GATE CSE 2014 Set 1 | Question: GA-5

Question

The roots of $ax^{2}+bx+c = 0$ are real and positive. $a, b$ and $c$ are real. Then $ax^{2}+b\mid x \mid + c =0$ has 

• A. no roots
• B. $2$ real roots
• C. $3$ real roots
• D. $4$ real roots

Comments

None of the answer is correct for all possible quadratic equations. Here is a counter example.

For $x^{2} - 2x + 1 = 0$, the only root is $1$ and it satisfies the criteria of being real & positive. Also the other criteria of  $a, b, c$ being real numbers is satisfied as well.

Now, the equation $x^2 -2|x| + 1 = 0$ can be defined as two equations $x^{2} - 2x + 1 = 0$, $x \geq 0$ and $x^{2} + 2x + 1 = 0$, $x < 0$ which only have a total of 2 roots which are $1$ and $-1$ respectively and not 4 roots.

All quadratic equations where the discriminant is zero and $b$ & $a$ are of same sign will only give 2 roots for such transformation.

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correct statement should be: $x^2- 2|x|+1 = 0$ has $2$ $\textbf{distinct}$ real roots in which root $”1”$ has multiplicity = $2$ and root $“-1”$ also has multiplicity as $2$ because as you have written, equation $x^2-2x+1=0$ for $x \geq 0$ has $2$ $\textbf{equal}$ real roots (discriminant=0) which has value = $“1”$ and $x^2+2x+1=0$ for $x < 0$ also has $2$ $\textbf{equal}$ real roots which has value = $“-1”$ So, total real roots are $4$.

This is based on the following statement from the Fundamental theorem of algebra:

“every non-zero, single-variable, degree n polynomial with complex coefficients has, counted with multiplicity, exactly n complex roots.” 

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@ankitgupta.1729 You are right. Thanks.

Answer 1

The second equation has both +ve and -ve values of roots of first equation as root.

so it will have 4 roots

Answer 2

None of the answer is correct for all possible quadratic equations. Here is a counter example.

For $x^{2} - 2x + 1 = 0$, the only root is $1$ and it satisfies the criteria of being real & positive. Also the other criteria of  $a, b, c$ being real numbers is satisfied as well.

But even then the two equations $x^{2} - 2x + 1 = 0$, $x \geq 0$ and $x^{2} + 2x + 1 = 0$, $x < 0$ only have a total of 2 roots which are $1$ and $-1$ and not 4 roots.

All quadratic equations where the determinant is zero and $b$ is positive will only give 2 roots for such transformation.