Index
A quadratic equation is written in the form:
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 \( \Delta \):
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\( \alpha) \) 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:
$$ P_2(X) = a(X - C_1)(X - C_2) $$
Furthermore, in the general case we will also have these two relationships between the coefficients and the roots:
Solving by finding roots and factorizing
A quadratic equation is written in the form:
We want to solve the equation :
So we start from the equation \( (1) \):
First of all, we factorize it by \( a \):
However, we notice that the expression in brackets has the same start as \(\left(X + \frac{b}{2a}\right)^2 \), because :
So that,
By rewriting the equation the other way around:
We can then transform \( (2) \) by injecting \( (3) \):
We then recognize the third quadratic remarkable identity which is:
With:
Which leads us to:
For the sake of simplicity, let us set down:
We now have:
So:
Starting from this result, there will be three cases to take into account.
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\( \alpha) \) 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.
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Sum of roots\(: X_1 + X_2 \)
$$ X_1 + X_2 = \frac{- b - \sqrt{\Delta}}{2a} + \frac{- b + \sqrt{\Delta}}{2a}$$$$ X_1 + X_2 = \frac{- 2b}{2a} $$$$ X_1 + X_2 = -\frac{b}{a} $$ -
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:
$$ X_1 X_2 = \frac{b^2 - (b^2 - 4ac)}{4a^2} $$$$ X_1 X_2 = \frac{ 4ac}{4a^2} $$$$ X_1 X_2 = \frac{c}{a} $$
Examples
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Solving a second degree polynomial
$$ P_2(X) = 2X^2 - 3X + 1 = 0 $$We calculate the discriminant \( \Delta \):
$$ \Delta = (-3) ^2 - 4 \times 2 \times 1 = 1$$\(\Delta\) is positive, so \(P_2(X) \) has two real roots \(X_1, X_2\):
$$ X_1 = \frac{- (-3) - \sqrt{1}}{2 \times 2} $$$$ X_1 = \frac{2}{4} $$$$ X_2 = \frac{- (-3) + \sqrt{1}}{2 \times 2} $$$$ X_2 = \frac{4}{4} $$$$ \mathcal{S} = \biggl \{X_{1} = \frac{1}{2} , \ X_{2} = 1 \biggr \} $$\(P_2(X) \) 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 \( X = 1\).
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Solving a fourth degree polynomial by a variable change
Let us solve this new equation:
$$ P_4(X) = 6X^4 - X^2 - 1 = 0 $$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 \(\varphi _1, \varphi _2\):
$$ \varphi _1 = \frac{- (-1) - \sqrt{25}}{2 \times 6} $$$$ \varphi _1= -\frac{1}{3} $$$$ \varphi _2 = \frac{- (-1) + \sqrt{25}}{2 \times 6} $$$$ \varphi _2 = \frac{1}{2 } $$$$ \mathcal{S} = \biggl \{\varphi_{1} =-\frac{1}{3} , \ \varphi_{2} = \frac{1}{2 } \biggr \} $$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,
$$ P_4(X) = 6 \left(X^2 + \frac{1}{3} \right)\left(X + \frac{1}{\sqrt2 } \right)\left(X - \frac{1}{\sqrt2} \right) $$$$ P_4(X) = 6 \left(X^2 + \frac{1}{3} \right)\left(X + \frac{\sqrt2}{2 } \right)\left(X - \frac{\sqrt2}{2} \right) $$
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