Quadratic Functions (2U, Term 1, Y12) (1 Viewer)

Makro

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I have a few questions, normally I'd just wait till school to ask, but I missed on two lessons so I don't feel falling behind some more.

Topic: Quadratic Functions
Exercise: Equations reducible to quadratics.

1. x4 - 7x2 - 18 = 0

2. (x2 - x)2 + (x2 - x) - 2 = 0 (giving exact values)

3. (x2 + 3x - 1)2 - 7 (x2 + 3x -1) + 10 = 0 ( 2 d.p.)

Full working out would be appreciated, I usually forget a lot of stuff, so if someone does provide an answer could you please write out most of the working out.

Thanks for you time.
 

dwarven

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1. x<sup>4</sup> - 7x<sup>2</sup> - 18 = 0

let x<sup>2</sup> = m

.'. m<sup>2</sup> -7m - 18 = 0
(m - 9)(m+2) = 0
m = 9 or m = -2
ie x<sup>2</sup> = 9, x<sup>2</sup> =/= -2

x = +/- 3
 

gurmies

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(x<SUP>2</SUP> - x)<SUP>2</SUP> + (x<SUP>2</SUP> - x) - 2 = 0

Let u = x^2 - x

u^2 + u - 2 = 0

(u-1)(u+2) = 0

u = 1 or/ u = -2

Now since u = x^2 - x,

x^2 - x = 1 (1) or/ x^2 - x = -2 (2)

x^2 - x -1 = 0

Using quadratic formula, x = (1 +- root (1 + 4))/2 = (1+-root(5))/2

Next case,

x^2 - x = -2

x^2 -x +2 = 0

since discriminant b^2 - 4ac < 0, there are no REAL solutions, only complex conjugates, but that's not important for 2 U.

Next question

(x2 + 3x - 1)^2 - 7 (x2 + 3x -1) + 10 = 0 ( 2 d.p.)

Let u = x^2 + 3x -1

u^2 - 7u + 10 = 0

(u-2)(u-5) = 0

u=2 or/ u=5

x^2 + 3x -1 = 2 or/ x^2 + 3x -1 = 5

x^2 + 3x -1 = 2

x^2 + 3x -3 = 0

x = (-3 +- root (9 + 12))/2 = (-3+-root(21))/2

x = 0.79 or/ x = -3.8

Now for u = 5

x^2 + 3x -1 = 5

x^2 + 3x -6 = 0

x = (-3 +- root(9 + 24))/2 = (-3 +- root(33))/2

x = 1.4 or/ x = -4.4
 
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Makro

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Thank you, the help was much appreciated. :) Why, yes I am doing maths at 3:40 AM on a Sunday morning.
 

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