Nature of Roots - Examples
Solved Example 1: What can you say about the roots of the equation \({x^2} + 2x - 4 = 0\)?
Solution: The discriminant is
\[D = {2^2} - 4\left( 1 \right)\left( { - 4} \right) = 20 > 0\]
Thus, the equation has real and distinct roots. Let us evaluate them using the quadratic formula:
\[\begin{align}&x = \frac{{ - b \pm \sqrt D }}{{2a}} = \frac{{ - 2 \pm \sqrt {20} }}{2}\\& = - 1 \pm \sqrt 5 \end{align}\]
Solved Example 2: If \(m{x^2} + 9x - 1 = 0\) has real roots, find the possible values of m.
Solution: For real roots, the discriminant of the given equation must be non-negative:
\[\begin{align}&D \ge 0\\&\Rightarrow \,\,\,{9^2} - 4\left( m \right)\left( { - 1} \right) \ge 0\\&\Rightarrow \,\,\,81 + 4m \ge 0\\&\Rightarrow \,\,\,m \le - \frac{{81}}{4}\end{align}\]
Solved Example 3: The equation
\[\left( {p + 1} \right){x^2} + 2\left( {p + 3} \right)x + \left( {p + 8} \right) = 0\]
has equal roots. Find the possible values of p.
Solution: The discriminant of the given equation must be 0:
\[\begin{align}&4{\left( {p + 3} \right)^2} - 4\left( {p + 1} \right)\left( {p + 8} \right) = 0\\&\Rightarrow \,\,\, - 3p + 1 = 0\\&\Rightarrow \,\,\,p = \frac{1}{3}\end{align}\]
Solved Example 4: Find the possible values of t if the roots of the equation \({x^2} + {t^2} = 6t + 8x\) are real.
Solution: In standard form, the given equation is
\[{x^2} - 8x + {t^2} - 6t = 0\]
For the roots to be real, the discriminant should be non-negative:
\[\begin{align}&{\left( { - 8} \right)^2} \ge 4\left( {{t^2} - 6t} \right)\\&\Rightarrow \,\,\,{t^2} - 6t \le 16\\&\Rightarrow \,\,\,\left( {t - 8} \right)\left( {t + 2} \right) \le 0\\&\Rightarrow \,\,\,t \le - 2\,\,\,{\rm{or}}\,\,\,t \ge 8\end{align}\]
Solved Example 5: Find the values which m can take if the roots of \(\left( {m - 3} \right){x^2} - 2mx + 5m = 0\) are real.
Solution: The discriminant must be non-negative:
\[\begin{align}&D = {\left( { - 2m} \right)^2} - 4\left( {m - 3} \right)\left( {5m} \right) \ge 0\\&\Rightarrow \,\,\,4{m^2} - 4\left( {5{m^2} - 15m} \right) \ge 0\\&\Rightarrow \,\,\,{m^2} - 5{m^2} + 15m \ge 0\\&\Rightarrow \,\,\,4{m^2} - 15m \le 0\\&\Rightarrow 4m\left( {m - \frac{{15}}{4}} \right) \le 0\\&\Rightarrow \,\,\,0 \le m \le \frac{{15}}{4}\end{align}\]
Solved Example 6: For what values of k does \(5{x^2} + kx + 5 = 0\) have real roots?
Solution: For real roots, the discriminant should be non-negative:
\[\begin{array}{l}D \ge 0\\ \Rightarrow \,\,\,{k^2} - 4\left( 5 \right)\left( 5 \right) \ge 0\\ \Rightarrow \,\,\,{k^2} - 100 \ge 0\\ \Rightarrow \,\,\,k \le - 10\,\,\,{\rm{or}}\,\,\,k \ge 10\end{array}\]
Solved Example 7: What is the nature of the roots of the following quadratic equation?
\[\left( {b - c} \right){x^2} + \left( {c - a} \right)x + \left( {a - b} \right) = 0\]
Solution: Let us represent the coefficients of this quadratic equation as follows:
\[b - c = A,\,\,\,c - a = B,\,\,\,a - b = C\]
We note that \(A + B + C = 0\). Now, the discriminant of the quadratic is
\[\begin{align}& D = {B^2} - 4AC = {\left( { - \left( {A + C} \right)} \right)^2} - 4AC\\&\qquad\qquad \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {A^2} + {C^2} - 2AC\\&\qquad\qquad \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {A - C} \right)^2} \ge 0 \end{align}\]
Clearly, the roots are real. The roots will be equal if \(A = C\), which means that \(b - c = a - b\) or \(b = \frac{{a + c}}{2}\).
