# 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 $$b < a < 0\;{\rm{or}}\;b > a > 0.$$

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