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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 $b=$ _____
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$x^2 -13x+36 = 0$

Roots = 5,6

Hence, $x=5,6$ will satisfy this given equation.

Putting, $x=5,$

$(5)_b * (5)_b - (13)_b*(5)_b+(36)_b=(0)_b$

Converting everything in base 10,

$25-(b+3)5+3b+6=0$

$b=8$
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roots are incorrect its should be 4 an 9 so i think question is mathematically in correct??
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roots are 4 and 9 in decimal. not in base b
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5 Answers

79 votes
79 votes
Best answer
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, $\color{blue}{b = 8}$
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4 Comments

how 

(5)b∗(6)b=(36)b means 30=3b+6

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@Ani_

We can think as follows :

If we convert 5 and 6 in base b to base 10 , they will be 5 and 6 only.

Now, 5+6=(11)10=(13)b

Converting (13)b to base 10,

11= b+3 => b=8

Similarly,

5*6=(30)10=(36)b

Converting (36)b to base 10,

30= 3b+6 => b=8

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very good explanation @VS

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nice explanation mcjoshi
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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.

3 Comments

x2-(b+3)x+(3b+6)=0.
how its came???
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simply take base b, and expand the coeffiecient in quadratic equation
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$x^2 − 13x + 36 = 0$

coefficients are in base b

$x^2 -(b + 3)x^1 + ( 3b + 6)x^0 = 0$

At x = 5

$25 – 5b – 15 + 3b + 6 = 0$

$b  = 8$

solving for x = 6 will give the same result.
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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.
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