plz only expert have to solve it 1- let a and b be positive integer show that root - Maths - Real Numbers

Question

plz only expert have to solve it
1- let a and b be positive integer show that root 2 always
lies between a upon b and a+2b upon a+b
2- factorize it
1- x2-x-1
2- x2+60x+36
plz factorize it as soon as possible
u have to direct factorise it not find the roots

Lalit Mehra · Accepted Answer

Given, a and b are positive integers $\frac{a}{b} - \frac{a + 2 b}{a + b} = \frac{a^{2} + a b - a b - 2 b^{2}}{b (a + b)} = \frac{a^{2} - 2 b^{2}}{b (a + b)} \frac{a}{b} - \frac{a + 2 b}{a + b} > 0 ∴ \frac{a^{2} - 2 b^{2}}{b (a + b)} > 0 \Rightarrow a^{2} - 2 b^{2} > 0 (∵ b (a + b) > 0) \Rightarrow a^{2} > 2 b^{2} \Rightarrow a > \sqrt{2} b \frac{a}{b} - \frac{a + b}{a + b} < 0 ∴ \frac{a^{2} - 2 b^{2}}{b (a + b)} < 0 \Rightarrow a^{2} - 2 b^{2} < 0 \Rightarrow a^{2} < 2 b^{2} \Rightarrow a < \sqrt{2} b So, \frac{a}{b} > \frac{a + 2 b}{a + b} if a > \sqrt{2} b and \frac{a}{b} < \frac{a + 2 b}{a + b} if a < \sqrt{2} b When a > \sqrt{2} b, \frac{a + 2 b}{a + b} < \frac{a}{b} We need to prove \frac{a + 2 b}{a + b} < \sqrt{2} < \frac{a}{b} a < \sqrt{2} b \Rightarrow a^{2} > 2 b^{2} \Rightarrow a^{2} + a^{2} > a^{2} + 2 b^{2} \Rightarrow 2 b^{2} + 2 b^{2} > (a^{2} + 2 b^{2}) + 2 b^{2} ({Adding 2 b}^{2} to both sides) \Rightarrow 2 (a^{2} + b^{2}) + 4 a b > a^{2} + 4 b^{2} + 4 a b (Adding 4 a b to both sides) \Rightarrow 2 (a^{2} + b^{2} + 2 a b) > (a + 2 b)^{2} \Rightarrow 2 (a + b)^{2} > (a + 2 b)^{2} \Rightarrow (\sqrt{2} (a + b))^{2} > (a + 2 b)^{2} \Rightarrow \sqrt{2} (a + b) < a + 2 b \Rightarrow \sqrt{2} > \frac{a + 2 b}{a + b} (1) a > \sqrt{2} b ∴ \frac{a}{b} > \sqrt{2} (2)$ From (1) and (2), we get $\frac{a + 2 b}{a + b} < \sqrt{2} < \frac{a}{b} (3)$ Similarity when a < 2b,we have $\frac{a}{b} < \sqrt{2} < \frac{a + 2 b}{a + b} (4)$ From (3) and (4), we have $\sqrt{2} lies between \frac{a}{b} and \frac{a + 2 b}{a + b}$ 2 x2 â€“ x â€“ 1 $= x^{2} (\frac{\sqrt{5} - 1}{2} - \frac{\sqrt{5} + 1}{2}) x - (\frac{\sqrt{5} - 1}{2}) (\frac{\sqrt{5} + 1}{2}) = x^{2} (\frac{\sqrt{5} - 1}{2}) x - (\frac{\sqrt{5} + 1}{2}) x - (\frac{\sqrt{5} - 1}{2}) (\frac{\sqrt{5} + 1}{2}) = [x + (\frac{\sqrt{5} - 1}{2})] - (\frac{\sqrt{5} + 1}{2}) [x + (\frac{\sqrt{5} - 1}{2})] = [x + (\frac{\sqrt{5} - 1}{2})] [x - (\frac{\sqrt{5} + 1}{2})]$ The equation x2 + 60x + 36 = 0 cannot be factorised without finding the roots of the given equation

plz only expert have to solve it 1- let a and b be positive integer show that root 2 always lies between a upon b and a+2b upon a+b 2- factorize it 1- x2-x-1 2- x2+60x+36 plz factorize it as soon as possible u have to direct factorise it not find the roots

plz only expert have to solve it

1- let a and b be positive integer show that root 2 always lies between a upon b and a+2b upon a+b

2- factorize it

1- x2-x-1

2- x2+60x+36

plz factorize it as soon as possible

u have to direct factorise it not find the roots