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Solve it, - Complex numbers and quadratic equations - JEE Main-4

Which of the following is condition of exactly one root common between ax^{2}+bx+c= 0 and cx^{2}+bx+a= 0

  • Option 1)

    a= c

  • Option 2)

    a+b+c= 0

  • Option 3)

    a-b-c= 0

  • Option 4)

    b-c+a= 0

 
Answers (1)
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\\*a^{2}+bx+c=0\\*cx^{2}+bx+a=0\\*\Rightarrow (a^{2}-c^{2})^{2}=(ab-bc)(ab-bc)\\*\Rightarrow (a-c)^{2}(a+c)^{2}=b^{2}(a-c)^{2}\\*\Rightarrow (a-c)^{2}\left [ (a+c)^{2}-b^{2} \right ]=0\\*\Rightarrow a=c\; \; or\; \; a+b+c=0\; \; \; or\; \; a-b+c=0

When a=c then both quadratics are same and have both roots common.

a+b+c=0 will give x=1 as common root.

a-b+c=0 will give x=-1 as common root.

In above options (A) & (B) are only condition of common roots in which (A) will be rejected as explained above.

 

Condition for one common root -

\left ( {a}' c-a{c}'\right )^{2}= \left ( b{c}' -{b}'c\right )\left ( a{b}' -{a}'b\right )

- wherein

ax^{2}+bx+c=0 &

a'x^{2}+b'x+c'=0

are the 2 equations

 

 


Option 1)

a= c

This is incorrect

Option 2)

a+b+c= 0

This is correct

Option 3)

a-b-c= 0

This is incorrect

Option 4)

b-c+a= 0

This is incorrect

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