Theorem: For a quadratic equation $ax^2+bx+c=0$, condition for the two roots to be:
$\qquad\quad\quad\;$ (i) real is $D = b^2-4ac\geq 0$
$\qquad\quad\quad\;$ (ii) equal is $D=0$
$\qquad\quad\quad\;$ (iii) positive is $ab<0\;\text{ and } \; ac>0$
$\qquad\quad\quad\;$ (iv) negative is $ab>0\;\text{ and }\; ac>0$
$\qquad\quad\quad\;$ (v) one root positive and one negative is $ac<0$
Prerequisites:
Quadratic Formula (proof)
Sum and product of roots (Proof)
Proof:
(i) Roots of the quadratic equation given by quadratic formula are:
$\qquad\quad \alpha_{1,2} = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$
For the roots to be real, the term inside square root should be non negative.
Hence, $D = b^2-4ac\geq 0$
(ii) Difference in both the roots is due to the square root term. Hence for equality $D$ should be equal to $0$.
$\therefore\quad\;\; \alpha_{1,2} = -\dfrac{b}{2a}$
(iii) For both the roots to be positive, both sum and the product of the roots should be positive.
$\therefore\quad\;\; -b/a > 0 \quad\text{ and }\quad c/a > 0$.
Multiplying with $a^2$,
$\Rightarrow\quad\;\; ab < 0 \quad \text{ and }\quad ac > 0$.
(iv) Similarly, for both the roots to be negative, sum of the roots should be negative but the product should be positive.
Hence, $ab>0\;\text{ and }\; ac>0$
(v) Again for one root to be positive and one negative, product of the roots should be negative.
Hence, $ac<0$
Showing posts with label Polynomials. Show all posts
Showing posts with label Polynomials. Show all posts
Sum And Product Of The Roots
Theorem: If $f(x)=a_{n}x^n + a_{n-1}x^{n-1} + \cdots + a_{0}x^0$ is a polynomial,$\\[12pt]$ then the normalized coefficients with alternate '$+$'and '$-$' signs, i.e., $-\dfrac{a_{n-1}}{a_{n}}, \dfrac{a_{n-2}}{a_{n}}, \cdots , (-1)^n\dfrac{a_{0}}{a_{n}}$ gives the different sums and products of the roots of the equation $f(x)=0$.
Prerequisites:
Factor Representation of a Polynomial (proof)
Proof:
Let the given polynomial be:
$\qquad\quad f(x)=a_{n}x^n + a_{n-1}x^{n-1} + \cdots + a_{0}x^0 \qquad\qquad\qquad\qquad\qquad\qquad\qquad\cdots (1)$
This can be written in the form of factor representation as:
$\qquad\quad f(x) = (x-\alpha_1)(x-\alpha_2) \cdots (x-\alpha_n) g(x)\qquad\qquad\qquad\qquad\qquad\qquad\;\;\cdots (2)$
Here $\;\alpha_1, \alpha_2, \cdots ,\alpha_n$ are the roots of the equation $f(x)=0$.
Also, $\;deg\big(g(x)\big) = n-n = 0$. This implies that $g(x)$ is a constant $= g$ (say).
Expanding equation $(2)$:
$\qquad\quad f(x) = gx^n - g\left(\alpha_1+\alpha_2+\cdots +\alpha_n\right)x^{n-1} + g\left(\alpha_1\alpha_2+\alpha_2\alpha_3+\cdots +\alpha_{n-1}\alpha_n\right)x^{n-2}\\
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad + \cdots + (-1)^n g\left(\alpha_1\alpha_2\cdots\alpha_n\right)$
Comparing coefficients with equation $(1)$,
$\qquad\quad g = \alpha_n\\
\qquad\quad\displaystyle\sum\limits_{i=1}^n \alpha_i = \alpha_1 +\alpha_2+\cdots +\alpha_n = - \dfrac{a_{n-1}}{a_{n}}\\
\\
\qquad\quad\displaystyle\sum\limits_{i\neq j} \alpha_i\alpha_j =\alpha_1\alpha_2+\alpha_2\alpha_3+\cdots +\alpha_{n-1}\alpha_n = \dfrac{a_{n-2}}{a_{n}}\\
\\
\qquad\quad\;\;\vdots\\[6pt]
\\
\qquad\quad\displaystyle\prod\limits_{i=1}^n \alpha_i = \alpha_1\alpha_2\cdots\alpha_n = (-1)^n\dfrac{a_{0}}{a_{n}}$
Hence the result.
