Example- 9

If \(\alpha ,\,\beta \) are the roots of \(a{x^2} + bx + c = 0\) and \(\alpha + \delta ,\,\,\beta + \delta \) are the roots of \(A{x^2} + Bx + C = 0\) . for some constant \(\delta \), then prove that

\[\frac{{{b^2} - 4ac}}{{{a^2}}} = \frac{{{B^2} - 4AC}}{{{A^2}}}\]

Solution: This result can be achieved by some straight forward algebraic manipulation.

\[\begin{align}&\alpha + \beta = - \frac{b}{a}\;\;\;\;\;\;\;\; \alpha \beta = \frac{c}{a}\\&\alpha + \beta + 2\delta = - \frac{B}{A} \;\;\;\;\;\;\;\; \left( {\alpha + \delta } \right)\left( {\beta + \delta } \right) = \frac{C}{A}\end{align}\]

\[\begin{align}&{\rm{L}}{\rm{.H}}{\rm{.S  }} = \frac{{{b^2} - 4ac}}{{{a^2}}} = {\left( {\frac{b}{a}} \right)^2} - 4\left( {\frac{c}{a}} \right)\\ &\qquad\quad= {\left( {\alpha + \beta } \right)^2} - 4\alpha \beta \\ &\qquad\quad= {\left( {\alpha - \beta } \right)^2}\end{align}\]

\[\begin{align}&{\rm{R}}{\rm{.H}}{\rm{.S }} = {\left( {\frac{B}{A}} \right)^2} - 4\frac{C}{A}\\ &\qquad\quad= {\left( {\alpha + \beta + 2\delta } \right)^2} - 4\left( {\alpha + \delta } \right)\left( {\beta + \delta } \right)\\ &\qquad\quad= {\left\{ {\left( {\alpha + \delta } \right) + \left( {\beta + \delta } \right)} \right\}^2} - 4\left( {\alpha + \delta } \right)\left( {\beta + \delta } \right)\\ &\qquad\quad= {\left\{ {\left( {\alpha + \delta } \right) - \left( {\beta + \delta } \right)} \right\}^2}\\ &\qquad\quad= {\left( {\alpha - \beta } \right)^2}\end{align}\]

Example- 10

Prove that the equation

\(\) \[\frac{1}{x} + \frac{1}{{x + a}} + \frac{1}{{x - b}} = 0 \,\,\,\,\,\,\,\,\left( {a,b > 0} \right)\]

has two roots, one each in the intervals \(\begin{align}&\left( { - \frac{{2a}}{3}, - \frac{a}{3}} \right)\end{align}\) and \(\begin{align}&\left( {\frac{b}{3},\frac{{2b}}{3}} \right)\end{align}\)

Solution: The given equation, when written in the form of a standard quadratic equation, becomes \(3{x^2} + 2\left( {a - b} \right)x - ab = 0\)

Let the left hand expression be denoted by \(f\left( x \right).\) To prove that a root of \(f\left( x \right)\) lies between \(\begin{align}{I}- \frac{{2a}}{3}\end{align}\) and \(\begin{align}{I}- \frac{a}{3},\end{align}\) the obvious thing to do is evaluate  \(\begin{align}&f\left( { - \frac{{2a}}{3}} \right)\end{align}\) and \(\begin{align}&f\left( { - \frac{a}{3}} \right)\end{align}\) and show that these are opposite signs, or, \(f\left( { - \frac{{2a}}{3}} \right) \cdot f\left( { - \frac{a}{3}} \right) < 0\)

\[\begin{align}&f\left( -\frac{2a}{3} \right)\cdot f\left( -\frac{a}{3} \right)=\left( \frac{4{{a}^{2}}}{3}-\frac{4a\left( a-b \right)-ab}{3} \right)\times \left( \frac{{{a}^{2}}}{3}-\frac{2a\left( a-b \right)}{3}-ab \right)\\&\qquad\qquad\qquad\qquad\;\;\;=-\frac{{{a}^{2}}b}{9}\left( a+b \right)<0\end{align}\]

