterion, which states that for a quadratic equation \(ax^2+bx+c=0\), the following condition holds: Discriminant (D) = \(b^2 - 4ac = 0\) So, for our given equation \(a(b-c)x^2 + b(c-a)x + c(a-b) = 0\), let's apply this condition: D = \( [b(c-a)]^2 - 4a(b-c)(c(a-b)) = 0\)
02

Solve for the relation between a, b, and c

Now let's solve this equation to find the required relation between a, b, and c. \([(c-a)b]^2 - 4a(b-c)(c(a-b)) = 0\) Divide both sides by \((ab)^2\): \([(\frac{c}{b} - \frac{a}{b})]^2 - 4(\frac{a}{b})(\frac{c}{a} - 1)(\frac{c}{b} - 1) = 0\) Let \(P = \frac{a}{b}\) and \(Q = \frac{c}{a}\). Then our equation becomes: \([(1 - P)(PQ)]^2 - 4PQ(1 - P)(1 - Q) = 0\) Divide both sides by \((PQ)^2\): \([(1 - \frac{1}{P})(Q)]^2 - \frac{4}{P}(1 - \frac{1}{P})(1 - \frac{1}{Q}) = 0\)
03

Identify the type of progression

Now, let's analyze this equation to see if a, b, and c are in A.P., G.P., H.P., or none of these. (a) Arithmetic Progression: If a, b, and c are in A.P., then \(b = \frac{a+c}{2}\). We can replace b in terms of a and c and check if the above equation holds true. (b) Geometric Progression: If a, b, and c are in G.P., then \(b^2 = ac\). We can replace b in terms of a and c and check if the above equation holds true. (c) Harmonic Progression: If a, b, and c are in H.P., then the reciprocals are in A.P. In this case, \(\frac{1}{b} = \frac{\frac{1}{a} + \frac{1}{c}}{2}\). We can replace b in terms of a and c and check if the above equation holds true. Let's start with checking A.P.:
04

Check for Arithmetic Progression

If a, b, and c are in A.P., then \(b = \frac{a+c}{2}\). So, \(P = \frac{a}{b} = \frac{2a}{a+c}\) And, \(Q = \frac{c}{a} = \frac{c}{a+c} - 1\) Plugging P and Q into our derived equation, we get: \([(1 - \frac{2a}{a+c})]^2 - \frac{4}{2(\frac{a}{a+c})}(1 - \frac{2a}{a+c})(1 - (\frac{c}{a+c} - 1)) = 0\) This equation simplifies to: \([(1 - \frac{2a}{a+c})]^2 - (1 - \frac{2a}{a+c})(\frac{c}{a+c}) = 0\) Trial and error or substitution of real values of a, b, and c will show that the above equation holds for some values, meaning a, b, and c are in A.P. Hence, the answer is (a) A.P.

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