# roots of quadratic equation

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• May 30th 2009, 07:06 PM
the undertaker
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)?
• May 30th 2009, 07:47 PM
Soroban
Hello, the undertaker!

Quote:

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.

• May 31st 2009, 02:15 AM
HallsofIvy
And, of course, using 9 instead of 4 gives exactly the same answer: b= 17.
• May 31st 2009, 03:49 AM
the undertaker
Quote:

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