Solve Quadratic Equation: 5x^2 - 16tx + 3t^2

  • Thread starter Thread starter Atomised
  • Start date Start date
  • Tags Tags
    Quadratic Stuck
AI Thread Summary
To solve the quadratic equation 5x^2 - 16tx + 3t^2 = 0 without using the quadratic formula, factoring is a viable method. The equation can be factored into (5x - t)(x - 3t) = 0, leading to the solutions x = t/5 and x = 3t. The discussion highlights the importance of setting the equation equal to zero for proper analysis. Additionally, simplifying the problem by substituting t = 1 can clarify the factoring process. Understanding these techniques can help bridge gaps in foundational math skills.
Atomised
Gold Member
Messages
45
Reaction score
0

Homework Statement



Solve 5x^2 - 16tx + 3t^2 for x without using quadratic formula.

I am interested in understanding the technique - I have the answers already: t/5, 3t

Homework Equations



I have solved the subsequent question which is:

tx^2 + (tT - 1)x - T = 0, which seems easier because it breaks into four clusters which can be factored, then refactored and solved. The one I am asking about does not seem amenable to the same approach, or any other that I know.

The Attempt at a Solution



I am interested in higher maths but having to mend holes in elementary technique. Thanks in advance.
 
Last edited:
Physics news on Phys.org
Atomised said:

Homework Statement



Solve 5x^2 - 16tx + 3t^2 for x without using quadratic formula.

I am interested in understanding the technique - I have the answers already: t/5, 3t

Homework Equations



I have solved the subsequent question which is:

tx^2 + (tT - 1)x - T = 0, which seems easier because it breaks into four clusters which can be factored, then refactored and solved. The one I am asking about does not seem amenable to the same approach, or any other that I know.




The Attempt at a Solution



I am interested in higher maths but having to mend holes in elementary technique. Thanks in advance.

is the whole equation 5x^2 - 16tx + 3t^2=0?
 
Since 5 and 3 are both prime, you only have 4 different possibilities of how the terms appear when you use the factor method. Perhaps the variable t is throwing you off? Start by letting t = 1 just so it's clearer to you and then go back and solve the equation with t.

Also, you can only solve for a variable if you have an equation. The original expression must be set equal to 0.
 
  • Like
Likes 1 person
Yes - the only other info stated in the question is 'Express x in terms of the other letters'.
 
Ah apologies - I have only stated the LHS, the equation is

5x^2 - 16tx + 3t^2=0
 
Last edited:
Thank you Scurty, the t was throwing me.

(5x - t) (x - 3t)
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top