Recommended:
Condition for roots of quadratic equation
Quadratic Formula
Factor Theorem
Prerequisites:
Factor Representation of a Polynomial (proof)
Proof:
Let the given polynomial be:
$\qquad\quad f(x)=a_{n}x^n + a_{n-1}x^{n-1} + \cdots + a_{0}x^0 \qquad\qquad\qquad\qquad\qquad\qquad\qquad\cdots (1)$
This can be written in the form of factor representation as:
$\qquad\quad f(x) = (x-\alpha_1)(x-\alpha_2) \cdots (x-\alpha_n) g(x)\qquad\qquad\qquad\qquad\qquad\qquad\;\;\cdots (2)$
Here $\;\alpha_1, \alpha_2, \cdots ,\alpha_n$ are the roots of the equation $f(x)=0$.
Also, $\;deg\big(g(x)\big) = n-n = 0$. This implies that $g(x)$ is a constant $= g$ (say).
Expanding equation $(2)$:
$\qquad\quad f(x) = gx^n - g\left(\alpha_1+\alpha_2+\cdots +\alpha_n\right)x^{n-1} + g\left(\alpha_1\alpha_2+\alpha_2\alpha_3+\cdots +\alpha_{n-1}\alpha_n\right)x^{n-2}\\
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad + \cdots + (-1)^n g\left(\alpha_1\alpha_2\cdots\alpha_n\right)$
Comparing coefficients with equation $(1)$,
$\qquad\quad g = \alpha_n\\
\qquad\quad\displaystyle\sum\limits_{i=1}^n \alpha_i = \alpha_1 +\alpha_2+\cdots +\alpha_n = - \dfrac{a_{n-1}}{a_{n}}\\
\\
\qquad\quad\displaystyle\sum\limits_{i\neq j} \alpha_i\alpha_j =\alpha_1\alpha_2+\alpha_2\alpha_3+\cdots +\alpha_{n-1}\alpha_n = \dfrac{a_{n-2}}{a_{n}}\\
\\
\qquad\quad\;\;\vdots\\[6pt]
\\
\qquad\quad\displaystyle\prod\limits_{i=1}^n \alpha_i = \alpha_1\alpha_2\cdots\alpha_n = (-1)^n\dfrac{a_{0}}{a_{n}}$
Hence the result.
Recommended:
Condition for roots of quadratic equation
Quadratic Formula
Factor Theorem
Factor Representation Of A Polynomial
Theorem: If $\alpha_1, \alpha_2, \cdots , \alpha_m$ are the roots of a polynomial equation $f(x)=0$ and $f(x)$ has a degree $n$ $(n>m)$, then $f(x)$ can be written as $(x-\alpha_1)(x-\alpha_2) \cdots (x-\alpha_m) g(x)$, such that $deg\big(g(x)\big) = n-m $
Prerequisites:
Factor Theorem (proof)
Proof:
If $\alpha_1$ is a root of the polynomial $f(x)$, then by Factor Theorem, $f(x)$ can be written as:
$\qquad\quad f(x) = q_1(x)(x-\alpha_1)\qquad\qquad\qquad\qquad\qquad\cdots (1)$
Now, since $\alpha_2$ is also a root of $f(x)$, hence $f(\alpha_2) = 0$
$\therefore\quad\;\;\; f(\alpha_2) = q_1(\alpha_2)(\alpha_2-\alpha_1)=0$
Since, $\alpha_2\neq\alpha_1$, therefore $q_1(\alpha_2) = 0$.
Now using Factor Theorem on $q_1(x)$:
$\qquad\quad q_1(x) = q_2(x)(x-\alpha_2)$
Substituting this in equation $(1)$,
$\qquad\quad f(x) = q_2(x)(x-\alpha_2)(x-\alpha_1)$
Continuing in this way, we obtain the desired result:
$\qquad\quad f(x) = (x-\alpha_1)(x-\alpha_2) \cdots (x-\alpha_m) g(x)$
Degree of g(x):
$\qquad\quad deg\big(LHS\big) = n$
$\qquad\quad\!\begin{align}deg\big(RHS\big) &= deg\big((x-\alpha_1)(x-\alpha_2) \cdots (x-\alpha_m)\big) + deg\big(g
(x)\big)\\
&= m + deg\big(g(x)\big)\\
&= deg\big(LHS\big) = n\end{align}\\
\Rightarrow\quad\;\; deg\big(g(x)\big) = n-m$
Hence the result.
Corollary:
(i) Maximum number of real roots of a polynomial of degree $n$ can be $n$.
$\quad$This follows since, $deg\big(g(x)\big) = n-m > 0$, implying $m < n$.
(ii) Total number of all real and complex roots of a polynomial of degree $n$ are $n$.
Recommended:
Sum and product of the roots
Condition for roots of quadratic equation
Quadratic Formula
Prerequisites:
Factor Theorem (proof)
Proof:
If $\alpha_1$ is a root of the polynomial $f(x)$, then by Factor Theorem, $f(x)$ can be written as:
$\qquad\quad f(x) = q_1(x)(x-\alpha_1)\qquad\qquad\qquad\qquad\qquad\cdots (1)$
Now, since $\alpha_2$ is also a root of $f(x)$, hence $f(\alpha_2) = 0$
$\therefore\quad\;\;\; f(\alpha_2) = q_1(\alpha_2)(\alpha_2-\alpha_1)=0$
Since, $\alpha_2\neq\alpha_1$, therefore $q_1(\alpha_2) = 0$.
