A quadratic expression can *always* be factorized, but the factorization process may be difficult if the zeroes of the expression are non-integer real numbers, or non-real numbers. In such cases, we can use the quadratic formula to determine the zeroes of the expression.

Consider an arbitrary quadratic equation:

\[a{x^2} + bx + c = 0,\,\,\,a \ne 0\]

To determine the roots of this equation, we proceed as follows:

\[\begin{align}&a{x^2} + bx = - c\\&\Rightarrow \,\,\,{x^2} + \frac{b}{a}x = - \frac{c}{a}\end{align}\]

Now, we express the left hand side as a perfect square, by introducing a new term on both sides:

\[{x^2} + \frac{b}{a}x + \underbrace {{{\left( {\frac{b}{{2a}}} \right)}^2}}_{{\rm{New}}\,{\rm{term}}} = - \frac{c}{a} + \underbrace {{{\left( {\frac{b}{{2a}}} \right)}^2}}_{{\rm{New term}}}\]

The left hand side is now a perfect square:

\[{\left( {x + \frac{b}{{2a}}} \right)^2} = - \frac{c}{a} + \frac{{{b^2}}}{{4{a^2}}} = \frac{{{b^2} - 4ac}}{{4{a^2}}}\]

This is good for us, because now we can take square roots to obtain:

\[\begin{align}&x + \frac{b}{{2a}} = \pm \frac{{\sqrt {{b^2} - 4ac} }}{{2a}}\\&\Rightarrow \,\,\,x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\end{align}\]

Thus, by completing the squares, we were able to isolate \(x\) and obtain the two roots of the equation. Let us apply this formula to a few examples.

**Solved Example 1:** Using the quadratic formula, find the roots of the equation \({x^2} - 7x + 6 = 0\).

**Solution:** Comparing the given equation to the standard quadratic form \(a{x^2} + bx + c = 0\), we have:

\[a = 1,\,\,b =\;\; - 7,\,\,c = 6\]

Now, we use the quadratic formula to find the roots:

\[\begin{align}&x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\&\Rightarrow \,\,\,x = \frac{{ - \left( { - 7} \right) \pm \sqrt {{{\left( { - 7} \right)}^2} - 4\left( 1 \right)\left( 6 \right)} }}{{2\left( 1 \right)}}\\& = \frac{{7 \pm \sqrt {49 - 24} }}{2} = \frac{{7 \pm \sqrt {25} }}{2}\\& = \frac{{7 \pm 5}}{2} = 1,\,\,6\end{align}\]

Thus, the two roots are \(x = 1\) and \(x = 6\). This can be verified by factorization as well. The given equation can be factorized as follows:

\[\begin{align}&{x^2} - 7x + 6 = 0\\&\Rightarrow \,\,\,{x^2} - 6x - x + 6 = 0\\&\Rightarrow \,\,\,x\left( {x - 6} \right) - 1\left( {x - 6} \right) = 0\\&\Rightarrow \,\,\,\left( {x - 1} \right)\left( {x - 6} \right) = 0\end{align}\]

**Solved Example 2:** Find the roots of the following equation in the variable *x*:

\[2p\left( {1 + {x^2}} \right) - \left( {1 + {p^2}} \right)\left( {x + p} \right) = 0\]

**Solution:** Writing the given equation in the standard quadratic form, we have (verify this):

\[2p{x^2} - \left( {1 + {p^2}} \right)x + p - {p^3} = 0\]

The coefficients are:

\[a = 2p,\,\,b = - \left( {1 + {p^2}} \right),\,\,c = p - {p^3}\]

Applying the quadratic formula, we have:

\[\begin{align}&\Rightarrow \,\,\,x = \frac{{\left( {1 + {p^2}} \right) \pm \sqrt {{{\left( {1 + {p^2}} \right)}^2} - 4\left( {2p} \right)\left( {p - {p^3}} \right)} }}{{2\left( {2p} \right)}}\\& = \frac{{\left( {1 + {p^2}} \right) \pm \sqrt {{p^4} + 2{p^2} + 1 - 8p\left( {p - {p^3}} \right)} }}{{4p}}\\&= \frac{{\left( {1 + {p^2}} \right) \pm \sqrt {9{p^4} - 6{p^2} + 1} }}{{4p}}\\& = \frac{{\left( {1 + {p^2}} \right) \pm \sqrt {{{\left( {3{p^2} - 1} \right)}^2}} }}{{4p}}\end{align}\]

