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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
4 months ago

Samyak Jain
333 Points
							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  $\dpi{100} \Rightarrow$ (ax3 + 2bx2 + 3cx + 4d) – (ax3 + bx2 + cx) = 0bx2 + 2cx + 4d = 0   ...(4) Thus, equations ax2 + bx + c = 0 and bx2 + 2cx + 4d = 0 have a common root $\alpha$. This impliesa$\alpha$2 + b$\alpha$ + c = 0   ...(i)      b$\alpha$2 + 2c$\alpha$ + 4d = 0   …(ii) Multiply (i) by 2c and (ii) by b and then subtract (ii) from (i) to get(2ac – b2)$\alpha$2 + (2c2 – 4bd) = 0   $\dpi{100} \Rightarrow$   $\alpha$2 = 2(c2 – 2bd) / (b2 – 2ac) $\dpi{100} \because$ $\alpha$ is non-zero, $\alpha$2 > 0,  i.e.  2(c2 – 2bd) / (b2 – 2ac)  > 0  or  2(c2 – 2bd)(b2 – 2ac) / (b2 – 2ac)2  > 0$\dpi{100} \therefore$  (c2 – 2bd)(b2 – 2ac)  >  0   provided  b2 $\dpi{80} \neq$ 2ac.

4 months ago
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• 101 Video Lectures
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• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions