Equation x^2

[danielvnl]

x²/3 - (28x)/3 + 9 = 0

(28/3 +_- sqrt( (28/3)^2 -4(1/3)(9))/2(1/3)

I think there aren't naturals roots.
I tried on solve this equation but I can't.
When I paste it in Wolfram it gave me the result as 1 and 27.

Why ??

 

[pka]

x²/3 - (28x)/3 + 9 = 0

(28/3 +_- sqrt( (28/3)^2 -4(1/3)(9))/2(1/3)
When I paste it in Wolfram it gave me the result as 1 and 27.

Why don't you rewrite first as $\displaystyle x^2-28x+27=0~?$

 

[danielvnl]

Thx:d

 

[HallsofIvy]

As pka said, multiplying by 3 gives $\displaystyle x^2- 28x+ 27= (x- 1)(x- 27)$

But the quadratic formula certainly ought to give them too!

The "discriminant" is $\displaystyle b^2- 4ac= \left(\frac{28}{3}\right)^2- 4\left(\frac{1}{3}\right)(9)= \frac{784}{9}- \frac{36}{3}= \frac{784}{9}- \frac{108}{9}$
$\displaystyle = \frac{676}{9}= \left(\frac{26}{3}\right)^2$

So $\displaystyle \frac{-b\pm\sqrt{b^2- 4ac}}{2a}= \frac{-\frac{28}{3}\pm\frac{26}{3}}{2(1/3)}$

Taking the "+", $\displaystyle x= (-\frac{28}{3}+ \frac{26}{3})(3/2)= \left(\frac{2}{3}\right)\left(\frac{3}{2}\right)= 1$.

Taking the '-", $\displaystyle x= (-\frac{28}{3}- \frac{26}{3})(3/2)= \left(\frac{54}{3}\right)\left(\frac{3}{2}\right)= 27$.