# Quadratics Set-2

- Live one on one classroom and doubt clearing
- Practice worksheets in and after class for conceptual clarity
- Personalized curriculum to keep up with school

**Example- 5**

Find the set of real values of the parameter *c* so that \(\begin{align}&\frac{{{x^2} + 2x + c}}{{{x^2} + 4x + 3c}}\end{align}\) can take all real values for \(x\in \mathbb{R}.\)

**Solution:** We want the given expression to assume all real values (for appropriate *x*), i.e, we want the range of the given expression to be \(\mathbb{R}.\)

\[\begin{align}&\qquad\qquad \frac{{{x^2} + 2x + c}}{{{x^2} + 4x + 3c}} = y\\\\\ \;\;\;\;&\Rightarrow\quad \left( {1 - y} \right){x^2} + \left( {2 - 4y} \right)x + c\left( {1 - 3y} \right) = 0.\end{align}\]

For *x* to be real, the *D* for this equation should be non-negative

\[\begin{array}{l} \Rightarrow 4{\left( {1 - 2y} \right)^2} \ge 4c\left( {1 - y} \right)\left( {1 - 3y} \right)\\\\ \Rightarrow 4{y^2} - 4y + 1 \ge c\left( {3{y^2} - 4y + 1} \right)\\\\ \Rightarrow \left( {4 - 3c} \right){y^2} + \left( {4c - 4} \right)y + \left( {1 - c} \right) \ge 0 \ldots (i)\end{array}\]

Now comes the crucial step. Since we want the range of *y* to be \(\mathbb{R}.\) , the constraint (i) should be satisfied by each real value of *y*. This means that the parabola for the left-hand side of (i) should not go below the axis for any value of *y*.

\[\begin{align}& \Rightarrow\quad \,{\rm{The \,\,discriminant \,\,for \,\,the\,\, left\,\, hand \,\,side \,\,of }}\left( {\,\,\rm{i}} \right){\rm{ \,\,cannot\,\, be\,\, positive}}\\\\ &\Rightarrow\quad D\,of{\rm{ }}\left( i \right) \le \,0\\ &\Rightarrow\quad 16{\left( {1 - c} \right)^2} \le 4\left( {1 - c} \right)\left( {4 - 3c} \right)\\\\ &\Rightarrow\quad 4{c^2} - 8c + 4 \le 3{c^2} - 7c + 4\\\\& \Rightarrow\quad {c^2} - c \le 0\\\\ &\Rightarrow\quad c\left( {c - 1} \right) \le 0\\\\ &\Rightarrow\quad 0 \le c \le 1\end{align}\]

You are urged to read this example carefully more than once until you understand it completely.

**Example- 6**

Find the minimum value of \(\begin{align}&f\left( x \right) = \frac{{\left( {x + a} \right)\left( {x + b} \right)}}{{\left( {x + c} \right)}}\end{align}\) where \(x > - \,c,\,\,\,a > c,\,\,\,b > c\)

**Solution:** Following the approach of the last two examples, we let

\[\begin{align}&\qquad\qquad y = \frac{{\left( {x + a} \right)\left( {x + b} \right)}}{{\left( {x + c} \right)}}\\ &\Rightarrow\quad {x^2} + \left( {a + b - y} \right)x + \left( {ab - cy} \right) = 0.\end{align}\]

Since *x* is real, \(D \ge 0\)

\[\begin{array}{l} \Rightarrow {\left( {a + b - y} \right)^2} - 4\left( {ab - cy} \right) \ge 0\\\\ \Rightarrow {y^2} - 2\left( {a + b - 2c} \right)y + {\left( {a + b} \right)^2} - 4ab \ge 0\\\\ \Rightarrow {y^2} - 2\left( {a + b - 2c} \right)y + {\left( {a - b} \right)^2} \ge 0 \ldots (i)\end{array}\]

The roots of this quadratic expression in *y* are

\[\begin{align}&{y_1},{y_2} = \frac{{2\left( {a + b - 2c} \right) \pm \sqrt {4{{\left( {a + b - 2c} \right)}^2} - 4{{\left( {a - b} \right)}^2}} }}{2}\\\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\;= \left( {a + b - 2c} \right) \pm \sqrt {4ab + 4{c^2} - 4c\left( {a + b} \right)} \\\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\;= a + b - 2c \pm 2\sqrt {ab + {c^2} - c\left( {a + b} \right)} \\\,\,\,\,\,\,\,\,\,\,\, &\qquad\;= a + b - 2c \pm 2\sqrt {\left( {a - c} \right)\left( {b - c} \right)} \\\,\,\,\,\,\,\,\,\,\,\, &\qquad\;= \left( {a - c} \right) + \left( {b - c} \right) \pm 2\sqrt {\left( {a - c} \right)\left( {b - c} \right)} \\\,\,\,\,\,\,\,\,\,\, &\qquad\;= {\left( {\sqrt {a - c} \pm \sqrt {b - c} } \right)^2}\end{align}\]

Let \({y_1} = {\left( {\sqrt {a - c} - \sqrt {b - c} } \right)^2}{\rm{and}}\,{y_2} = {\left( {\sqrt {a - c} + \sqrt {b - c} } \right)^2}\)

The solution for *y* in (i) is

\(y \le {y_1}\,{\rm{and }}y \ge {y_2}\)

We can discard the solution \(y \le {y_1}\) (why ?) so that \(y \ge {y_2}.\) The minimum value of *y* is *y* 2 or

\[{y_{\min }} = {\left( {\sqrt {a - c} + \sqrt {b - c} } \right)^2}\]

**Example- 7 **

Find the values of *a* for which the inequality \({x^2} + \left| {x - a} \right| - 3 < 0\) is satisfied by at least one negative *x*.

