# 52. Quadratic Equations Solutions Part 7¶

1. $$f(11) = 1331 - 1210 - 121 - 100 < 0$$

$$f(12) = 1728 - 1440 - 132 - 100 > 0$$

$$f'(x) = 3x^2 - 20x - 11$$

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 $$11$$.

2. $$f(0) = a_n$$ 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.

3. 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 $$\gamma.$$

4. 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 $$-1, 0, 4$$

$$\Rightarrow a = 0, a = 3, a = 128$$

$$\therefore 0\le a \le 3$$

5. This problem is similar to 226 and can be solved similarly.

6. $$f'(x) = 5(x - 1)^4 + 7(x + 2)^6 + 9(7x - 5)^8$$ can never be zero for any real value of $$x$$ therefore the given equation can have only one real real root at maximum.

7. 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.

1. $$\sum_{i = 1}^n(x - a_i)^2 = nx^2 - 2(a_1 + a_2 + ... + a_n)x + (a_1^2 + a_2^2 + ... + a_n^2)$$

This is a quadratic equation in $$x$$ and coefficient of $$x^2$$ is $$n > 0,$$ therefore this quadratic equation will have least value at $$x = \frac{a_1 + a_2 + ... + a_n}{n}$$

2. 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 $$\frac{4}{15}.$$

3. 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 $$x.$$

4. 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.

5. 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 $$2 - \sqrt{5}<m<-2+\sqrt{5}$$

6. 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 $$\frac{5}{2}<a<\infty$$

7. This problem is similar as last one and has been left as an exercise.

8. 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.

9. This inequality is system of two inequalities. First we do one.

1. $$x^2 - \frac{3}{16} > 0 \Rightarrow x < -\frac{\sqrt{3}}{4}$$ or $$x > \frac{\sqrt{3}}{4}$$

2. $$x > 1$$

3. $$x^2 - \frac{3}{16}> x^4 \Rightarrow 16x^4 - 16x^2 + 3 < 0$$

$$(4x^2 -3)(4x^2 - 1) < 0 \Rightarrow \frac{1}{4}<x^2\frac{3}{4}$$

$$x^2 > \frac{1}{4} \Rightarrow x < -\frac{1}{2}$$ 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.

10. 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 $$\sqrt{7}\le m < \infty$$

11. 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 $$x$$

1. $$D\leq 0 \Leftrightarrow 16 - 4(5 - a)^2 \leq 0 \Leftrightarrow (7 - a)(a - 3)\leq 0$$

$$\Leftrightarrow a\leq 3$$ or $$a\geq 7$$

2. $$5 - a > 0 \Leftrightarrow a < 5$$

Combining from (i) we have $$a\leq 3$$

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 $$2 < a \leq 3$$

12. 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.