Roots, Coefficients, and Discriminants of Quadratic Expressions

Roots And Coefficients

In a quadratic equation, the relation between its roots and coefficients is not negligible. In lower secondary, knowing how to use and to apply the Viete Theorem is more than enough.

The Viete Theorem states that if $x_1,x_2$ are the real roots of the equation $ax^2+bx+c=0 (a\neq0)$ , then:

$\displaystyle{x_1+x_2=-\frac{b}{a}}$
$\displaystyle{x_1x_2=\frac{c}{a}}$

Proof: (need not know)

By the factor theorem, the equation $ax^2+bx+c=0(a\neq0)$ has roots $x_1,x_2$ iff

$ax^2+bx+c=a(x-x_1)(x-x_2), \forall x \in R$ .

Expanding the R.H.S. gives us $ax^2-a(x_1+x_2)x+ax_1x_2$ , thus by the comparing the coefficients yields us the theorem.

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There is also an inverse theorem, basically it states that for any 2 real numbers $\alpha, \beta$ , the equation

$x^2-(\alpha+\beta)x+\alpha\beta =0$

has $\alpha, \beta$ as its two real roots.

The inverse theorem basically says that the Viete theorem works vice versa.

Viete theorem is useful in the sense that if you are given the values of $a,b,c$ in a quadratic equation, not only the roots $x_1,x_2$ can be solved (by using simultaneous equations), many expressions in $x_1,x_2$ can be given by $a,b,c$ , provided the expression is a function of $x_1+x_2$ and $x_1x_2$ .

For example, the following expressions are often used and solved by Viete theorem:

i) $\displaystyle{x_1^2 + x_2^2=(x_1+x_2)^2-2x_1x_2=\frac{b^2}{a^2}-\frac{2c}{a}}$
ii) $\displaystyle{(x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2=\frac{\Delta}{a^2}}$
iii) $\displaystyle{\frac{1}{x_1}+\frac{1}{x_2}=\frac{x_1+x_2}{x_1x_2}=-\frac{b/a}{c/a}=-\frac{b}{c}}$
iv) $\displaystyle{x_1^3+x_2^3=(x_1+x_2)(x_1^2-x_1x_2+x^2_2)=(x_1+x_2)[(x_1+x_2)^2-3x_1x_2]=\left(-\frac{b}{a}\right)^3+3\frac{bc}{a^2}}$

By using these expressions, it is possible to establish new equations to solve for $x$ .

Example 1

Given that $m$ is a real number not less than $-1$ , such that the equation in $x$ :

$x^2 +2(m-2)x+m^2-3m+3=0$

has two distinct real roots $x_1, x_2$ .

If $x_1^2+x_2^2=6$ , find the value of $m$ .
Find the maximum value of $\displaystyle{\frac{mx_1^2}{1-x_1}+\frac{mx_2^2}{1-x_2}}$ .

Solution 1.1

The equation has 2 distinct real roots implies that $\Delta > 0$ , thus

$\Delta = 4(m-2)^2-4(m^2-3m+3)=-4m+4>0$

Therefore, we obtain $-1\leq m<1$ .

By Viete Theorem, $x_1+x_2=-2(m-2)$ and $x_1x_2=m^2-3m+3$ , thus

$x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=4(m-2)^2-2(m^2-3m+3)=2m^2-10m+10$ .

Therefore $2m^2-10m+10=6$ , i.e. $m^2-5m+2=0$ . By applying the quadratic formula, we obtain

$\displaystyle{m=\frac{5\pm\sqrt{17}}{2}}$

Since $-1\leq m<1$ ,

$\displaystyle{m=\frac{5-\sqrt{17}}{2}}$ .

Solution 1.2

$\displaystyle{\frac{mx_1^2}{1-x_1}+\frac{mx_2^2}{1-x_2} = \frac{m[x_1^2(1-x_2)+x_2^2(1-x_1)]}{(1-x_1)(1-x_2)}}$

$\displaystyle{=\frac{m[x_1^2+x_2^2-x_1x_2(x_1+x_2)]}{x_1x_2-(x_1+x_2)+1}}$

$\displaystyle{=\frac{m[(2m^2-10m+10)+2(m^2-3m+3)(m-2)]}{m^2-3m+3+2(m-2)+1}}$

$\displaystyle{=\frac{m(2m^3-8m^2+8m-2)}{m^2-m}=2(m^2-3m+1)}$

By completing the squares,

$\displaystyle{2(m^2-3m+1)=2\left(m-\frac{3}{2}\right)^2-\frac{5}{2} \leq 2\left(-1-\frac{3}{2}\right)^2-\frac{5}{2} =10}$ , since $-1\leq m < 1$ .

Thus the maximum value of $\displaystyle{\frac{mx_1^2}{1-x_1}+\frac{mx_2^2}{1-x_2}}$ is $10$ .

Extended Viete Formula

If $\alpha, \beta$ and $\gamma$ are the roots of the cubic equation $ax^3+bx^2+cx+d=0$ , then

$\displaystyle{\sum\alpha = \alpha +\beta+\gamma=-\frac{b}{a}}$ , $\displaystyle{\sum\alpha\beta=\alpha\beta+\beta\gamma+\gamma\alpha=\frac{c}{a}}$ , $\displaystyle{ \alpha\beta\gamma=-\frac{d}{a}}$ .