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Roots, Coefficients and Discriminants of Quadratic Equations

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.

##

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 .

  1. If x_1^2+x_2^2=6 , find the value of m .
  2. 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}} .

If \alpha,\beta,\gamma and \delta are the roots of the quartic equation ax^4+bx^3+cx^2+dx+e=0, then

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

Example 2

It is known that the roots of the equation

x^5+3x^4-4044118x^3-12132362x^2-12132363x-2011^2=0

are all integers. How many distinct roots does the equation have?

Solution 2

Sum of roots = \displaystyle{\alpha+\beta+\gamma+\delta+\epsilon = -\frac{b}{a} = -3} .

Product of roots = \displaystyle{\alpha\beta\gamma\delta\epsilon=-\frac{f}{a}=2011^2} .

Since 2011 is a prime, we can conclude from the simultaneous equations that:

\alpha=2011, \beta=-2011, \gamma=\delta=\epsilon=-1 .

Therefore, there are 3 distinct roots.

Cheers,
etzhky =)

Roots and Discriminants

Roots are the solutions to a quadratic equation while the discriminant is a number that can be calculated from any quadratic equation.

Discriminant of a quadratic equation =  = \Delta

Nature of the solutions :

1)  , two real solutions

2)  , one real solutions

3)  , no real solutions

Example 1

   has at least a real root, find the range of m.

i) When  , then , so there is a real root .

ii) When , then , so there is a real root .

iii) When m^2-1\neq0 ,  then

\Delta=4(m+2)^2-4(m^2-1)=16m+20\geq0 ,

which implies that \displaystyle{m\geq-\frac{5}{4}} and m\neq\pm1 .

Thus, the range of m is \displaystyle{m\geq-\frac{5}{4}} .

Cheers,
etzhky

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