For example, let's consider the quadratic equation $-3x^2 + 6x -2 = 0$.

Multiplying both sides by $-1$, we get the equation $3x^2 - 6x +2 = 0$.

The graph of the above equations are different even though they have the same roots. The graph of the first equation is an inverted parabola while the graph of the second equation is an upright parabola (because of the sign of the coefficient of the $x^2$ term).

If we divide both sides of the original equation with $-3$, we get $x^2 - 2x + 2/3 = 0$. The graph of this equation is different from the first 2 equations. It has different roots as well.

Does this imply that multiplying or dividing both sides of a quadratic equation by a constant changes its characteristics? If so, does it also imply that we should not multiply or divide both sides of a quadratic equation by a constant when we try to find its roots? In general, when are we permitted to multiply or divide both sides of an equation by a constant?

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    $\begingroup$ The graph of the equation $\textbf{y} = 3x^2 - 6x + 2$ is an upright parabola. This is very different from the equation $0 = 3x^2 - 6x + 2$. $\endgroup$ – user296602 Jul 18 '16 at 7:45
  • $\begingroup$ You are correct. I see the problem with my question/logic. I should have written $y = -3x^2 + 6x - 2$, which makes the 2nd and 3rd equations as $-y = 3x^2 - 6x + 2$ and $y/-3 = x^2 - 2x - 2/3$ thus preserving the equality. My revised question would be what kind of algebraic manipulations are allowed in general while solving a quadratic equation? Thanks $\endgroup$ – Algebra2 Student Jul 18 '16 at 7:59
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    $\begingroup$ You seem to be mixing between functions and equations: a function has a graph, an equation has solutions. More precisely: a function can be illustrated with a graph, an equation may have one or more solutions. $\endgroup$ – barak manos Jul 18 '16 at 8:21
  • $\begingroup$ I agree that I have mixed the functions and equations in my question above. So, let me rephrase my question as follows. Why are the roots of the 3 quadratic equations $−3x^2+6x−2=0, 3x^2−6x+2=0, x^2−2x+2/3=0$ different even though the only difference between them are the constant with which they are multiplied? $\endgroup$ – Algebra2 Student Jul 19 '16 at 3:12

If we are searching the roots of an equation, as: $ax^2+bx+c=0$ and $k\ne 0$ is a constant, than, multiplying both sides by $k$ we have: $$ kax^2+kbx+kc=k(ax^2+bx+c)=k\cdot 0=0 $$ and, since $k\ne 0$, we must have $ax^2+bx+c=0$ and this means that the roots of the two equations are the same. ( note that we can do the same for a polynomial of any degree).

This means that we can multiply an equation by a constant $\ne 0$ and we always find the same roots.

Your observation that the multiplication by a negative constant inverts the parabola, is important when we are solving an inequality. E.g.:

$ax^2+bx+c >0 \quad \iff \quad kax^2+kbx+kc > 0 $ iff $k>0$

but $ax^2+bx+c >0 \quad \iff \quad kax^2+kbx+kc < 0 $ if $k<0$

  • $\begingroup$ The graphs and roots of the equations (not inequalities) $−3x^2+6x−2=0$ and $x^2−2x+2/3=0$ are not identical. $k$ in this case is $-3$. Why is it so? $\endgroup$ – Algebra2 Student Jul 19 '16 at 3:04
  • $\begingroup$ The two equations have the same roots: $x=1\pm \frac{\sqrt{3}}{3}$. Obviously the graph are not identical, and one is positive where the other is negative and negative where the other is positive, and this is the pictorial representation of the necessity to reverse the inequality. $\endgroup$ – Emilio Novati Jul 19 '16 at 8:38

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