Prove that there are two solutions of the equation $x^2 + kx = 3 - k $

The obvious solution would be finding the value of $b^2 - 4ac$

$x^2 + kx + k - 3=0$

$b^2 - 4 ac$

=$k^2 - 4(1)(k-3)$

=$k^2 - 4k + 12$

From here, $b^2 - 4ac$ have a algebraic value. If it is a constant, I would already be able to prove that the equation have $2$ real roots for every value of "$k$". So, how do I go about doing this question when I can't get $b^2 - 4ac$ to be a constant value?


Picking up from where you left off, $$k^2 - 4k + 12 \equiv (k-2)^2 + 8 \geq 0 +8 > 0$$

since squares are non-negative. Then it follows that your discriminant is positive for all values of $k \in \mathbb{R}$ and hence the quadratic equation always has two roots.

  • $\begingroup$ Ohhh.... One thing I had noticed too is that $k^2 - 4k + 12$ no matter what value of "k" I choose, I get a positive nunber. $\endgroup$ – ministic2001 Jul 10 '16 at 23:01
  • 1
    $\begingroup$ Yep, the technique of completing the square to show that an expression is always positive comes in useful a lot. $\endgroup$ – Zain Patel Jul 10 '16 at 23:02

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