retagged by
14,268 views
39 votes
39 votes
Consider the quadratic equation $x^2-13x+36=0$ with coefficients in a base $b$. The solutions of this equation in the same base $b$ are $x=5$ and $x=6$. Then $b=$ _____
retagged by

5 Answers

Best answer
80 votes
80 votes
Let $ax^2 + bx +c = 0$ be a quadratic equation, then

Sum of roots $= \frac{-b}{a}$ and product of roots $= \frac{c}{a}$

$\color{red}{(5)_{ b} + (6)_{b} = (13)_{b} \Rightarrow b = 8}$

and $ (5)_{b}*(6)_{b} = (36)_{b}$ means $30 = 3b+ 6$. So, $\color{blue}{b = 8}$
selected by
13 votes
13 votes
we can directly put x=5 or x=6, in the given equation by expanding its base.

 

x2−13x+36=0 to x2-(b+3)x+(3b+6)=0.

substituting x=5 or x=6 gives b=8.
12 votes
12 votes

Answer is 8

The 2 roots of the eqtn are 5 and 6

The eqtn x2-13x+36 will be similar to (x-5)(x-6)

So, x2-13x+36 is similar to x2-11x+30

This tells us that (13)b = 11

=> 1*b1 + 3*b0 = 11

=> b+3 = 11

=>b = 8

Verify this by putting b=8 in (36)b = 30

3 votes
3 votes
Ans is base 8. This is the explanation. The coefficients in given eqn are in base b. A quadratic eqn can be represented as x2 - (sum of roots) + product of roots. Therefore here sum of roots is 13(in base b) and product of roots is 36(in base b). Now you can represent 13 in base 10 as 1*b^1 + 3*b^0 and  3*b^1+6*b^0 in base 10. Therefore 13 is b+3 in base 10 and 36 is 3b+ 6 in base 10.

From roots as x = 5, 6 we can know one thing that since the roots are still in base b , therefore, the max value (that we know now from current scenario) is that the base is atleast greater than or it might be equal to 7 as for a base b the max value is b-1.

Now the roots are given as x = 5, 6. Therefore we can form the eqn from given roots using sum of roots and product of roots. The eqn is x2 - 11x + 30. Now b+3 = 11 and 3b+ 6 = 30. Solving any of these you will get b = 8 and hence the answer.

You can even try to convert 13 in base 8 to 11 in base 10 and 36 in base 8 to 30 in base 10.
Answer:

Related questions

30 votes
30 votes
3 answers
2
go_editor asked Sep 28, 2014
5,154 views
A non-zero polynomial $f(x)$ of degree 3 has roots at $x=1$, $x=2$ and $x=3$. Which one of the following must be TRUE? $f(0)f(4)< 0$$f(0)f(4) 0$$f(0)+f(4) 0$$f(0)+f(4)< 0...
37 votes
37 votes
4 answers
3
Kathleen asked Sep 14, 2014
7,584 views
A polynomial $p(x)$ satisfies the following:$p(1) = p(3) = p(5) = 1$ $p(2) = p(4) = -1$The minimum degree of such a polynomial is$1$$2$$3$$4$