Examples On Basic Quadratic Problems Set-3

Example-  9

What can be said about the roots of this equation?

\(\left( {a + b - 2c} \right){x^2} + \left( {b + c - 2a} \right)x + \left( {c + a - 2b} \right) = 0\)

Solution:   If we denote the coefficients, in this equation as  \(A, B\; and\; C \) respectively, we see that

\(A + B + C = 0\)

Thus, for discriminant D, we have

\[\begin{array}{l}D = {B^2} - 4AC = {\left( {A + C} \right)^2} - 4AC\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\; = {\left( {A - C} \right)^2} \ge 0\end{array}\]

      The roots are real.

Example- 10

Show that \({e^{\sin x}} - {e^{ - \sin x}} = 4\)  has no real solution.

Solution:   We observe that.  \({\left( {{e^{\sin x}}} \right)_{\min }} = {e^{ - 1}} = \frac{1}{e}\;{\rm{and}}\;{\left( {{e^{\sin x}}} \right)_{\max }} = {e^1} = e\)

Denoting \({e^{\sin x}}\) by \(y,\) the equation becomes

\[\begin{align}\,y - \frac{1}{y} = 4 \quad&\Rightarrow\quad {y^2} - 4y - 1 = 0 \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\Rightarrow\quad y = \frac{{4 \pm \sqrt {20} }}{2} = 2 + \sqrt 5 ,\;2 - \sqrt 5 \end{align}\]

Since \(2 + \sqrt 5  > e\)  and \(2 - \sqrt 5  < 0,\;y\)  can attain none of these values. Thus, no real solution exist for the given equation.

Example- 11

Let. Prove that \(f(x)\)  if is an integer whenever  \(x\)  is an integer, then \(2A,\;A + B\)  and  \(C\)  are all integers. Prove the converse also

Solution:   First, we assume that \(f(x)\) is an integer whenever x is an integer. Thus,

\[\begin{align}f(0)\; \text{is an integer}\quad&\Rightarrow \quad  \text{C is an integer}\\
f(1)\; \text {is an integer}    \quad&\Rightarrow  \quad  A + B + C \text {and hence}\\
&\qquad\;\;A + B,\text {are integers}\\f( - 1)\; \text {is an integer}  \quad&\Rightarrow\quad \,A - B + C \text {is an integer}\\&\Rightarrow\quad\left( {A + B + C} \right) + \left( {A - B + C} \right) = 2A + 2C \; \text {is an integer, and hence   is an integer}\end{align}\]

We have thus proved that \(2A,\;A + B\) and  \(C\)  are integers

Now, we prove the converse. We assume that \(2A,\;A + B\)  and C are integer, and we have to prove that if \(x\) is an integer, then \(f (x) \) is an integer. For that, we assume two cases:

\(\bf \text{x is even, say}\;2n\)	\(\bf \text {x is odd, say}\; 2n + 1\)
\(\begin{array}{l}f(x) = A{x^2} + Bx + C\\\,\,\,\,\,\,\,\,\,\,\,\, = \left( {2A} \right)\left( {2{n^2}} \right) + \left( {2B} \right)n + C\end{array}\)	\(\begin{array}{l}f(x) = A{x^2} + Bx + C\\\,\,\,\,\,\,\,\,\,\,\, = A\left( {4{n^2} + 4n + 1} \right) + B\left( {2n + 1} \right) + C\\\,\,\,\,\,\,\,\,\,\,\, = \left( {2A} \right)\left( {2{n^2} + 2n} \right) + \left( {2B} \right)n + \left( {A + B} \right) + C\end{array}\)
which is an integer	which is an integer

Thus \(f\left( x \right)\) , is an integer in both cases.

Example- 12

Solve the equation. \(\left( {12x - 1} \right)\left( {6x - 1} \right)\left( {4x - 1} \right)\left( {3x - 1} \right) = 5\)

Solution:   The trick involved is to somehow reduce it to the standard quadratic form. To do that, we manipulate the left side (LHS) as follows:

\[\begin{align}&LHS = \left( {\left( {12x - 1} \right)\left( {3x - 1} \right)} \right)\left( {\left( {6x - 1} \right)\left( {4x - 1} \right)} \right)\\\\\,\,\,\,\,\,\,\,\,\,\, &\qquad\;= \left( {36{x^2} - 15x + 1} \right)\left( {24{x^2} - 10x + 1} \right)\\\\\,\,\,\,\,\,\,\,\,\,\, &\qquad\;= 6\left( {12{x^2} - 5x + \frac{1}{3}} \right)\left( {12{x^2} - 5x + \frac{1}{2}} \right)\end{align}\]

Now we can use the substitution \(\begin{align}&12{x^2} - 5x + \frac{1}{3} = y\end{align}\) , so that the given equation reduces to

\[\begin{align}&6y\left( {y + \frac{1}{6}} \right) = 5    \Rightarrow  y\left( {6y + 1} \right) - 5 = 0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\quad\Rightarrow  y =  - 1,\;\frac{5}{6}\end{align}\]

The next step is to solve for \(x:\)

\[12{x^2} - 5x + \frac{1}{3} =  - 1 \qquad or\qquad 12{x^2} - 5x + \frac{1}{3} = \frac{5}{6}\]

 \[ \Rightarrow  12{x^2} - 5x + \frac{4}{3} = 0 \qquad or \qquad 12{x^2} - 5x - \frac{1}{2} = 0\]

 No real solution  \(\begin{align} \Rightarrow  x =  - \frac{1}{{12}},\;\frac{1}{2}\end{align}\)

TRY YOURSELF - I

1.   If the roots \(\left( {1 + {m^2}} \right){x^2} - 2\left( {1 + 3m} \right)x + \left( {1 + 8m} \right) = 0\) of are equal, find the set of possible values of \(m\).

2.   If the roots of \(\left( {{a^2} + {b^2}} \right){x^2} + 2\left( {bc + ad} \right)x + {c^2} + {d^2} = 0\) the equation are equal, prove that \(\begin{align}&\frac{2}{b} = \frac{1}{a} + \frac{1}{c}\end{align}\)

3.   If the roots of \(\left( {{a^2} + {b^2}} \right){x^2} + 2\left( {bc + ad} \right)x + {c^2} + {d^2} = 0\)  are real, prove that they must be equal.

4.   Solve for \(\begin{align}x:\;\sqrt {\frac{{2{x^2} + x + 2}}{{{x^2} + 3x + 1}}}  + 2\sqrt {\frac{{{x^2} + 3x + 1}}{{2{x^2} + x + 2}}}  = 3\end{align}\)

5.   Solve for  \(x:\;\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right)\left( {x + 4} \right) = 120\)

6.   Solve for \(x :\;x\left( {x - 1} \right)\left( {x + 1} \right)\left( {x + 2} \right) + 1 = 0\)

7.   If  \(x = 1\) is a root of the quadratic equation \(a{x^2} + bx + c = 0\)  with real coefficients, prove that \(4a{x^2} + 3bx + 2c = 0\)  must have real roots.

8.   If \(a{x^2} + \left( {b - c} \right)x + a - b - c = 0\)  has unequal real roots for all, prove that either \(b < a < 0\;{\rm{or}}\;b > a > 0.\)