+0

# help plz hectiar

+2
127
6

The solutions of x(x-3)=1$may be expressed in the form$\frac{a+\sqrt{b}}{c}$and$\frac{a-\sqrt{b}}{c}$, where$a$,$b$, and$c$are prime numbers. Find$abc\$.

Jun 1, 2019

#1
+25
-1

ok ima try to help though im not hectiar. you dont have to listen, just im doing my best

The solutions of x(x-3)=1 may be expressed in the form $$\frac{a+\sqrt{b}}{c}$$ and $$\frac{a-\sqrt{b}}{c}$$, where a, b, and c are prime numbers. Find abc.

Ok. So if i understand, you can use the quadractic formula. distribute, and you get x^2-3x=1

simplify: x^2-3x-1=0

quadractic formulas it: you get a=3, b=13, c=2

i think....... multiply and you get 3*13*2

78? right? help me out people....

78

Jun 1, 2019
#2
+25
-7

im sorry. Im not hectair..........

ProffesorNobody  Jun 1, 2019
edited by ProffesorNobody  Jun 1, 2019
#3
+8756
+3

I got the same answer as you......you can stop crying! You know how to do these questions too.

hectictar  Jun 1, 2019
#4
+25
-1

:D yayyyyyyy i got something right!

ProffesorNobody  Jun 1, 2019
#5
+7712
+2

$$\begin{array}{rcll} x(x-3) &=& 1\\ x^2 - 3x &=& 1\\ x^2 - 3x - 1 &=& 0\\ x &=& \dfrac{3\pm \sqrt{3^2 + 4}}{2}\\ x &=& \dfrac{3\pm\sqrt{13}}{2}\\ abc &=&3\cdot 13 \cdot 2 \\ abc &=& 78 \end{array}$$

.
Jun 1, 2019
#6
+25
-5

............ ok nice wonderful!

ProffesorNobody  Jun 1, 2019