**Solution:** Although this example can be solved analytically, we will follow a graphical approach here and the reader will see just how powerful thinking in terms of graphs is. The given inequality can be written as:

\[\left| {x - a} \right| < 3 - {x^2}\]

This should be satisfied for at least one negative value of *x*. Equivalently stated, the graph of \(\left| {x - a} \right|\) should lie *beneath* the graph of \(\left( {3 - {x^2}} \right)\) for at least one negative value of *x*.

The graph of \(3 - {x^2}\) is fixed:

The graph of \(\left| {x - a} \right|\) is a left or right-shifted version of the graph of \(\left| x \right|\) , depending on the value of *a*.

We need *a* to be such a value so that the graph of \(\left| {x - a} \right|\) goes beneath that of \(\left( {3 - {x^2}} \right)\) for some negative *x*.

Imagine that we start with large negative values of *a* so that the graph is left shifted to a large extent.

As we decrease the magnitude of *a*, the mod-graph shifts to the right. At some point (some particular value of *a*, *a* = \(a{{~}_{1}}\)), the right arm of the mod-graph just touches the parabola (becomes a tangent to it)

As we further right-shift the mod-graph, the right arm of the mod graph goes below the parabola for some negative values of *x*.

In fact, after a point, the left arm also goes below the parabola for certain negative values of *x* (you are urged to visualise all this). The left arm remains below the parabola for certain negative *x* values until the other extreme is achieved (at another value of *a*, *a* =\(a{{~}_{2}}\), when the left arm passes through the vertex of the parabola)

Hence, when \({a_1} < a < {a_2},\) at least one of the arms of the mod-graph will be below the parabola for some negative values of *x*. We can, using coordinate geometry, evaluate \(a{{~}_{1}}\) and \(a{{~}_{2}}\) to be \(- \begin{align}\frac{{13}}{4}\end{align}\) and 3 respectively. Therefore

\[a \in \left( { - \frac{{13}}{4},3} \right)\]

The description above is extremely long but you must realise that this elaborate discussion is for the sake of your understanding; while actually solving this problem in an exam all this discussion would have to be done purely in the mind!

**Example- 8**

For what values of the parameter *a* does the equation \({x^4} + 2a{x^3} + {x^2} + 2ax + 1 = 0\) have at least two distinct negative roots.

**Solution:** This is a fourth degree polynomial.

Notice that the coefficients are “symmetric about the middle”, so that some useful manipulation can be done:

We divide the equation by *\({{x}^{2}}\)* to get

\[\begin{align}{} &{x^2} + 2ax + 1 + \frac{{2a}}{x} + \frac{1}{{{x^2}}} = 0\\ &\Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2a\left( {x + \frac{1}{x}} \right) + 1 = 0\\ &\Rightarrow {\left( {x + \frac{1}{x}} \right)^2} - 2 + 2a\left( {x + \frac{1}{x}} \right) + 1 = 0\\ &\Rightarrow {y^2} + 2y - 1 = 0\,\,\,\,{\rm{Where\,\, }}y = x + \frac{1}{x}\end{align}\]

The roots for this equation are:

\[y = - a \pm \sqrt {1 + {a^2}} \]

Let

\[\begin{align}& x + \frac{1}{x} = - a + \sqrt {1 + {a^2}} \ldots (i)\\ &\Rightarrow {x^2} + \left( {a - \sqrt {1 + {a^2}} } \right)x + 1 = 0\\ &\Rightarrow S = \sqrt {1 + {a^2}} - a > 0\\ &\quad\; P = {\rm{ }}1{\rm{ }} > {\rm{ }}0\end{align}\]

so that both roots of (i) are positive

Therefore, for at least two negative roots, both roots of

\[x + \frac{1}{x} = - a - \sqrt {1 + {a^2}} \ldots (ii)\]

should be negative

\[\begin{array}{l} \Rightarrow {x^2} &+ \left( {a + \sqrt {1 + {a^2}} } \right)x + 1 = 0 \ldots (iii)\\S=& - \left( {a + \sqrt {1 + {a^2}} } \right) < 0\\P =& 1 > 0\end{array}\]

Hence, both the roots will be negative. But we have to still ensure that *a* satisfies the basic constraint *D* > 0 for equation (iii)

\[\begin{align}&D > 0 \qquad\qquad\Rightarrow {\left( {a + \sqrt {1 + {a^2}} } \right)^2} > 4\\ &\qquad\;\;\; \qquad\qquad\Rightarrow \left( {a + \sqrt {1 + {a^2}} + 2} \right)\left( {a + \sqrt {1 + {a^2}} - 2} \right) > 0\end{align}\]

Since the first term is always positive, the inequality above reduces to

\[\sqrt {1 + {a^2}} > 2 - a\]

This can easily be solved graphically (or analytically) to get

\[a > \frac{3}{4}\]

- Live one on one classroom and doubt clearing
- Practice worksheets in and after class for conceptual clarity
- Personalized curriculum to keep up with school