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If the equations ax^3+2bx^2+3cx+4d=0 and ax^2+bx+c=0 have a non-zero common root then prove that (c^2-2bd)(b^2-2ac)>_ 0

If the equations ax^3+2bx^2+3cx+4d=0 and ax^2+bx+c=0 have a non-zero common root then prove that (c^2-2bd)(b^2-2ac)>_ 0
 

Grade:11

1 Answers

Samyak Jain
333 Points
4 years ago
Let the non-zero common root of ax3 + 2bx2 + 3cx + 4d = 0  ..(1) and ax2 + bx + c = 0  ..(2)  be \alpha.
If \alpha is a root of ax2 + bx + c = 0 , it will also a root of x.(ax2 + bx + c) = x.0  i.e. 
ax3 + bx2 + cx = 0    ...(3).
 
Then \alpha is also a root of (1) – (3) = 0  \Rightarrow (ax3 + 2bx2 + 3cx + 4d) – (ax3 + bx2 + cx) = 0
bx2 + 2cx + 4d = 0   ...(4)
 
Thus, equations ax2 + bx + c = 0 and bx2 + 2cx + 4d = 0 have a common root \alpha. This implies
a\alpha2 + b\alpha + c = 0   ...(i)      b\alpha2 + 2c\alpha + 4d = 0   …(ii)
 
Multiply (i) by 2c and (ii) by b and then subtract (ii) from (i) to get
(2ac – b2)\alpha2 + (2c2 – 4bd) = 0   \Rightarrow   \alpha2 = 2(c2 – 2bd) / (b2 – 2ac)
 
\because \alpha is non-zero, \alpha2 > 0,  i.e.  2(c2 – 2bd) / (b2 – 2ac)  > 0  or  2(c2 – 2bd)(b2 – 2ac) / (b2 – 2ac)2  > 0
\therefore  (c2 – 2bd)(b2 – 2ac)  >  0   provided  b2 \neq 2ac.

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