Q

# 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)

$b-c+a= 0$

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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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