Solve the quadratic equation $x^2 – 5x + 6 = 0 $

algebra-precalculusquadratics

How to solve the quadratic equation $x^2 – 5x + 6 = 0 $?

There are two ways to solve it :

a) $x^2 – 3x-2x+6$

$x(x-3) -2(x-3)$

$x=2,3$

b) $x^2 – 6x+1x+6$

$x(x-6) + 1(x+6) = 0$.

Now , here also I used the same method. But the neither am I able the quadratic equation nor is the method working here.

Also , a quadratic equation cannot have more than $2$ roots. So , either $6$ or $-6$ can’t be the answer. Then , why do I even reach this approach ? Why is it coming different here ?

I think this approach is wrong because :

In this approach , what matters is that in the starting $3 \cdot 2$ ( this $-5x$ which we shortened up into $-3x-2x$) should be equal to $6$, i.e. our constant in the quadratic equation.

Even in 2nd case, $6 \cdot 1 = 6$.

But the problem is the signs: $-3 \cdot -2 = +6$ but not $(-6)(1)=+6$. Therefore , this approach is wrong.

Am I correct?

Best Answer

Compare $(a-r_1)(x-r_2) = x^2 - (r_1+r_2)x + r_1r_2$ to $(a-r_1)(x-r_2) = x^2 - 5x + 6$.

$$(x - r_1)(x - r_2) = \left\{\begin{array}{ccc} x^2 & -5x & +6 \\ x^2 & -(r_1+r_2)x & +r_1r_2 \end{array}\right\}$$

You can see that you need to find $r_1$ and $r_2$ such that $r_1+r_2 = 5$ and $r_1 r_2 = 6$

Note that the coefficient of $x^2$ is $1$. When it is not, then you will need to do a bit more work.

There are several ways to do this. The better you are at arithmetic, the easier this will be.

  • METHOD $1:$ Make a list of the pairs of factors that multiply to make $6$. \begin{array}{ccc} 1 & 6 \\ 2 & 3 \end{array}

Then append a third column that consists of their sums. \begin{array}{cc|c} 1 & 6 & 7 \\ 2 & 3 & 5 \end{array}

We see that $r_1=2$ and $r_2=3$ is what we are looking for.

  • METHOD $2:$

This method works. But it requires that you be pretty good at arithmetic and at solving simple linear equations. Let $r_1 = u - v$ and let $r_2 = u+v$

Then $5 = r_1 + r_2 = (u-v) + (u+v) = 2u$. So $u = \dfrac 52$. So

$6 = r_1 r_2 = \left(\dfrac 52 - v\right)\left(\dfrac 52 + v\right) = \dfrac{25}{4}-v^2$. Solve for $v$.

\begin{align} \dfrac{25}{4}-v^2 &= 6 \\ v_2 - \dfrac{25}{4} &= -6 \\ 4v^2 - 25 &= -24 \\ 4v^2 &= 1 \\ v^2 &= \dfrac 14 \\ v &= \dfrac 12 \end{align}

So $r_1 = u - v = \dfrac 52 - \dfrac 12 = 2$

and $r_2 = u + v = \dfrac 52 + \dfrac 12 = 3$

  • METHOD $3$: Completing the square.

\begin{align} x^2 - 5x + 6 &= 0 \\ x^2 - 5x &= -6 &\text{Move the constant term over to the other side.} \\ & & \text{Take half of $5$ and square it. $\left(\dfrac 52 \right)^2 = \dfrac{25}{4}$} \\ x^2 - 5x + \dfrac{25}{4} &= \dfrac{25}{4} - 6 &\text{Add that number to both sides.} \\ \left(x - \frac 52 \right)^2 &= \frac 14 &\text{Simplify} \\ x - \frac 52 &= \pm \dfrac 12 \\ x &= \frac 52 \pm \frac 12 \\ x &= 2, 3 \end{align}

Related Question