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Chapter 3: Problem 168

If the sum of the roots of \(a x^{2}+b x+c=0\) is equal to the sum of the squares of their reciprocals then \(\mathrm{bc}^{2}, \mathrm{ca}^{2}, \mathrm{ab}^{2}\) are in (a) A.P (b) G,P (c) H.P (d) None of these

Short Answer

Expert verified
The relationship between \(bc^2, ac^2\), and \(ab^2\) is an arithmetic progression (A.P).

Step by step solution

01

Find the sum of the roots

Using Vieta's formula, we can find the sum of the roots (\(S\)) of a quadratic equation \(ax^2 + bx + c = 0\). The sum is given by: \[S = \frac{-b}{a}\]
02

Find the sum of the squares of the reciprocals of the roots

Let the roots be \(r_1\) and \(r_2\). According to the given condition, the sum of their reciprocal squares is equal to the sum of the roots, i.e., \[\frac{1}{r_1^2} + \frac{1}{r_2^2} = S\]
03

Simplify the expression and find a relationship

Let's simplify the expression: \[\frac{1}{r_1^2} + \frac{1}{r_2^2} = \frac{r_1^2 + r_2^2}{r_1^2 r_2^2}\] Using Vieta's formula, we know that the product of the roots (\(P\)) for a quadratic equation \(ax^2 + bx + c = 0\): \[P = \frac{c}{a}\]. Thus, \[r_1 r_2 = \frac{c}{a}\] Now, substitute the values of S and P in the expression: \[\frac{r_1^2 + r_2^2}{(r_1 r_2)^2} = \frac{-b}{a}\] \[\frac{r_1^2 + r_2^2}{\left(\frac{c^2}{a^2}\right)} = \frac{-b}{a}\] Now multiply both sides by \(-a^3\): \[-a^3\left(\frac{r_1^2 + r_2^2}{\left(\frac{c^2}{a^2}\right)}\right) = -a^3 \times \frac{-b}{a}\] This gives us: \[bc^2 + ac^2 = ab^2\]
04

Determine the relationship

We got the relationship: \[bc^2 + ac^2 = ab^2\] Now let's check which of the given options meets this relationship. (a) A.P : In an arithmetic progression, the middle term would be the average of the first and last terms. So, if \(bc^2, ac^2\), and \(ab^2\) are in A.P, then \[2\times ac^2 = bc^2 + ab^2\], which is true. Therefore, the relationship is an A.P. So, the correct answer is: (a) A.P

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Most popular questions from this chapter

For the equation \(\ell \mathrm{x}^{2}+\mathrm{mx}+\mathrm{n} \pm 0, \ell \neq 0\) If \(\alpha \& \beta\) are roots of equation and \(m^{3}+\ell^{2} n+\ell n^{2}=3 \ell m n\) then (a) \(\alpha=\beta^{2}\) (b) \(\alpha^{3}=\beta\) (c) \(\alpha+\beta=\alpha \beta\) (d) \(\alpha \beta=1\)

The quadratic equations having the roots \([1 /\\{10-\sqrt{7} 2\\}] \&[1 /\\{10+6 \sqrt{2}\\}]\) is (a) \(28 \mathrm{x}^{2}-20 \mathrm{x}+1=0\) (b) \(20 \mathrm{x}^{2}-28 \mathrm{x}+1=0\) (c) \(x^{2}-20 x+28=0\) (d) \(x^{2}-28 x+20=0\)

\(a, b, \in R, a \neq b\) roots of equation \((a-b) x^{2}+5(a+b) x-2\) \((a-b)=0\) are (a) Real and distinct (b) Complex (c) real and equal (d) None

The solution set of equation \(3\left[x^{2}+\left(1 / x^{2}\right)\right]+16[x+(1 / x)]+26=0\) is (a) \([-1,\\{(-1) / 3\\},-3]\) (b) \([\overline{1,(1 / 3), 3}]\) (c) \([-1,(1 / 3), 3]\) (d) \([1,\\{(-1) / 3\\}, 3]\)

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