Similarly

\[\begin{align}&f\left( \frac{b}{3} \right)\cdot f\left( \frac{2b}{3} \right)=\left( \frac{{{b}^{2}}}{3}+\frac{2b\left( a-b \right)-ab}{3} \right)\times \left( \frac{4{{b}^{2}}}{3}+\frac{4b\left( a-b \right)}{3}=ab \right)\\&\qquad\qquad\qquad\quad=-\frac{-a{{b}^{2}}}{9}\left( a+b \right)<0\end{align}\]

Hence, the given statement has been proved

Example- 11

If a, b, c, d be real numbers in G. P and u, v, w satisfy the equations

\(u + 2v + 3w = 6 \,\,\,\,\,\,\,\,\,4u + 5v + 6w = 12\,\,\,\,\,\,\,\,\,\,\,6u + 9v = 4\)

then show that the roots of the equations

\(\begin{align}&\left( {\frac{1}{u} + \frac{1}{v} + \frac{1}{w}} \right){x^2} + \left\{ {{{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2} + {{\left( {d - b} \right)}^2}} \right\}x + \left( {u + v + w} \right) = 0\end{align}\)

and \(20{x^2} + 10{\left( {a - d} \right)^2}x - 9 = 0\) are reciprocal to each other.

Solution: To simplify this (enormous!) equation, we first simply solve the three given equation for u, v, w to get

\[u = - 1/3\,\;\;\;\;v = 2/3\,\;\;\;\;w = 5/3\]

Now, since a, b, c, d are in G. P, we let \(b = ar,\,\,c = a{r^2},\,\,d = a{r^3}\)

The coefficient of x in the given equation is

\[\begin{array}{l}{\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} + {\left( {d - b} \right)^2}\\ = {a^2}\left\{ {{{\left( {r - {r^2}} \right)}^2} + {{\left( {{r^2} - 1} \right)}^2} + {{\left( {{r^3} - r} \right)}^2}} \right\}\\ = {a^2}{\left( {r - 1} \right)^2}{\left( {{r^2} + r + 1} \right)^2}\\ = {a^2}{\left( {{r^3} - 1} \right)^2}\\ = {\left( {d - a} \right)^2}\end{array}\]

Therefore, the given equation reduces to (verify)

\[ - 9{x^2} + 10{\left( {d - a} \right)^2}x + 20 = 0\]

If \(\alpha ,\beta \) are its roots, then:

\[\alpha + \beta = \frac{{10}}{9}{\left( {d - a} \right)^2},\,\,\,\alpha \beta = - \frac{{20}}{9}\]

Now let us find the equation whose roots are \(\begin{align}&\frac{1}{\alpha },\frac{1}{\beta }:\end{align}\)

\[\begin{align}&\frac{1}{\alpha } + \frac{1}{\beta }\,\,\,\,\,\,\,\,\, = \frac{{\alpha + \beta }}{{\alpha \beta }}\,\,\,\,\,\,\,\,\, = - \frac{1}{2}{\left( {d - a} \right)^2}\\\\&\frac{1}{\alpha } \cdot \frac{1}{\beta }\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{\alpha \beta }}\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - \frac{9}{{20}}\end{align}\]

The equation is

\[{x^2} + \frac{1}{2}{\left( {d - a} \right)^2} - \frac{9}{{20}} = 0\]

or \(20{x^2} + 10{\left( {a - d} \right)^2} - 9 = 0\)

This is the second given equation. Hence the roots of these two given equations are reciprocal of each other

Example- 12

If \(\beta \) is such that \(\sin 2\beta \ne 0,\) then show that the value of the expression \(\begin{align}&\frac{{{x}^{2}}+2x\cos 2\alpha +1}{{{x}^{2}}+2x\cos 2\beta +1}\left( x\in \mathbb{R} \right)\end{align}\) always lies between \(\begin{align}&\frac{{{{\cos }^2}\alpha }}{{{{\cos }^2}\beta }}\,\,{\rm{and}}\,\,\frac{{{{\sin }^2}\alpha }}{{{{\sin }^2}\beta }}\end{align}\)

Solution: This is again a question of finding the range of the given expression.

