Question

If (x-3) and [ x - 1/3] and both are factors of ax 2 + 5x + b , then show that a = b.

Answer 1

ax²+5x+b = 0

(x–3)(x–⅓) = x²–(3+⅓)x+1
= x²–(10/3)x+1
x²–(10/3)x+1 = 0

Sum of roots = –5/a = 10/3
a = –5×(3/10) = –3/2

Product of roots = b/a = 1
a = b

Answer 2

To Prove:

If $(x-3) \text { and }\left[x-\frac{1}{3}\right]$ and both are factors of a $x^{2}+5 x+b$, then show that a = b.

Solution:  

Given: $(x-3) \text { and }\left[x-\frac{1}{3}\right]$ are factors of $a x^{2}+5 x+b$

If $(x-3) \text { and }\left[x-\frac{1}{3}\right]$ are factors of $a x^{2}+5 x+b$, then their product will give the said expression. Hence,  

$(x-3) \times\left(x-\frac{1}{3}\right)$

$\begin{array}{l}{x^{2}-\frac{x}{3}-3 x+1=0} \\ \\ {3 x^{2}-x-9 x+3=0} \\ \\ {3 x^{2}-10 x+3=0}\end{array}$

Comparing the equation $3 x^{2}-10 x+3=0\ with\ a x^{2}+5 x+b$, we can say that

a = 3 & b = 3

a = b

Hence proved.