Solving quadratic equations

A quadratic equation is under the form:

Solving by finding roots and factorizing

Solutions for \( X \) which \( P_2(X) = 0 \) are called the roots of the polynomial.

They make it possible to obtain a factorized form.

Three cases should be considered after calculating the discriminant \( \Delta \):

• \( \alpha) \) if \( \textcolor{#456E3D}{\Delta >0}\):two distinct roots \( X_1, \ X_2 \)
$$ X_1 = \frac{- b - \sqrt{\Delta}}{2a} $$
$$ X_2 = \frac{- b + \sqrt{\Delta}}{2a} $$

And \( P_2(X) \) can be factorized like this:

$$ P_2(X) = a(X - X_1)(X - X_2) $$


• \( \beta) \) if \( \Delta =0\): one double root \( X_0 \)
$$ X_0 = \frac{- b}{2a} $$

And \( P_2(X) \) can be factorized like this:

$$ P_2(X) = a(X - X_0)^2 $$


• \( \gamma) \) if \( \Delta < 0\): two conjugate complex roots \( C_1, \ C_2 \)
$$ C_1 = \frac{- b - i\sqrt{|\Delta|}}{2a} $$
$$ C_2 = \frac{- b + i\sqrt{|\Delta}|}{2a} $$

And \( P_2(X) \) can be factorized like this:

$$ P_2(X) = a(X - C_1)(X - C_2) $$

Link between coefficients and roots

Furhtermore, in the general case we will also have these two relationships between the coefficients and the roots:

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Demonstration

Solving by finding roots and factorizing

We start from the equation \( (1) \):

First of all, we factorize it by \( a \):

We can then transform \( (2) \) by injecting \( (3) \):

Which leads us to:

We now have:

So:

Starting from this result, there will be three cases to take into account.

• \( \alpha) \) if \( \textcolor{#456E3D}{\Delta >0}\):two distinct roots \( X_1, \ X_2 \)

Then \( \sqrt{\Delta} \) exists and the solutions are directly given by:

$$ X_1 = \frac{- b - \sqrt{\Delta}}{2a} $$
$$ X_2 = \frac{- b + \sqrt{\Delta}}{2a} $$

Thereby, the polynomial \(P_2(X)\) admits the following factorization:

$$ P_2(X) = a(X - X_1)(X - X_2) $$


• \( \beta) \) if \( \Delta =0 \): one double root \( X_0 \)

Then \( \sqrt{\Delta} = 0 \) and the root is double:

$$ X_0 = \frac{- b}{2a} $$

Then the factorization of \(P_2(X)\) becomes:

$$ P_2(X) = a(X - X_0)^2 $$


• \( \gamma) \) if \( \Delta < 0\): two conjugate complex roots \( C_1, \ C_2 \)

Then \( \sqrt{\Delta} \) is not defined on \( \mathbb{R} \). On the other hand, it can exists in the complex set \( (\mathbb{C}) \).

To solve a such equation in \( \mathbb{C} \) of type:

$$ Y= \sqrt{-\alpha } \Longrightarrow Y^2 = -\alpha $$

We do have the following solutions:

$$ \mathcal{S} = \left \{Y_{1} = i \sqrt{ |\alpha |} , \ Y_{2} = -i \sqrt{ |\alpha |} \right \} $$

In our case, the solution \( \mathcal{S} \) will become:

$$ \left \{ \Delta = i \sqrt{ |\Delta |} , \ \Delta = -i \sqrt{ |\Delta |} \right \} $$

We will then have two complex roots:

$$ C_1 = \frac{- b - i\sqrt{|\Delta|}}{2a} $$
$$ C_2 = \frac{- b + i\sqrt{|\Delta}|}{2a} $$

And the factorization will remain of the same form as for the case where \( \Delta > 0 \):

$$ P_2(X) = a(X - C_1)(X - C_2)$$

Link between coefficients and roots

From the two general formulas for the roots seen above,

We can calculate their sum and product.

