59. Quadratic Equations Solutions Part 7#
\(f(11) = 1331 - 1210 - 121 - 100 < 0\)
\(f(12) = 1728 - 1440 - 132 - 100 > 0\)
\(f'(x) = 3x^2 - 20x - 11\)
Clearly, \(f'(x) > 0~\forall~x > 12\)
Therefore, \(f(x)\) will have its greatest root between \(11\) and \(12\). Hence, integral part of greatest root is \(11\).
\(f(0) = a_n\) and \(f(1) = a_0 + a_1 + ... + a_n\) which are both odd numbers.
\(f(2k)=a_n+2(a_{n - 1}k)+2(2a_{n - 2}k^2)+\ldots +2(2^{n-1}k^na_n)\) is odd for all \(k\) and
\(f(2k+1)=\sum_{k=1}^na_k\sum_{i=1}^k{k\choose i}(2k)^i\)
Since all the inner summands are even except when \(i = 0\) we see that
\(f(2k+1)=2N+\sum_{k=1}^na_k\)
which is even plus add i.e. odd for all values of \(k\).
Thus, for no integer the function is zero hence the function has no integral roots.
Let \(g(x) = e^xf(x)\) then \(g"(x) = e^x[f(x) + 2f'(x) + f"(x)]\)
Therefore, roots of \(g"(x) = 0\) will be same as \(f(x) + 2f'(x) + f"(x) = 0\) as \(e^x \ne 0\)
But since \(e^x \ne 0, g(x)\) will have same roots as \(f(x).\)
Since \(f(x) = 0\) has three roots \(\alpha < \beta < \gamma,\) therefore equation \(g(x) = 0\) will have same roots and \(g'(x) = 0\) will have a root between \(\alpha\) and \(\beta\) and other roots between \(\beta\) and \(\gamma\) and hence \(g"(x)\) will have a root between \(\alpha\) and \(\gamma.\)
If this degree four polynomial has four real roots, then there is a local minimum between the first two and between the last two roots, and a local maximum between the second and hird root. The location of the extrema does not change with \(a\), only their \(y\) values. Thus we are looking for \(a\) such that the two local minima have negative \(y\) values and the local maximum has positive \(y\) value.
Locate the local extrema as roots of the derivative \(4x^3 - 12x^2 - 16x\) which has roots as \(-1, 0, 4\)
\(\Rightarrow a = 0, a = 3, a = 128\)
\(\therefore 0\le a \le 3\)
This problem is similar to 226 and can be solved similarly.
\(f'(x) = 5(x - 1)^4 + 7(x + 2)^6 + 9(7x - 5)^8\) can never be zero for any real value of \(x\) therefore the given equation can have only one real real root at maximum.
Let \(x + 2 = t^2\) then \(\sqrt{2(t^2 + 1)} - t = 3 \Rightarrow 2t^2 + 2 = 9 + 6t + t^2\)
\(\Rightarrow t = 7, -1\)
Now it is trivial to calculate the values. 308 through 317 have been left as exercises.
\(\sum_{i = 1}^n(x - a_i)^2 = nx^2 - 2(a_1 + a_2 + ... + a_n)x + (a_1^2 + a_2^2 + ... + a_n^2)\)
This is a quadratic equation in \(x\) and coefficient of \(x^2\) is \(n > 0,\) therefore this quadratic equation will have least value at \(x = \frac{a_1 + a_2 + ... + a_n}{n}\)
Let the quotient be \(\frac{n}{n^2 - 1},\) where \(n\) is a natural number.
According to question,
\(\frac{n + 2}{n^2 - 1 + 2} > \frac{1}{3 \Rightarrow n^2 - 3n - 5 < 0}\)
\(\Rightarrow \frac{3}{2} - \frac{\sqrt{29}}{2} < n < \frac{3}{2} + \frac{\sqrt{29}}{2}\)
Also, \(0 < \frac{n - 3}{n^2 - 1 - 3} < \frac{1}{10}\)
\(\Rightarrow -2 < n < 2\) or \(3 < n < \infty\) or \(n > 2\)
\(n = 4\) is the only natural number satisfying all the conditions. Therefore, quotient is \(\frac{4}{15}.\)
Let \(f(x) = ax^2 + bx + c\) then \(g(x) = ax^2 + (2a + b)x + c + b + 2a\)
Given, \(ax^2 + bx + c > 0\) for all \(x\) \(\therefore b^2 - 4ac < 0\) and \(a > 0\)
Discriminant of \(g(x) = 0\)
\(D = (b + 2a)^2 - 4a(c + b + 2a) = b^2 - 4ac - 4a^2 < 0\) and \(a > 0\)
Hence, \(g(x) > 0\) for all \(x.\)
Clearly, \(f(x) \ge 0\) for all \(x\) then \(D \le 0\) because coefficient of \(x^2\) is positive.
