Irrational Quartic

Algebra Level 2

( x − 1 ) ( x − 2 ) ( 3 x − 2 ) ( 3 x + 1 ) = 21 (x-1)(x-2)(3x-2)(3x+1)=21

How many irrational roots exists for the above equation ?


The answer is 2.

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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3 solutions

Anand Raj
Nov 5, 2014

Solving,

[(x-1)(3x-2)] and [(x-2)(3x+1)] separately and Substituting "A" For What You Know What I Mean We Get

(A-2)(A+2) = 21

Which gives A= 5 or -5 Which gives us 4 equations ....... Solving We Get 2 Complex And 2 Irrational Roots

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No, I don't know what you mean. A=? I did it by expanding the expression and then completing the square, and I got ( 3 x 2 − 5 x ) = ± 5 (3x^2-5x)=\pm 5 and then solved the four equations obtained.

Prasun Biswas - 6 years, 6 months ago

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( x − 1 ) ( 3 x − 2 ) = 3 x 2 − 5 x + 2 (x-1)(3x-2)=3x^2-5x+2 , ( x − 2 ) ( 3 x + 1 ) = 3 x 2 − 5 x − 2 (x-2)(3x+1)=3x^2-5x-2

Then he substituted A = 3 x 2 − 5 x A=3x^2-5x

Pranjal Jain - 6 years, 5 months ago
William Isoroku
Dec 3, 2014

Simple, just graph it!

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Even simpler it can have 4 irrational roots at max... Irrational roots always exist pairs so it would be 2 or 4.. I cannotuunderstand why it isn't 4

Siddhartha Kapoor - 6 years, 3 months ago
Jesse Nieminen
Aug 17, 2015

( x − 1 ) ( x − 2 ) ( 3 x − 2 ) ( 3 x + 1 ) = 21 {(x-1)(x-2)(3x-2)(3x+1)=21} ⇒ 9 x 4 − 30 x 3 + 25 x 2 − 25 = 0 \Rightarrow {9{x}^{4} - 30{x}^{3} + 25{x}^{2} - 25 = 0} Using Lodovico Ferrari's Method, we can find the roots of this polynomial: ⇒ x = 5 ± 85 6 ∨ 5 ± i 35 6 {\Rightarrow x = \frac{5 \pm \sqrt{85} }{6} \vee \frac{5 \pm i\sqrt{35} }{6} } Since two of these roots are irrational the answer is 2 \boxed{2} .

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Can you explain what is the Lodovico Ferrari's method, and how you applied it?

9 x 4 − 30 x 3 + 25 x 2 − 25 = 0 {9x^4 - 30x^3 + 25x^2 - 25 = 0}

Divide by 9 \text{Divide by 9}

x 4 − 10 3 x 3 + 25 9 x 2 − 25 9 = 0 x^4 - \frac{10}{3} x^3 + \frac{25}{9}x^2 - \frac{25}{9} = 0

Getting rid of x 3 \text{Getting rid of } x^3

5 6 = 10 3 4 \frac{5}{6} =\frac{\frac{10}{3}}{4} x = y + 5 6 x = y +\frac{5}{6}

Ferrari’s method is not even needed here. \text{Ferrari's method is not even needed here.}

y 4 − 2518 y 2 − 2975 1296 = 0 y^4 - {25}{18} y^2 - \frac{2975}{1296} = 0

Using Quadratic Formula we get: \text{Using Quadratic Formula we get:}

y = ± 85 6 ∨ ± i 35 6 y = \frac{\pm\sqrt{85}}{6} \vee \frac{\pm i\sqrt{35}}{6}

x = 5 ± 85 6 ∨ 5 ± i 35 6 x = \frac{5 \pm \sqrt{85}}{6} \vee \frac{5 \pm i\sqrt{35}}{6}

I’m too lazy to explain the Ferrari’s Method, but here is an url where explanation can be found: \text{I'm too lazy to explain the Ferrari's Method, but here is an url where explanation can be found:}

http://mathforum.org/library/drmath/view/51899.html

Jesse Nieminen - 5 years, 9 months ago

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