# Math Help - roots of quadratic equation

1. ## roots of quadratic equation

Mary saw this notation:

3x^2 - 25x + 66 = 0 --> x1=4, x2=9

Mary thought it seemed to be incorrect.

She the realised that a number system other than decimal was used to determine the roots of the quadratic equation. What is the base of that number system (with an explantion pls)?

2. Hello, the undertaker!

Mary saw this notation: . $3x^2 - 25x + 66 = 0,\;\;x \:=\:4,9$

Mary thought it seemed to be incorrect.

She then realised that a number system other than decimal was used.
What is the base of that number system (with an explantion pls)?
Let the base be $b.$

Then $25$ represents: . $2b+5$
and $66$ represents: . $6b+6.$

Then the equation is: . $3x^2 - (2b+5)x + (6b+b) \:=\:0$

Since $x = 4$ is a root: . $3\cdot4^2 - (2b+5)\cdot4 + (6n+6) \:=\:0$

. . and we have: . $48 - 8b - 20 + 6b + 6 \:=\:0 \quad\Rightarrow\quad -2b \:=\:-34 \quad\Rightarrow\quad b \;=\:17$

The problem was written in base-seventeen.

3. And, of course, using 9 instead of 4 gives exactly the same answer: b= 17.

4. Originally Posted by Soroban
Hello, the undertaker!

Let the base be $b.$

Then $25$ represents: . $2b+5$
and $66$ represents: . $6b+6.$

Then the equation is: . $3x^2 - (2b+5)x + (6b+b) \:=\:0$

Since $x = 4$ is a root: . $3\cdot4^2 - (2b+5)\cdot4 + (6n+6) \:=\:0$

. . and we have: . $48 - 8b - 20 + 6b + 6 \:=\:0 \quad\Rightarrow\quad -2b \:=\:-34 \quad\Rightarrow\quad b \;=\:17$

The problem was written in base-seventeen.
thanks a lot
but wt does x2=9 have to do with the problem, because i dont think u included it