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What is the radix of the numbers if the solution to the quadratic equation $x^2-10x+26=0$ is $x=4$ and $x=7$?

  1. $8$
  2. $9$
  3. $10$
  4. $11$
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2 Answers

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If you solve the quadratic equation, roots will be $5+i, 5-i$, but those are in base $10$

We know sum of roots of quadratic equation = $\frac{-b}{a} = 10$

$\therefore$ $(4)_{b} +(7)_{b} = (10)_{b}$, since $4$ and $7$ are the roots of the equation

$4*b^0+7*b^0=1*b^1+0*b^0$

$b=11$

Option D) is correct

2 votes
2 votes
In this case, it would be sufficient to take one solution  and put it into the equation. Say we do it with x=4. Convert the coefficients into base 10. Let the radix be r.

$x^2 – 10x + 26 =0 $

$ (4)^2 – ( 1 \times r^{1} + 0 \times r^{0} )\times 4 + (2 \times r^{1} + 6\times r^{0}) =0 $

$ 16 -4r + 2r +6 = 0 $

$ 2r =22 $

$ r=11$
Answer:

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