Answer to Question #163839 in Algebra for Gartiffer

Question #163839

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?


1
Expert's answer
2021-02-28T16:59:49-0500

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

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

= 210<n<210-2\sqrt{10} < n < 2\sqrt{10}


14. Let y=x+1x²+1y = \frac{x+1}{x²+1}

Make x the subject

yx²+y=x+1yx²+y = x+1

yx²x+y1=0yx²-x+y-1=0

From the general quadratic formula

ax²+bx+c=0ax²+bx+c=0

We can say that

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


from the quadratic formula

x=b±b²4ac2ax=\frac{-b±\sqrt{b²-4ac}}{2a}

x=(1)±(1)²4(y)(y1)2yx=\frac{-(-1)±\sqrt{(-1)²-4(y)(y-1)}}{2y}

x=1±14(y²y2yx=\frac{1±\sqrt{1-4(y²-y}}{2y}

x=1±14y²+4y2yx=\frac{1±\sqrt{1-4y²+4y}}{2y}


For the above equation to be defined, then 14y²+4y01-4y²+4y≥0


-4y² + 4y + 1 ≥ 0

4y² - 4y - 1 ≤ 0

Divide through by 4

y²y140y²-y-\frac{1}{4} ≤0

y²y+1414140y²-y+\frac{1}{4}-\frac{1}{4}-\frac{1}{4}≤0

y²y+14120y²-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

=> 22y12-\sqrt{2} ≤ 2y-1≤\sqrt{2}

22y1-\sqrt{2}≤2y-1

2y122y-1≥-\sqrt{2}

y122y≥\frac{1-\sqrt{2}}{2}


=>2y122y-1≤\sqrt{2}

y2+12y ≤ \frac{\sqrt{2}+1}{2}


The range of values is

[122,0)(0,2+12][\frac{1-\sqrt{2}}{2},0) \bigcup (0, \frac{\sqrt{2}+1}{2}]


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