# Examples On Basic Quadratic Problems Set-2

Example- 5

(a)  For what positive values of  $$‘a’$$ will $$\left( {x - 3} \right)\left( {x + 2} \right) + a = 0$$ have integral roots?

(b)  If  $$a, b, c$$ are odd integers, prove that cannot have rational roots.

Solution:   (a)  To have integral roots, we require the discriminant to be necessarily a perfect square.

\begin{align}&\qquad\left( {x - 3} \right)\left( {x + 2} \right) + a = 0\\\\&\Rightarrow{x^2} - x - \left( {6 - a} \right) = 0\\\\ &\Rightarrow D = 25 - 4a\end{align}

D is a perfect square when $$a = 4, 6$$

For $$a = 4$$$$D = 9$$  and the roots are

\begin{align}&{x_1},{x_2} = \frac{{1 \pm 3}}{2} = - 1,2\end{align}which are integral

For  $$a = 6$$$$D = 1$$  and the roots are

\begin{align}&{x_1},{x_2} = \frac{{1 \pm 1}}{2} = 0,1\end{align} which are again integral

Hence, the required values for $$a$$ are $$4, 6$$

(b) To have rational roots, we require the D to be perfect square; i.e

${b^2} - 4ac = {q^2}$

where  $$a, b, c$$  are all odd integers and hence  $$q$$ is also odd (verify)

Hence we assume  $$a, b, c, q$$ to be of the form $$2t + 1,\,2u + 1,\,2v + 1,\,2w + 1$$ respectively

Now

\begin{align}&{b^2} - 4ac = {q^2}\\\\ \Rightarrow\qquad &\left( {b + q} \right)\left( {b - q} \right) = 4ac\\\\ \Rightarrow\qquad &\left( {u + w + 1} \right)\left( {u - w} \right) = \left( {2t + 1} \right)\left( {2v + 1} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {by\;{\rm{}}substitution}\right)\end{align}

Notice now that the left hand side is always even, no matter what integral values $$u$$ and $$w$$ take, while the right hand side is odd. This leads to a contradiction in our original assumption of letting   $${b^2} - 4ac = {q^2}$$ Hence, no such $$q$$ exists.

$$\Rightarrow$$ The equation does not have rational roots

Example- 6

Prove that $${\left( {{a_1}{b_1} + {a_2}{b_2} + .... + {a_n}{b_n}} \right)^2} \le \left( {a_1^2 + a_2^2 + .....a_n^2} \right)\left( {b_1^2 + b_2^2 + .....b_n^2} \right)$$  where are $${a_i},{b_i}$$ real numbers.

Solution:  This widely used inequality can be proved in a variety of ways. Here, we will use a quadratic equation approach to prove this.

If we can somehow make this inequality correspond to the form of the discriminant $${b^2} - 4ac,$$ our task could be accomplished.

Consider the following expression:

$S\left( x \right) = {\left( {{a_1}x + {b_1}} \right)^2} + {\left( {{a_2}x + {b_2}} \right)^2} + ..... + {\left( {{a_n}x + {b_n}} \right)^2}$

Obviously $$S\left( x \right) \ge 0$$, for all x.

Now rearranging $$S\left( x \right)$$ in the form of a standard quadratic expression, we get:

$S\left( x \right) = \left( {a_1^2 + a_2^2 + .... + a_n^2} \right){x^2} + 2\left( {{a_1}{b_1} + {a_2}{b_2} + .... + {a_n}{b_n}} \right)x + \left( {b_1^2 + b_2^2 + ....b_n^2} \right)$

Since $$S\left( x \right) \ge 0$$  for all $$x,$$ the parabola for lies $$S\left( x \right)$$ entirely above the x-axis. Hence, the discriminant for $$S\left( x \right)$$ cannot be positive (since if D is positive, $$S\left( x \right)$$ will have real roots and the parabola will go below the x-axis for some x)

Therefore,

\begin{align}D &\le 0\\\Rightarrow\quad{\left( {{a_1}{b_1} + {b_2}{b_2} + .... + {a_n}{b_n}} \right)^2} &\le \left( {a_1^2 + a_2^2 + .... + a_n^2} \right)\left( {b_1^2 + b_2^2 + .... + b_n^2} \right)\end{align}

You are urged to find out the condition when the equality will hold.

Example- 7

What can you say about the roots of the equation

$f\left( x \right) = \left( {x - a} \right)\left( {x - b} \right) + \left( {x - b} \right)\left( {x - c} \right) + \left( {x - c} \right)\left( {x - a} \right)?$

Solution:   $$f\left( x \right)$$ can be rearranged into the form of a standard quadratic equation:

$f\left( x \right) = 3{x^2} - 2\left( {a + b + c} \right)x + \left( {ab + bc + ca} \right)$

The D is given by:

\begin{align}&D = 4{\left( {a + b + c} \right)^2} - 12\left( {ab + bc + ca} \right)\\\\\,\,\,\, &\;\;\;= 4\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)\\\\\,\,\,\, &\;\;\;= 4\left\{ {\frac{1}{2}\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right]} \right\}\\\\\,\,\, &\;\;\;= 2\left\{ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right\} \ge 0\end{align}

Hence $$f\left( x \right)$$ , has real roots which are equal only when $$D = 0$$ or when $$a = b = c$$

Example-  8

Find the values that $$m$$ can take if the roots of the equation $$\left( {m - 3} \right){x^2} - 2mx + 5m = 0$$ are real.

Solution:   For real roots $$D \ge 0$$,

\begin{align}&\Rightarrow 4{m^2} - 20m\left( {m - 3} \right) \ge 0\\\\ &\Rightarrow - 16{m^2} + 60m \ge 0\\\\ &\Rightarrow 4{m^2} - 15m \le 0\\\\ &\Rightarrow m\left( {m - 15/4} \right) \le 0\\\\ &\Rightarrow m \ge 0\,\,{\rm{and}}\,\,m \le \frac{{15}}{4}\,\,{\rm{or}}\,\,m \in \left[ {0,\frac{{15}}{4}} \right]\end{align}

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