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Does x^2+px+q = 0 have a root?

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Does x^2+px+q = 0 have a root?  [#permalink]

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New post 29 Aug 2018, 01:54
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[Math Revolution GMAT math practice question]

Does \(x^2+px+q = 0\) have a root?

\(1) p<0\)
\(2) q<0\)

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Does x^2+px+q = 0 have a root?  [#permalink]

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New post 29 Aug 2018, 03:40
MathRevolution wrote:
[Math Revolution GMAT math practice question]

Does \(x^2+px+q = 0\) have a root?

\(1) p<0\)
\(2) q<0\)


Note:-
1) In GMAT, roots of QE mean REAL roots.
2) A quadratic equation posses real roots when 1) D=0 2) D>0
3) A quadratic equation posses imaginary when D<0
Where D is the discriminant=\(p^2-4*1*q=p^2-4q\)

Polarity of \(p^2-4q\) is the key decision maker:
1) When p=0, then polarity of D depends on the polarity of q. In order to have obtain real roots, q must be -ve or zer0. (a) Say q=-5, then D=\(p^2-4q\)=0-4*(-5)=20>0, b)say q=0, then D=\(p^2\)-4q=0-4*(0)=0)
2) When p is +ve or -ve, \(p^2\) is always positive, so, q must be -ve so that D>0.

\(1) p<0\)

a) When q>0 and \(p^2>4q\), roots are imaginary.
b) When q=0 or q<0, then D>0. real roots .
Insufficient.

\(2) q<0\)
Irrespective of sign and value of p, when q<0, \(p^2-4q>0\) Or, D>0. Hence, roots are real.
So QE has roots.
sufficient.

Ans. (B)
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Re: Does x^2+px+q = 0 have a root?  [#permalink]

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New post 29 Aug 2018, 05:05
MathRevolution wrote:
[Math Revolution GMAT math practice question]

Does \(x^2+px+q = 0\) have a root?

\(1) p<0\)
\(2) q<0\)
Answer is B
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Re: Does x^2+px+q = 0 have a root?  [#permalink]

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New post 29 Aug 2018, 05:58
Finds a solution on website Finactax.
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Re: Does x^2+px+q = 0 have a root?  [#permalink]

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New post 29 Aug 2018, 06:07
MathRevolution wrote:
[Math Revolution GMAT math practice question]

Does \(x^2+px+q = 0\) have a root?

\(1) p<0\)
\(2) q<0\)

\(?\,\,\,:\,\,\,\,\Delta = \left( {{\text{}}{b^2} - 4ac{\text{}}} \right) = {p^2} - 4q\,\,\,\mathop \geqslant \limits^? \,\,\,\,0\,\,\)

\(\left( 1 \right)\,\,\,\left\{ \begin{gathered}
\,Take\,\,\left( {p,q} \right) = \left( { - 1,0} \right)\,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\
\,Take\,\,\left( {p,q} \right) = \left( { - 1,1} \right)\,\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \hfill \\
\end{gathered} \right.\)

\(\left( 2 \right)\,\,\,q < 0\,\,\,\,\mathop \Rightarrow \limits^{{\text{FOCUS!}}} \,\,\, - 4q > 0\,\,\,\,\mathop \Rightarrow \limits^{{{\text{p}}^{\text{2}}}\,\, \geqslant \,\,0} \,\,\,{p^2} - 4q > 0\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle\)


Conclusion: the correct answer is (B).

The above follows the notations and rationale taught in the GMATH method.
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Re: Does x^2+px+q = 0 have a root?  [#permalink]

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New post 29 Aug 2018, 06:08
adityagupta27 wrote:
Finds a solution on website Finactax.


Hi adityagupta27,
Could you please elaborate how Finactax is related to explanation of the question ?
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Re: Does x^2+px+q = 0 have a root?  [#permalink]

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New post 31 Aug 2018, 01:11
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

The discriminant of the equation is \(p^2 – 4q\). If the discriminant is greater than or equal to zero, then the quadratic equation has roots.
The question asks if \(p^2-4q ≥ 0\) or not.

Since \(p^2 ≥ 0\), if q < 0, then \(p^2-4q ≥ 0\). Thus, condition 2) is sufficient.

Condition 1)
If \(p = -1\) and \(q = 0\), then the discriminant is positive and the equation has \(2\) roots, which are \(0\) and \(1\). So, the answer is ‘yes’.
If \(p = -1\) and \(q = 1\), then the discriminant is negative and the equation has no real roots. So, the answer is ‘no’.
Since we don’t have a unique solution, condition 1) is not sufficient.

Therefore, B is the answer.
Answer: B
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Re: Does x^2+px+q = 0 have a root?   [#permalink] 31 Aug 2018, 01:11
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