Call the left hand side f(x). By testing out values (or by graphing the parabola), it's easy to see there is one solution to f(x)=0 with x less than 1, and one with x greater than 1. f(0) = 1 > 0 f(1) = -4 < 0 f(5) = -4 < 0 f(6) = 1 > 0 Because f(x) is a continuous function, there has to be one solution to f(x)=0 with 0
Before watching the video clip: The original equation is: x^2 - 6x + 1 = 0 x = 0 is not a solution. Therefore, by dividing by x we get: x + 1/x = 6 = (√x - 1/√x)^2 + 2 Therefore: √x - 1/√x = ± 2
Let √x-1/√x=y Squaring both sides, y²=x+(1/x)-2 From the given equation x²-6x+1=0 Thus x²+1=6x Dividing by x on both sides, x+(1/x)=6 Put up,y²=6-2=4=±2
I am very proud of myself that I was able to simplify the rt(3+2rt2) in my head thanks to your videos!! Thanks, Sy!
Call the left hand side f(x). By testing out values (or by graphing the parabola), it's easy to see there is one solution to f(x)=0 with x less than 1, and one with x greater than 1.
f(0) = 1 > 0
f(1) = -4 < 0
f(5) = -4 < 0
f(6) = 1 > 0
Because f(x) is a continuous function, there has to be one solution to f(x)=0 with 0
Before watching the video clip:
The original equation is:
x^2 - 6x + 1 = 0
x = 0 is not a solution. Therefore, by dividing by x we get:
x + 1/x = 6 = (√x - 1/√x)^2 + 2
Therefore: √x - 1/√x = ± 2
Let √x-1/√x=y
Squaring both sides,
y²=x+(1/x)-2
From the given equation
x²-6x+1=0
Thus x²+1=6x
Dividing by x on both sides,
x+(1/x)=6
Put up,y²=6-2=4=±2
Thx so much dear teacher 🙏
You are very welcome
Went straight to method 2. The negative solution is surely OK as the first equation is based on the square of the second.
x^2 - 6x +1 = 0 (x not null) hence x + 1/x = 6. Next, (sqr(x) - sqr(1/x))^2 = x + 1/x - 2 = 6-2 = 4. Hence, sqr(x) - sqr(1/x) = +/- 2. With f(x) = x^2 - 6x +1 and f'(x) = 2x-6 min f(x) at x=3 hence 2 solutions -1 < x < 0 and 5 < x < 6 thus 2 solutions sqr(x) - sqr(1/x) = +/- 2.
whats the app youre using to make the videos?
He’s using notability.
Thanks! @@scottleung9587
without watching
√x - (1/√x)
= √[√x - (1/√x)]²
= √[x - 2 + (1/x)]
= √[(1/x)(x² - 2x + 1 - 4x + 4x)]
= √[(1/x)(x² - 6x + 1 + 4x)]
= √[(1/x)(4x)]
= 2
Let t=x^(1/2).
t⁴-6t²+1=0
(t²-2t-1)(t²+2t-1)=0
t=sqrt2-1 or t=sqrt2+1
Therefore 2 or -2.
From the thumbnail:
x² - 6x + 1 = 0
√x - 1/√x = ???
x² - 6x + 1 = 0
x² - 6x + 9 = 8
(x - 3)² = 8
x - 3 = ±2√2
x = 3 ± 2√2
x = 1 + 2 ± 2√2
x = (1 ± √2)²
√x = | 1 ± √2 |
√x = √2 ± 1
Note that (√2 + 1)(√2 − 1) = 2 − 1 = 1 ==>
√x - 1/√x = (√2 + 1) − (√2 − 1) = 2
OR
√x - 1/√x = (√2 − 1) − (√2 + 1) = −2
I got 2 but not -2.