Solving $3t^2-\frac{12}{3}t+\frac{4}{3}=0$

Question

I need to to solve: $$3t^2-\frac{12}{3}t+\frac{4}{3}=0$$

The solution manual factorizes this to $\dfrac{1}{3}(3t-2)^2$. How can you do this easily?

Why not use quadratic fomula – Aditya Hase Dec 08 '14 at 11:29 — Aditya Hase, Dec 08 '14 at 11:29
Start multiplying everything by $3$. It looks nicer. – Claude Leibovici Dec 08 '14 at 11:45 — Claude Leibovici, Dec 08 '14 at 11:45

score 2 · Answer 1 · edited Dec 08 '14 at 11:32

2

Take $\dfrac13$ common, to get this -

$\dfrac13\left({9t^2 - 12t + 4}\right)$ Using $(a + b)^2 = a^2 + b^2 + 2ab$, we can write the latter bracket as $(3t -2)^2$

edited Dec 08 '14 at 11:32

Aditya Hase

8,851

answered Dec 08 '14 at 11:30

Ayan

101

1

Refer to http://math.stackexchange.com/help/notation to see how to use MathJax here. – AlexR Dec 08 '14 at 11:31
@Integrator I didn't vote FYI. – AlexR Dec 08 '14 at 11:32
Yes thank you, I will go to that link right away. – Ayan Dec 08 '14 at 11:33
1

@Integrator Very funny: The help page features the formula from your answer ^^ – AlexR Dec 08 '14 at 11:34

score 1 · Accepted Answer · edited Apr 13 '17 at 12:20

1

Hint: for $ax^2+bx+c=0$ Stick to $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

Interesting thing is, It always works!

edited Apr 13 '17 at 12:20

Community

1

answered Dec 08 '14 at 11:30

Aditya Hase

8,851

Side note: Use $\sqrt{-x} = i\sqrt{x}$ for negative discriminant to obtain the $\mathbb C$-factorisation for an irreducible polynomial (over $\mathbb R$). – AlexR Dec 08 '14 at 11:32
Maybe somebody denies existence of the roots of $x^2+1$. – AlexR Dec 08 '14 at 13:01
I know mate, I just made fun of him/her :) – AlexR Dec 08 '14 at 13:09

score 0 · Answer 3 · answered Dec 08 '14 at 15:39

A problem like this is easy if you make a substitution, for example. $$3t^2-\frac{12}{3}t+\frac{4}{3}=0$$ $$\frac{9}{3}t^2-\frac{12}{3}t+\frac{4}{3}=0$$ $$\frac{1}{3}(9t^2-12t+4)=0$$ $$\frac{1}{3}((3t)^2-4(3t)+4)=0$$ Now, let $s=3t$ $$\frac{1}{3}(s^2-4s+4)=0$$ $$\frac{1}{3}(s-2)^2=0$$ Substituting back we have $$\frac{1}{3}(3t-2)^2=0$$

Solving $3t^2-\frac{12}{3}t+\frac{4}{3}=0$

3 Answers3