Examples On Basic Quadratic Problems Set-2

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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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