Question

13.2x²+ nx + 5 = 0 has imaginary roots. The range of values of n is?

14.The range of values of, (x + 1)/ (x² + 1)  X e R is?

Answer 1

13. For any quadratic equation to have imaginary roots, b² – 4ac must be less than 0

b² – 4ac < 0

a = 2 ; b = n ; c = 5

n² – 4(2)(5) < 0

n² – 40 < 0

n² < 40

=> $-\sqrt{40} < n < \sqrt{40}$

= $-2\sqrt{10} < n < 2\sqrt{10}$

14. Let $y = \frac{x+1}{x²+1}$

Make x the subject

$yx²+y = x+1$

$yx²-x+y-1=0$

From the general quadratic formula

$ax²+bx+c=0$

We can say that

a = y ; b = -1; c = y-1

from the quadratic formula

$x=\frac{-b±\sqrt{b²-4ac}}{2a}$

$x=\frac{-(-1)±\sqrt{(-1)²-4(y)(y-1)}}{2y}$

$x=\frac{1±\sqrt{1-4(y²-y}}{2y}$

$x=\frac{1±\sqrt{1-4y²+4y}}{2y}$

For the above equation to be defined, then $1-4y²+4y≥0$

-4y² + 4y + 1 ≥ 0

4y² - 4y - 1 ≤ 0

Divide through by 4

$y²-y-\frac{1}{4} ≤0$

$y²-y+\frac{1}{4}-\frac{1}{4}-\frac{1}{4}≤0$

$y²-y+\frac{1}{4}-\frac{1}{2}≤0$

Multiply through by 4

(4y² – 4y + 1) – 2 ≤ 0

(2y – 1)² – 2 ≤ 0

(2y – 1)² ≤ 2

=> $-\sqrt{2} ≤ 2y-1≤\sqrt{2}$

$-\sqrt{2}≤2y-1$

$2y-1≥-\sqrt{2}$

$y≥\frac{1-\sqrt{2}}{2}$

=>$2y-1≤\sqrt{2}$

$y ≤ \frac{\sqrt{2}+1}{2}$

The range of values is

$[\frac{1-\sqrt{2}}{2},0) \bigcup (0, \frac{\sqrt{2}+1}{2}]$