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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

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

For the equation \(\mathrm{x}^{2}+\mathrm{k}^{2}=(2 \mathrm{k}+2) \mathrm{x}, \mathrm{k} \in \mathrm{R}\) roots are complex then (a) \(\overline{k=\\{(-1) / 2\\}}\) (b) \(k>\\{(-1) / 2\\}\) (c) \(k>\\{(-1) / 2\\}\) (d) \(\\{(-1) / 2\\}<\mathrm{k}<0\)

If the sum of the two roots of the equation \([1 /(x+a)]+[1 /(x+b)]=(1 / k)\) is zero then their Product is (a) \(\overline{(1 / 2)\left(a^{2}+b^{2}\right)}\) (b) \(\\{(-1) / 2\\}(a+b)^{2}\) (c) \([\\{a+b\\} / 2]^{2}\) (d) None

Construct the quadratic equation whose roots are three times the roots of \(5 \mathrm{x}^{2}-3 \mathrm{x}+3=0\) (a) \(5 x^{2}-9 x+27=0\) (b) \(5 x^{2}+9 x+27=0\) (c) \(5 x^{2}-9 x-27=0\) (d) \(5 \mathrm{x}^{2}+9 \mathrm{x}-27=0\)

If \(\tan A \& \tan B\) are roots of \(x^{2}-p x+q=0\) then the value of \(\cos ^{2}(\mathrm{~A}+\mathrm{B})=\) (a) \(\left[(1-q)^{2} /\left\\{p^{2}+(1-q)^{2}\right\\}\right]\) (b) \(\left[p^{2} /\left\\{p^{2}+(1-q)^{2}\right\\}\right]\) (c) \(\left[(1-q)^{2} /\left(p^{2}-q^{2}\right)\right]\) (d) \(\left[\mathrm{p}^{2} /\left(\mathrm{p}^{2}+\mathrm{q}^{2}\right)\right]\)
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