Solved Example 8: What can you say about the nature of roots of the following equation?
\[{\left( {x - 1} \right)^2} + {\left( {x - 2} \right)^2} + {\left( {x - 3} \right)^2} = 0\]
Solution: The sum of three square terms can be zero only if all the three are zero simultaneously. However, this is not possible for the equation above. For example, if the first term is 0 (for \(x = 1\)), then the second and third terms cannot be zero. Thus, directly by observation, we can say that this equation will have no real roots.
Let us verify this. Expanding the brackets and writing the equation in standard quadratic form, we obtain:
\[3{x^2} - 12x + 14 = 0\]
The discriminant is \(D = {12^2} - 4 \times 3 \times 14 = - 24\), which is negative, confirming our observation.
Solved Example 9: What can you say about the nature of roots of the following equation?
\[2\left( {{a^2} + {b^2}} \right){x^2} + 2\left( {a + b} \right)x + 1 = 0\]
Solution: Let us evaluate the discriminant of this equation:
\[\begin{align}&D = 4{\left( {a + b} \right)^2} - 8\left( {{a^2} + {b^2}} \right)\\&= 4\left\{ {{{\left( {a + b} \right)}^2} - 2\left( {{a^2} + {b^2}} \right)} \right\}\\& = 4\left\{ { - {a^2} - {b^2} + 2ab} \right\}\\& = - 4{\left( {a - b} \right)^2}\end{align}\]
Clearly, the discriminant is negative, which means that the given equation has non-real roots.
Solved Example 10: Determine the values of \(k\) for which the equation
\[\frac{{{x^2} + x + 2}}{{3x + 1}} = k\]
has real roots.
Solution: First, we rearrange the given equation and write in the standard quadratic form:
\[\begin{align}&{x^2} + x + 2 = 3kx + k\\&\Rightarrow \,\,\,{x^2} + \left( {1 - 3k} \right)x + \left( {2 - k} \right) = 0\end{align}\]
For this equation to have real roots, its discriminant must be non-negative:
\[\begin{align}&D \ge 0\\& \Rightarrow \,\,\,{\left( {1 - 3k} \right)^2} - 4\left( 1 \right)\left( {2 - k} \right) \ge 0\\&\Rightarrow \,\,\,9{k^2} - 6k + 1 - 8 + 4k \ge 0\\&\Rightarrow \,\,\,9{k^2} - 2k - 7 \ge 0\\&\Rightarrow \,\,\,\left( {9k + 7} \right)\left( {k - 1} \right) \ge 0\\&\Rightarrow \,\,\,k \le - \frac{7}{9}\,\,\,{\rm{or}}\,\,\,k \ge 1\,\end{align}\]
Solved Example 11: (a) Find all integers \(a\) such that the quadratic expression
\[\left( {x - a} \right)\left( {x - 12} \right) + 2\]
can be factored as \(\left( {x + p} \right)\left( {x + q} \right)\), where \(p,\,\,q\) are integers.
(b) For what positive values of a will \(\left( {x - 3} \right)\left( {x + 2} \right) + a = 0\) have integer roots?
Solution: (a) Read the problem statement carefully. It is equivalent to saying that the quadratic equation
\[\left( {x - a} \right)\left( {x - 12} \right) + 2 = 0\]
should have integer roots. Writing this in the standard form, we have:
\[{x^2} - \left( {a + 12} \right)x + 12a + 2 = 0\]
Using the quadratic formula, the roots of this equation can be written as
\[\begin{align}&\alpha ,\,\,\beta = \frac{{\left( {a + 12} \right) \pm \sqrt {{{\left( {a + 12} \right)}^2} - 4\left( {12a + 2} \right)} }}{2}\\& \Rightarrow \,\,\,\alpha ,\,\,\beta = \frac{{\left( {a + 12} \right) \pm \sqrt {{a^2} - 24a + 136} }}{2}\end{align}\]
For \(\alpha ,\,\,\beta \) to be integers, the expression under the square-root sign must firstly be a perfect square, and the overall expression should then also come out to be an integer.