We let

\[\begin{align}&\qquad\qquad\qquad y = \frac{{{x^2} + 2x\cos 2\alpha + 1}}{{{x^2} + 2x\cos 2\beta + 1}}\\ &\Rightarrow \left( {1 - y} \right){x^2} + 2\left( {\cos 2\alpha - y\cos 2\beta } \right)x + \left( {1 - y} \right) = 0\end{align}\]

Since x is real, the D for this equation is non-negative

\[\begin{align}&D \ge 0 \quad\Rightarrow\quad 4{\left( {{{\cos }^2}\alpha - y{{\cos }^2}\beta } \right)^2} \ge 4{\left( {1 - y} \right)^2}\\ &\qquad\;\;\quad\Rightarrow\quad {\cos ^2}2\alpha + {y^2}{\cos ^2}2\beta - 2y{\cos ^2}\alpha {\cos ^2}\beta \ge {y^2} + 1 - 2y\\ &\qquad\;\;\quad\Rightarrow\quad \left( {{{\cos }^2}2\beta - 1} \right){y^2} + 2y\left( {1 - \cos 2\alpha \cos 2\beta } \right) + \left( {{{\cos }^2}2\alpha - 1} \right) \ge 0\, \ldots (i)\end{align}\]

The roots of this expression are

\[\begin{align}&{y_1},{y_1}\,\,\,\,\, = \frac{{\left( {\cos 2\alpha \cos 2\beta - 1} \right) \pm \sqrt {{{\left( {1 - \cos 2\alpha \cos 2\beta } \right)}^2} - \left( {{{\cos }^2}2\alpha - 1} \right)\left( {{{\cos }^2}2\beta - 1} \right)} }}{{\left( {{{\cos }^2}2\beta - 1} \right)}}\\ &\qquad\quad= \frac{{\left( {\cos 2\alpha \cos 2\beta - 1} \right) \pm \sqrt {{{\cos }^2}2\alpha {{\cos }^2}2\beta - 2\cos 2\alpha \cos 2\beta } }}{{{{\cos }^2}2\beta - 1}}\\ &\qquad\;\;\;\;= \frac{{\left( {\cos 2\alpha \cos 2\beta - 1} \right) \pm \left( {\cos 2\alpha - \cos 2\beta } \right)}}{{{{\cos }^2}2\beta - 1}}\\ &\qquad\quad= \frac{{\left( {\cos 2\beta + 1} \right)\left( {\cos 2\alpha - 1} \right)}}{{{{\cos }^2}2\beta - 1}},\frac{{\left( {\cos 2\beta - 1} \right)\left( {\cos 2\alpha + 1} \right)}}{{{{\cos }^2}2\beta - 1}}\\ &\qquad\quad= \frac{{\cos 2\alpha - 1}}{{\cos 2\beta - 1}},\,\,\frac{{\cos 2\alpha + 1}}{{\cos 2\beta + 1}}\\ &\qquad\;\;\;\;= \frac{{{{\sin }^2}\alpha }}{{{{\sin }^2}\beta }},\frac{{{{\cos }^2}\alpha }}{{{{\cos }^2}\beta }}\end{align}\]

Notice that in (i), the coefficient of y2 is negative so its solution will be all y lying between the roots. Hence,

y lies between \(\begin{align}&\frac{{{{\sin }^2}\alpha }}{{{{\sin }^2}\beta }}\,\,{\rm{and}}\,\,\frac{{{{\cos }^2}\alpha }}{{{{\cos }^2}\beta }}\end{align}\)

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