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What can be the value of $k$ for which the equation $9x^2+2kx-1=0$ has real roots?

Things should be known

  • When the quadratic equation has real roots, then $d=b^2-4ac \ge 0$ .

    Where $a$, $b$ and $c$ are the constant terms of a quadratic equation $ax^2+bx+c=0$.

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It is perhaps better to say: when $d=b^2-4ac>0$, the quadratic equation has two distinct real roots, and a repeated one when $d=0$. Anyway, here $d\geq 36>0$ for every $k$. –  1015 Jan 7 '13 at 23:29
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2 Answers

If $b^2 - 4ac \geq 0$, then this gives you $4k^2 + 36 \geq 0$. What can you say about $k$ from here, considering that all the numbers appearing in the inequality are non-negative?

In general, what can you say about the roots of a quadratic $ax^2 + bx + c$, where $a > 0$ and $c \leq 0$ or where $a < 0$ and $c \geq 0$?

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I am not sure where exactly you had a problem solving this question given the information you already know. –  Rankeya Jan 5 '13 at 8:25
    
I had also did the same my final answer that arrived was $k \le \pm3$. Is that correct? –  Alpha Jan 5 '13 at 8:26
    
Note that the inequality $4k^2 + 36 \geq 0$ consists of only non-negative numbers, so it true for any $k$. –  Rankeya Jan 5 '13 at 8:28
    
What about the repeted roots? –  Nancy Rutkowskie Jan 5 '13 at 8:43
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Dear @Nancy R: The OP's quadratic cannot give us any repeated roots. That aside, I am not sure he mentions anything about repeated roots in his question. –  Rankeya Jan 5 '13 at 8:46
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Besides to @Rankeya's answer, you can see that: $$9x^2+2kx=1\Leftrightarrow 9x^2+2kx+\frac{k^2}{9}=1+\frac{k^2}{9}\\\Leftrightarrow \left(3x+\frac{k}{3}\right)^2=\frac{k^2}{9}+1\Leftrightarrow 3x+\frac{k}{3}=\pm\sqrt{\frac{k^2}{9}+1}$$ or $$3x=\pm\sqrt{\frac{k^2}{9}+1}-\frac{k}{3}$$ which shows that you always have two distinct solutions for $x$ for all $k$.

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What about the repeted roots? –  Nancy Rutkowskie Jan 5 '13 at 8:42
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