Is there an easier way to do it?

  • Thread starter Thread starter davedave
  • Start date Start date
AI Thread Summary
The equation x(x-1)(x-2)(x-3)=840 can be approached by finding factors of 840 or graphing both sides to identify the intersection at x=7. While trial and error is one method, it can be cumbersome. A more efficient approach involves rewriting the equation as a biquadratic by substituting y = x-1.5, leading to a polynomial that simplifies the problem and reveals integer roots, specifically x=7 or x=-4. Overall, while there are multiple methods to solve the equation, leveraging its symmetry can provide a clearer path to the solution.
davedave
Messages
50
Reaction score
0
Consider the equation x(x-1)(x-2)(x-3)=840

One way to do it is to find factors of 840. By trial and error, 7 which is a factor of 840 satisfies the equation.

Another way is to graph both sides of the equation and find the intersection which is 7.

Is there an easier way to find the answer without using the graphing calculator and the trial and error method?
 
Mathematics news on Phys.org
One way to do it is to find factors of 840.

Only if the equation was helpfully constructed to have integer roots - otherwise factors won't help.

Generally speaking, fourth degree equations are hard to solve. In this particular case, the left hand side has a nice symmetry that allows you to rewrite it as a biquadratic equation: substitute y = x-1.5, express the left hand side as a polynomial, and you get a nice biquadratic equation with roots y = +/- 5.5 and therefore x=7 or x=-4.
 
Notice that 292 = 841

If we set B = (x-3)(x-2)(x-1)x

(x-2)(x-1) = x2 - 3x + 2

(x-3)x = x2 - 3x

(x-2)(x-1) - (x-3)x = 2

A = (x-2)(x-1) - 1 = x2 - 3x + 1

A + 1 = (x-2)(x-1)

(x-3)x = A - 1

B = A2 - 1

(x-3)(x-2)(x-1)x + 1 = (x2 - 3x + 1)2 = 292

We can now solve for x.
 
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Back
Top