\(4(a_1b_1 + a_2b_2 + ... + a_nb_n)^2 \le 4(a_1^2 + a_2^2 + ... + a_n^2)(b_1^2 + b_2^2 + ... + b_n^2)\)
Hence, proven.
Given equation is \([(x+1)(x+m)][x(x+m+1)] = m^2\)
\((x^2 + (m + 1)x + m)(x^2 + (m + 1)x) = m^2\)
\((y + m)y = m^2\) where \(y = x^2 + (m + 1)x\)
\(y^2 + my - m^2 = 0 \therefore y = \frac{-m \pm \sqrt{5}m}{2}\)
\(\Rightarrow x^2 + (m + 1)x = \frac{-m \pm \sqrt{5}m}{2}\)
These two equations must have real roots only then original will have four. Now it is trivially obtained that
\(-\infty<m<-2-\sqrt{5}\) or \(2 + \sqrt{5}<m<\infty\) or \(2 - \sqrt{5}<m<-2+\sqrt{5}\)
Given equation is \(x^4 + (a - 1)x^3 + x^2 + (a - 1)x + 1 = 0\)
Dividing both sides from \(x^2\) we have
\(x^2 + (a - 1)x + 1 + (a - 1)\frac{1}{x} + \frac{1}{x^2} = 0\)
\(\Rightarrow \left(x+\frac{1}{x}\right)^2 - 2x\frac{1}{x} + (a - 1)\left(x + \frac{1}{x}\right) + 1 = 0\)
\(\Rightarrow y^2 + (a - 1)y - 1 = 0\) where \(y = x + \frac{1}{x}\)
\(\therefore y = \frac{-(a - 1)\pm \sqrt{(a - 1)^2 + 4}}{2}\)
\(\Rightarrow x + \frac{1}{x} = \frac{-(a - 1)\pm \sqrt{(a - 1)^2 + 4}}{2}\)
\(2x^2 + [(a - 1) - \sqrt{(a - 1)^2 + 4}]x + 2 = 0\) and
\(2x^2 + [(a - 1) + \sqrt{(a - 1)^2 + 4}]x + 2 = 0\) are two equations we have.
Let \(\alpha, \beta\) are roots for first and \(\gamma, \delta\) are roots for second.
\(\therefore \alpha + \beta = -\frac{[(a - 1) - \sqrt{(a - 1)^2 +4}]}{2}\) and \(\alpha\beta = 1\)
\(\therefore \gamma + \delta = -\frac{[(a - 1) + \sqrt{(a - 1)^2 +4}]}{2}\) and \(\gamma\delta = 1\)
\(\because \sqrt{(a - 1)^2 + 4}> a - 1\)
\(\alpha + \beta > 0\) and \(\alpha\beta > 0\) hence both roots cannot be negative.
Thus, \(\gamma\) and \(\delta\) have to be negative as per requirement from question. They have to be real and distinct because \(\gamma + \delta < 0\) and \(\gamma\delta >0\) do not imply that \(\gamma<0\) and \(\delta < 0\).
For both these to be negative discriminant has to be positive.
Thus, \([(a - 1) + \sqrt{(a - 1)^2 + 4}] - 16 > 0\)
Solving this we get \(\frac{5}{2}<a<\infty\)
This problem is similar as last one and has been left as an exercise.
Given system of equations can be written as:
\(ax_1^2 + (b - 1)x_1 + c = x_2 - x_1\)
\(ax_2^2 + (b - 1)x_2 + c = x_3 - x_2\)
\(\ldots\)
\(ax_n^2 + (b - 1)x_n + c = x_1 - x_n\)
Let \(f(x) = ax^2 + (b - 1)x + c\) then we have
\(f(x_1) = x_2 - x_1\)
\(f(x_2) = x_3 - x_2\)
\(\ldots\)
\(f(x_n) = x_1 - x_n\)
Case I: When \((b - 1)^2 - 4ac < 0\):
In this case all of \(f(x_1), f(x_2), \ldots, f(x_n)\) will have same sign as that of \(a\).
\(\therefore f(x_1) + f(x_2) + \ldots + f(x_n) \ne 0.\)
Hence, the system of equation have no solution.
Case II: When \((b - 1)^2 - 4ac = 0\)
In this case all of \(f(x_1), f(x_2), \ldots, f(x_n) \ge 0\) or \(f(x_1), f(x_2), \ldots, f(x_n) \le 0\)
But \(f(x_1) + f(x_2) + \ldots + f(x_n) = 0\) implies
\(f(x_1) = f(x_2) = \ldots = f(x_n) = 0\)
Then, we have \(f(x_k)=0\Rightarrow ax_k^2 + (b - 1)x_k + c = 0\)
\(\Rightarrow x_k = \frac{1 - b}{2a}\)
Case III: When \((b - 1)^2 - 4ac > 0:\)
Roots are \(\alpha,\beta = \frac{1 - b + \sqrt{(1 - b)^2 - 4ac}}{2a}\)
Now it is trivial to find rest from sign scheme.