1. Sum of roots\(: X_1 + X_2 \)

$$ X_1 + X_2 = \frac{- b - \sqrt{\Delta}}{2a} + \frac{- b + \sqrt{\Delta}}{2a}$$
$$ X_1 + X_2 = \frac{- 2b}{2a} $$
$$ X_1 + X_2 = -\frac{b}{a} $$

2. Roots product\(: X_1 X_2 \)

$$ X_1 X_2 = \biggl( \frac{- b - \sqrt{\Delta}}{2a} \biggr) \biggl( \frac{- b + \sqrt{\Delta}}{2a} \biggr)$$
$$ X_1 X_2 = \frac{b^2 - b\sqrt{\Delta} + b\sqrt{\Delta}- \Delta }{4a^2} $$
$$ X_1 X_2 = \frac{b^2 - \Delta }{4a^2} $$

But, \(\Delta = b^2 - 4ac \), so:

$$ X_1 X_2 = \frac{b^2 - (b^2 - 4ac)}{4a^2} $$
$$ X_1 X_2 = \frac{ 4ac}{4a^2} $$
$$ X_1 X_2 = \frac{c}{a} $$

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Examples

1. Solving a second degree polynomial

$$ P_2(X) = 2X^2 - 3X + 1 $$

We calculate the discriminant \( \Delta \):

$$ \Delta = (-3) ^2 - 4 \times 2 \times 1 = 1$$

So, \(P_2(X) \) has two real roots \(X_1, X_2\):

$$ X_1 = \frac{- (-3) - \sqrt{1}}{2 \times 2} $$
$$ X_2 = \frac{- (-3) + \sqrt{1}}{2 \times 2} $$
$$ X_1 = \frac{2}{4} $$
$$ X_2 = \frac{4}{4} $$
$$ \mathcal{S} = \biggl \{X_{1} = \frac{1}{2} , \ X_{2} = 1 \biggr \} $$

\(P_2(X) \) can be factorized:

$$ P_2(X) = 2 \left(X- \frac{1}{2} \right)\Bigl(X -1 \Bigr) $$

\(P_2(X) \) is the polynomial which is worth \(0\) at \( X = \frac{1}{2}\) et \( X = 1\).




2. Solving a fourth degree polynomial by a variable change

Let us solve this new equation:

$$ P_4(X) = 6X^4 - X^2 - 1 $$

When we have this type of situation, we do a variable change:

Let us set a new variable down:

$$ \varphi = X^2$$

Then the polynomial \(P_4(X)\) becomes a second degree polynomial:

$$ P_4(X) \Longrightarrow P_2( \varphi) = 6\varphi^2 - \varphi - 1 $$

We can now calculate the discriminant \( \Delta \):

$$ \Delta = (-1) ^2 - 4 \times (-1) \times 6 = 25$$

Then \(P_2(\varphi ) \) has two real roots \(\varphi _1, \varphi _2\):

$$ \varphi _1 = \frac{- (-1) - \sqrt{25}}{2 \times 6} $$
$$ \varphi _2 = \frac{- (-1) + \sqrt{25}}{2 \times 6} $$
$$ \varphi _1= -\frac{1}{3} $$
$$ \varphi _2 = \frac{1}{2 } $$
$$ \mathcal{S} = \biggl \{\varphi_{1} =-\frac{1}{3} , \ \varphi_{2} = \frac{1}{2 } \biggr \} $$

And \(P_2(\varphi) \) can be factorized:

$$ P_2(\varphi) = 6 \left(\varphi + \frac{1}{3} \right)\left(\varphi - \frac{1}{2 } \right) $$

We finally replace \(\varphi\) by its initial value, \( \varphi = X^2\) :

$$ P_4(X) = 6 \left(X^2 + \frac{1}{3} \right)\left(X^2 - \frac{1}{2 } \right) $$

We can further break it down:

But, we know from the third quadratic remarkable identity that:

$$ A^2 - B^2 = (A-B)(A+B)$$

So,

$$ P_4(X) = 6 \left(X^2 + \frac{1}{3} \right)\left(X + \frac{1}{\sqrt2 } \right)\left(X - \frac{1}{\sqrt2} \right) $$
$$ P_4(X) = 6 \left(X^2 + \frac{1}{3} \right)\left(X + \frac{\sqrt2}{2 } \right)\left(X - \frac{\sqrt2}{2} \right) $$