Now, we take the square root to obtain the two roots:

\[\begin{align}&x = \frac{{\left( {1 + {p^2}} \right) \pm \left( {3{p^2} - 1} \right)}}{{4p}}\\& = \frac{{4{p^2}}}{{4p}},\,\,\,\,\frac{{2 - 2{p^2}}}{{4p}}\\& = p,\,\,\,\frac{{1 - {p^2}}}{{2p}}\end{align}\]

**Solved Example 3:** Solve the following equation:

\[\left( {12x - 1} \right)\left( {6x - 1} \right)\left( {4x - 1} \right)\left( {3x - 1} \right) = 5\]

**Solution:** This is not a quadratic equation. If you expand the left side by multiplying, you will obtain a four-degree equation. However, by a little manipulation, you can reduce this equation to a quadratic equation. The trick is to multiply the first and the fourth brackets, and the second and the third brackets:

\[\begin{align}&\left\{ {\left( {12x - 1} \right)\left( {3x - 1} \right)} \right\}\left\{ {\left( {6x - 1} \right)\left( {4x - 1} \right)} \right\} = 5\\&\Rightarrow \,\,\,\left\{ {36{x^2} - 15x + 1} \right\}\left\{ {24{x^2} - 10x + 1} \right\} = 5\end{align}\]

Why is this helpful? Well, take out 3 common from the first bracket, and 2 common from the second bracket:

\[6\left\{ {12{x^2} - 5x + \frac{1}{3}} \right\}\left\{ {12{x^2} - 5x + \frac{1}{2}} \right\} = 5\]

This is great, because now we can substitute the common expression in both the brackets as a new variable. Substitute \(12{x^2} - 5x = y\) to obtain:

\[6\left( {y + \frac{1}{3}} \right)\left( {y + \frac{1}{2}} \right) = 5\]

This is now a quadratic equation, and you know how to solve this! Writing in standard form, we have:

\[\begin{align}&6{y^2} + 5y - 4 = 0\\&\Rightarrow \,\,\,y = \frac{{ - 5 \pm \sqrt {25 + 96} }}{{12}} = \frac{{ - 5 \pm 11}}{{12}}\\&= \frac{1}{2},\,\, - \frac{4}{3}\end{align}\]

But we are not done yet! We have to find the values of corresponding values of *x*:

If \(y = \frac{1}{2}\):

\[\begin{align}&12{x^2} - 5x = \frac{1}{2}\\&\Rightarrow \,\,\,24{x^2} - 10x - 1 = 0\\&\Rightarrow \,\,\,\left( {12x + 1} \right)\left( {2x - 1} \right) = 0\\&\Rightarrow \,\,\,x = - \frac{1}{{12}},\,\,\frac{1}{2}\end{align}\]

If \(y = - \frac{4}{3}\):

\[\begin{align}&12{x^2} - 5x = - \frac{4}{3}\\&\Rightarrow \,\,\,36{x^2} - 15x + 4 = 0\end{align}\]

You can verify that this equation has non-real zeroes. Thus, there are only two real roots of the original equation: \(x = - \frac{1}{{12}},\,\,\frac{1}{2}\).

**Solved Example 4:** Find all the possible values of the variable *a* for which the following quadratic equations have at least one common root:

\[\begin{align}&{x^2} + ax + 1 = 0\\&{x^2} + x + a = 0\end{align}\]

**Solution:** We can make the observation that if \(a = 1\), the two equations are identical and will have both the roots common. Let us try to find other values of *a* for which the equations have a common root.

