In general, let us consider the expression \(P_n(x)=a_0x^n+a_1x^{n-1}+…a_n\), where \(a_k\) is real constant for all \(k\) and \(a_0 \ne 0\). This is called a polynomial in \(x\) since it’s the variable, and the natural number \(n\), which is the highest degree of \(x\) in the expression, is termed to be the degree of the polynomial. More specifically, the expression for \(n=2\) is called a quadratic expression.
We define a quadratic in the general form by \(ax^2+bx+c\) for reals \(a,b,c\). Thus, as an example the quadratic \(x^2+5x+6\) has \(a=1\), \(b=5\) & \(c=6\).
A polynomial equation of degree \(n\) can have at most \(n\) roots (\(\alpha\) is a root of \(f(x)\) iff \(f(\alpha )=0\)). Among the real roots, we have a simple way to determine the maximum number of positive roots of an equation. This is called Descartes Rule of Signs, which states
The maximum number of positive roots of an equation f(x)=0 is the number of sign changes from positive to negative and from negative to positive in the coefficients of f(x)
Thus for \(f(x)=2x^2+4x-4\), we have 1 sign change in the equation \(f(x)=0\), and hence we can conclude that it will have maximum 1 positive root.
Ex-1
If \(x=1\) and \(x=2\) are roots of the equation \(f(x)=x^3+ax^2+bx+c=0\) and \(a+b=1\), find the value of \(b\).
Solution:
We have, from our earlier discussion, \(f(1)=0\) & \(f(2)=0\). From the first equation we have \(a+b+c=-1\), and since \(a+b=1\), we conclude that \(c=-2\). Next using the fact that \(f(2)=0\) we have \(8+4a+2b-2=0\). That gives
\(6+2a+2(a+b)=0 \iff 6+2a+2=0 \iff a=-4\). Thus \(b=5\).
2.1 Roots of a Quadratic Equation
For \(f(x)=ax^2+bx+c\), the roots of \(f(x)=0\) are defined by
We derive these roots through a method called Sridhar Acharya Method of solving quadratic equations. The proof is as follows:
We have
Multiplying both sides by \(4a\) we have
Adding \(b^2\) to both sides gives
\((2ax+b)^2=b^2-4ac\). Deriving the roots from tis should be trivial, which includes taking the square root on both sides and getting the value of \(x\).
2.2 Conditions on roots
We call \(b^2-4ac\) as the Discriminant, commonly referred to as \(D\). Note that for roots to be real, \(D \ge 0\). For distinct roots, we must have \( D >0\) (why?).
We call a number \(k\) as rational if it can be represented in the form \(k=\frac{p}{q}\) where \((p,q) \in Z\) and \(q \ne 0\). Thus we must note that even if \(a,b,c\) are all rationals, the roots of \(f(x)=0\) may not be rational. For the roots to be rational we need \(\sqrt{b^2-4ac}\) to be rational, which means \(b^2-4ac=\frac{p^2}{q^2}\) for some integers \(p,q\).
As discussed above, let \(\alpha\) & \(\beta\) be the roots of the quadratic equation \(ax^2+bx+c=0\). We note the folowing 2 results:
We have \(f(x)=ax^2+bx+c=0\). Dividing both sides by \(a\), we have \(x^2+\frac{b}{a}x+\frac{c}{a}=0\). Thus, we can say any general quadratic equation can be written as:
\(x^2-(\)sum of the roots\()x+(\)product of the roots\()=0\)
In general for any polynomial equation \(P(x)=a_0x^n+a_1x^{n-1}+..a_n=0\) with roots \(\alpha_i\) we can say
Trivially for any polynomial \(P(x)=a_0x^n+a_1x^{n-1}+..a_n\) with roots(also called zeros) \(\alpha_i\), we have \(P(x)=a_0(x-\alpha_1)(x-\alpha_2)…(x-\alpha_n)\).
Ex-2
If \(\alpha, \beta\) be the roots of the equation \(ax^2+bx+c=0\), find the equation with roots \(\alpha + \frac{1}{\beta}\) & \(\beta + \frac{1}{\alpha}\).
Solution:
Sum of the roots = \(\alpha+\beta+\frac{1}{\alpha}+\frac{1}{\beta} = (\alpha+\beta)+\frac{\alpha+\beta}{\alpha \beta} =-\frac{b(a+c)}{ac} \).
Product of the roots = \(\alpha\beta + \frac{1}{\alpha\beta}+2=\frac{(c+a)^2}{ac}\).
Thus the equation is:
which on simplification boils down to
Test Your Concepts:
If \(\alpha, \beta\) are roots of the equation \(ax^2+bx+c=0\), find the value of \(\alpha^4+\beta^4\). Hence find the equation with roots \(\alpha^4,\beta^4\).
2.3 Condition for common roots
Let’s say we have 2 quadratic equations \(ax^2+bx+c=0\) & \(a’x^2+b’x+c’=0\). If these 2 equations must have both the roots common, then the ratios between the coefficients of \(x,x^2\) and the constant must be same. Thus
Now let’s assume that they have only one root in common, let that root be \(\alpha\). We then have
\(a\alpha^2+b\alpha+c=0\) & \(a’\alpha^2+b’\alpha+c’=0\).
From Cramer’s Rule it follows that
\(\Rightarrow \alpha = \frac{a’c-ac’}{ab’-a’b}\) and \( \alpha^2 = \frac{bc’-b’c}{ab’-a’b}\). Thus we can say the condition for 1 common root is
Ex-3
If the equations \(x^2-ax+b=0\) & \(x^2-cx+d=0\) have 1 root in common and the second equation has equal roots, then show that \(ac=2(b+d)\)
Solution:
For the second equation we must have \(D=0\), thus \(c^2-4d=0 \Rightarrow c^2=4d\). Also we can say that \(x=\frac{c}{2}\) is the equal root of the equation. Now since this is the common root, it must satisfy the first equation
Thus \(\frac{c^2}{4}-a\left( \frac{c}{2}\right)+b=0 \Rightarrow c^2+4b=2ac \Rightarrow 4d+4b=2ac\). The conclusion thus follows.
Test Your Concepts:
If the ratio of roots of the equation \(x^2+px+q=0\) be equal to the ratio of roots of \(x^2+bx+c=0\), then show that \(p^2c-b^2q=0\).
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