UGC NET CSE | October 2020 | Part 2 | Question: 8

Question

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

• A. $8$
• B. $9$
• C. $10$
• D. $11$

Comments

sum of roots of quadratic equation (ax^2+bx+c=0) = -b/a

here 4+7 =(10) assume base is r convert both sides into decimal

4+7=r +0

r=11

Answer 1

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

Answer 2

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$