This inequality is system of two inequalities. First we do one.
\(x^2 - \frac{3}{16} > 0 \Rightarrow x < -\frac{\sqrt{3}}{4}\) or \(x > \frac{\sqrt{3}}{4}\)
\(x > 1\)
\(x^2 - \frac{3}{16}> x^4 \Rightarrow 16x^4 - 16x^2 + 3 < 0\)
\((4x^2 -3)(4x^2 - 1) < 0 \Rightarrow \frac{1}{4}<x^2\frac{3}{4}\)
\(x^2 > \frac{1}{4} \Rightarrow x < -\frac{1}{2}\) or \(x > \frac{1}{2}\)
and \(x^2 < \frac{3}{4} \Rightarrow -\frac{\sqrt{3}}{2}< x < \frac{\sqrt{3}}{2}\)
Second system of inequality is left as an exercise.
Given \(\log_{\frac{1}{2}}x^2 \ge \log_{\frac{1}{2}}(x + 2)\)
\(\Rightarrow x^2 \le x + 2\) [ \(\because\) base \(< 1\) ]
\(\Rightarrow x^2 - x - 2 \le 0\)
\(\Rightarrow -1 \le x \le 2, x \ne 0\)
[For logarithms to be defined \(x\ne 0\) and \(x>-2\)]
Again, \(49x^2 - 4m^4 < 0\)
\(-\frac{2}{7}m^2\le x \le \frac{2}{7}m^2\)
According to question \([-1, 2] \subset \left[-\frac{2}{7}m^2, \frac{2}{7}m^2\right]\)
Now it is trivial to solve for various values of \(m\) which reader can compute to be \(-\infty<m \le -\sqrt{7}\) or \(\sqrt{7}\le m < \infty\)
Given, \(1 + \log_5(x^2 + 1)\geq \log_5(ax^2 + 4x + a)\)
\(\Leftrightarrow \log_55 + \log_5(x^2 + 1)\geq \log_5(ax^2 + 4x + a)\)
\(\Leftrightarrow \log_55(x^2 + 1)\geq \log_5(ax^2 + 4x + a)\)
\(\Leftrightarrow 5(x^2 + 1)\geq ax^2 + 4x + a\)
\(\Leftrightarrow (5 - a)x^2 - 4x + 5 -a \geq 0~~\forall\) real \(x\)
\(D\leq 0 \Leftrightarrow 16 - 4(5 - a)^2 \leq 0 \Leftrightarrow (7 - a)(a - 3)\leq 0\)
\(\Leftrightarrow a\leq 3\) or \(a\geq 7\)
\(5 - a > 0 \Leftrightarrow a < 5\)
Combining from (i) we have \(a\leq 3\)
But for \(\log_5(ax^2 + 4x + a)\) to be defined
\(ax^2 + 4x + a > 0~\forall\) real \(x\)
\(D < 0 \Leftrightarrow 16 - 4a^2 < 0 \Leftrightarrow a < -2\) or \(a > 2\)
and \(a > 0\)
Combining from (ii) we have \(2 < a \leq 3\)
Let \(a - c = \alpha, b - c = \beta\) and \(c + x = u\) then for \(\sqrt{a - c}\) and and \(\sqrt{b - c}\) to be real \(\alpha \geq 0\) and \(\beta \geq 0\). Also as, \(x > -c \therefore u > 0\)
Let \(y = \frac{(a + x)(b + x)}{c + x} = \frac{(u + \alpha)(u + \beta)}{u}\)
\(y = u + \frac{\alpha\beta}{u} + \alpha + \beta\)
\(u^2 + (\alpha + \beta - y)u + \alpha\beta = 0\)
Since \(u\) is real \(\therefore (\alpha + \beta - y)^2 - 4\alpha\beta \geq 0\)
\(y^2 -2(\alpha + \beta)y + (\alpha - \beta)^2 \geq 0\)
Roots are \(y = (\sqrt{\alpha} - \sqrt{\beta})^2, (\sqrt{\alpha} + \sqrt{\beta})^2\)
\(y \leq (\sqrt{\alpha} - \sqrt{\beta})^2\) and \(y \geq (\sqrt{\alpha} + \sqrt{\beta})^2\)
But \(y \leq (\sqrt{\alpha} - \sqrt{\beta})^2 \Leftrightarrow y - (\alpha + \beta) + 2\sqrt{\alpha\beta}\leq 0\) which is not possible as \(y - \alpha - \beta = u + \frac{\alpha\beta}{u}>0\)
Hence, least value of y = \((\sqrt{a - c} + \sqrt{b - c})^2\)
Rest of the problems have been left as exercise.