Question

please solve this.
very urgent
I want both of them​

Answer 1

1) Given equation

(x-1)² - 3x + 4 = 0

(x-1)² - 3x + 3 + 1 = 0

(x-1)² - 3(x-1) + 1 = 0

Let (x-1) = t

So equation becomes

t² - 3t + 1 = 0

$t = \frac{3+\sqrt{9-4} }{2} \ \ \ OR \ \ \ t = \frac{3-\sqrt{9-4} }{2}\\\\t = \frac{3+\sqrt{5} }{2} \ \ \ OR \ \ \ t = \frac{3-\sqrt{5} }{2} \\ \\t = 2.6 \ \ \ OR \ \ \ t = 0.4$

Since x = t + 1

x = 2.6 + 1     OR     x = 0.4 + 1

x = 3.6     OR     x = 1.4

2)

$x-\frac{18}{x} = 6\\ \\x^2-18 = 6x\\\\x^2-6x-18 = 0\\\\x = \frac{6+\sqrt{36+72} }{2} \ \ \ OR \ \ \ x = \frac{6-\sqrt{36+72} }{2}\\\\x = 8.2 \ \ \ OR \ \ \ x = 2.2$