Let the common root of the two equations be *t. *Since *t* must satisfy both the equations, we have:

\[\left. \begin{array}{l}{t^2} + at + 1 = 0\\{t^2} + t + a = 0\end{array} \right\} ...\left( 1 \right)\]

If we treat this system of equations as a linear system in the variables \({t^2}\)and *t*, we have:

\[\begin{align}&\frac{{{t^2}}}{{{a^2} - 1}} = \frac{t}{{1 - a}} = \frac{1}{{1 - a}}, \,\,a \ne 1\\&\Rightarrow \,\,\,{t^2} = - \left( {a + 1} \right),\,\,\,t = 1\\&\Rightarrow \,\,\, - \left( {a + 1} \right) = 1\\&\Rightarrow \,\,\,a = - 2\end{align}\]

*Verification*: For this value of *a, *the two equations are:

\[\begin{align}&{x^2} - 2x + 1 = 0\\&{x^2} + x - 2 = 0\end{align}\]

If you solve these two equations, you will find that \(x = 1\) is a common root for both.

An alternate method could have been to equate the two expressions in (1):

\[\begin{align}&{t^2} + at + 1 = {t^2} + t + a = 0\\&\Rightarrow \,\,\,t\left( {a - 1} \right) = \left( {a - 1} \right)\\&\Rightarrow \,\,\,t = 1\end{align}\]

Thus, the common root has the value\(t = 1\). Substitution this back into the first equation of (1), we have:

\[1 + a + 1 = 0\,\,\, \Rightarrow \,\,\,a = - 2\]

**Solved Example 5:** Find the condition(s) such that the following equations may have a common root:

\[\begin{align}&{x^2} + bx + c = 0\\&b{x^2} + cx + 1 = 0\end{align}\]

**Solution:** Let *k* be a common root of the two equations. Since it must satisfy both the equations, we have:

\[\begin{align}&{k^2} + bk + c = 0\\&b{k^2} + ck + 1 = 0\end{align}\]

If we treat this as a linear system in \({k^2}\) and *k*, we have (using the cross-multiplication rule):

\[\begin{align}&\frac{{{k^2}}}{{b - {c^2}}} = \frac{k}{{bc - 1}} = \frac{1}{{c - {b^2}}}\\&\Rightarrow \,\,\,{k^2} = \frac{{b - {c^2}}}{{c - {b^2}}},\,\,\,k = \frac{{bc - 1}}{{c - {b^2}}}\\&\Rightarrow \,\,\,\frac{{b - {c^2}}}{{c - {b^2}}} = {\left( {\frac{{bc - 1}}{{c - {b^2}}}} \right)^2}\\&\Rightarrow \,\,\,\left( {b - {c^2}} \right)\left( {c - {b^2}} \right) = {\left( {bc - 1} \right)^2}\end{align}\]

Now, we expand this and rearrange the terms:

\[\begin{align}&bc - {b^3} - {c^3} + {b^2}{c^2} = {b^2}{c^2} - 2bc + 1\\&\Rightarrow \,\,\,{b^3} + {c^3} - 3bc + 1 = 0\\&\Rightarrow \,\,\,{\left( {b + c} \right)^3} - 3bc\left( {b + c} \right) - 3bc + 1 = 0\\&\Rightarrow \,\,\,1 + {\left( {b + c} \right)^3} - 3bc\left( {b + c + 1} \right) = 0\\&\Rightarrow \,\,\,{\left( {b + c + 1} \right)^3} - 3\left( {b + c} \right)\left( {b + c + 1} \right)\\& \qquad\qquad\qquad\quad - 3bc\left( {b + c + 1} \right) = 0\\&\Rightarrow \,\,\,{\left( {b + c + 1} \right)^3} - 3\left( {b + c + 1} \right)\left( {b + c + bc} \right) = 0\\&\Rightarrow \,\,\,\left( {b + c + 1} \right)\left\{ {{{\left( {b + c + 1} \right)}^2} - 3\left( {b + c + bc} \right)} \right\} = 0\end{align}\]

Thus, we have two possible conditions either of which is sufficient for the two equations to have a common root.

### Condition-1

\[b + c + 1 = 0\]

### Condition-2

\[\begin{align}&{\left( {b + c + 1} \right)^2} - 3\left( {b + c + bc} \right) = 0\\&\Rightarrow \,\,\,{\left( {b + c} \right)^2} + 1 + 2\left( {b + c} \right) - 3\left( {b + c + bc} \right) = 0\\&\Rightarrow \,\,\,{b^2} + {c^2} + 1 = bc + b + c\end{align}\]