Thus, we first want \({a^2} - 24a + 136\) to be a perfect square. We have:
\[\begin{align}&{a^2} - 24a + 136 = {\left( {a - 12} \right)^2} - 8\\&\Rightarrow \,\,\,{\left( {a - 12} \right)^2} - \left( {{a^2} - 24a + 136} \right) = 8\end{align}\]
On the left side above, both the terms are perfect squares, and their difference is 8. The first few perfect squares are:
0, 1, 4, 9, 16, 25, 36, …
Which pair of perfect squares has a difference of 8? It is easy to see that there is only one such pair, namely (1,9). Thus, we immediately conclude that:
\[\begin{align}&{\left( {a - 12} \right)^2} = 9\\&\Rightarrow \,\,\,a - 12 = \pm 3\\&\Rightarrow \,\,\,a = 9,\,\,15\end{align}\]
For these two values of a, the equation will have integer roots. Let us verify this.
If \(a = 9\), the equation is:
\[\begin{align}&\left( {x - 9} \right)\left( {x - 12} \right) + 2 = 0\\&\Rightarrow \,\,\,{x^2} - 21x + 110 = 0\\&\Rightarrow \,\,\,\left( {x - 11} \right)\left( {x - 10} \right) = 0\end{align}\]
The roots are \(11,\,\,10\).
If \(a = 15\), the equation is:
\[\begin{array}{l}\left( {x - 15} \right)\left( {x - 12} \right) + 2 = 0\\ \Rightarrow \,\,\,{x^2} - 27x + 182 = 0\\ \Rightarrow \,\,\,\left( {x - 13} \right)\left( {x - 14} \right) = 0\end{array}\]
The roots in this case are \(13,\,\,14\).
(b) We follow the same approach as in part-(a). Rewriting the equation in standard form, we have:
\[{x^2} - x - \left( {6 - a} \right) = 0\]
The discriminant in this case is \(D = 25 - 4a\)(verify this!). The positive values of a for which D is a perfect square are \(a = 4,6\).
If \(a = 4\), then \(D = 9\)and the roots are:
\[x = \frac{{1 \pm 3}}{2} = - 1,\,\,2\]
If \(a = 6\), then \(D = 1\) and the roots are:
\[x = \frac{{1 \pm 1}}{2} = 0,\,\,1\]
For both these values of a, the roots are integers.
Solved Example 12: If \(a,\,\,b,\,\,c\) are odd integers, prove that \(a{x^2} + bx + c = 0\) cannot have rational roots.
Solution: We can assume that:
\[\begin{array}{l}a = 2p + 1\\b = 2q + 1\\c = 2r + 1\end{array}\]
\(p,\,\,q,\,\,r\) are some integers. We will proceed by the method of contradiction. Suppose that the roots are rational. Then, the discriminant D must be a perfect square, say \({k^2}\). We have:
\[{k^2} = {b^2} - 4ac\]
Note that k must be odd (why?). Assume that \(k = 2s + 1\). Now, we can write this as follows:
\[\begin{array}{l}{b^2} - {k^2} = 4ac\\ \Rightarrow \,\,\,\left( {b + k} \right)\left( {b - k} \right) = 4ac\\ \Rightarrow \,\,\,\left( {2q + 1 + 2s + 1} \right)\left( {2q - 2s} \right) = 4ac\\ \Rightarrow \,\,\,4\left( {q + s + 1} \right)\left( {q - s} \right) = 4ac\\ \Rightarrow \,\,\,\left( {q + s + 1} \right)\left( {q - s} \right) = ac\end{array}\]
The right hand side is odd. Now, observe the left hand side. What can you say about its parity? Note that the sum of the terms in the two brackets is \(2q + 1\), i.e., an odd number, which means that two terms are of opposite parities, so that one of them must be even. Thus, the left side is even!
We have arrived at a contradiction. This means that our original assumption that D is a perfect square is invalid. D is not a perfect square, and hence the roots of the equation are irrational.