Now using Factor Theorem on $q_1(x)$:
$\qquad\quad q_1(x) = q_2(x)(x-\alpha_2)$
Substituting this in equation $(1)$,
$\qquad\quad f(x) = q_2(x)(x-\alpha_2)(x-\alpha_1)$
Continuing in this way, we obtain the desired result:
$\qquad\quad f(x) = (x-\alpha_1)(x-\alpha_2) \cdots (x-\alpha_m) g(x)$
Degree of g(x):
$\qquad\quad deg\big(LHS\big) = n$
$\qquad\quad\!\begin{align}deg\big(RHS\big) &= deg\big((x-\alpha_1)(x-\alpha_2) \cdots (x-\alpha_m)\big) + deg\big(g
(x)\big)\\
&= m + deg\big(g(x)\big)\\
&= deg\big(LHS\big) = n\end{align}\\
\Rightarrow\quad\;\; deg\big(g(x)\big) = n-m$
Hence the result.
Corollary:
(i) Maximum number of real roots of a polynomial of degree $n$ can be $n$.
$\quad$This follows since, $deg\big(g(x)\big) = n-m > 0$, implying $m < n$.
(ii) Total number of all real and complex roots of a polynomial of degree $n$ are $n$.
Recommended:
Sum and product of the roots
Condition for roots of quadratic equation
Quadratic Formula
Euclidean Polynomial Division
Theorem: Given two polynomials $f(x)$ and $g(x)$, with $g(x) \neq 0$, there exists unique polynomials $q(x)$ and $r(x)$ such that $f(x)=q(x)g(x)+r(x)$ with $deg\big(r
(x)\big) < deg\big(g(x)\big)$.
Proof:
Existence:
Let $\quad f(x) = a_{n}x^n + a_{n-1}x^{n-1} + \cdots + a_{0}x^0$
and $\quad g(x) = b_{m}x^m + b_{m-1}x^{m-1} + \cdots + b_{0}x^0$
If $\quad n<m$
Then take $q(x) = 0$ and $r(x) = f(x)$
Hence $f(x)=q(x)g(x)+r(x)$ with $deg\big(r(x)\big) = n < deg\big(g(x)\big)$.
If $\quad n\geq m$
Let $\;\;q_{1}(x) = \dfrac{a_{n}}{b_{m}}x^{n-m}$
Then $r_{1}(x) = f(x) - q_{1}(x)g(x)$
Now, if $deg\big(r_{1}(x)\big)< deg\big(g(x)\big)$, then $r(x) = r_{1}(x)$ and $q(x) = q_{1}(x)$. Otherwise:
$\qquad\quad r_{1}(x) = f(x) - q_{1}(x)g(x) = c_{n-1}x^{n-1} + c_{n-2}x^{n-2} + \cdots + c_{0}x^0$
Take $\quad q_{2}(x) = q_{1}(x) + \dfrac{c_{n-1}}{b_{m}}x^{n-m-1}$
This procedure can be continued till degree of $f(x) - q_{k}(x)g(x) < m$. This value of $q_{k}(x) = q(x)$ and corresponding $r_{k}(x) = f(x) - q_{1}(x)g(x) = r(x)$ with $deg\big(r(x)\big) < n = deg\big(g(x)\big)$.
Uniqueness:
If possible, let there exist two sets of polynomials $q_{1}(x)$, $r_{1}(x)$ and $q_{2}(x)$, $r_{2}(x)$ $\left(q_{1}(x)\neq q_{2}(x)\text{ and } r_{1}(x)\neq r_{2}(x)\right)$ such that $f(x)=q_{1}(x)g(x)+r_{1}(x)$ and $f(x)=q_{2}(x)g(x)+r_{2}(x)$ with $deg\big(r_{1}(x)\big)< deg\big(g(x)\big)$ and $deg\big(r_{2}(x)\big)< deg\big(g(x)\big)$.
$\therefore\quad\;\; f(x)=q_{1}(x)g(x)+r_{1}(x) = q_{2}(x)g(x)+r_{2}(x)$
$\Rightarrow\quad\;\; \left(q_{1}(x) - q_{2}(x)\right)g(x) + \left(r_{1}(x)-r_{2}(x)\right) = 0$
$\Rightarrow\quad\;\; \left(q_{1}(x) - q_{2}(x)\right)g(x) = \left(r_{1}(x)-r_{2}(x)\right)$
Since, $deg\big(g(x)\big) = n$, therefore, degree of LHS $>n$. But the degree of RHS$<n$. Hence there is a contradiction. Therefore $q(x)$ and $r(x)$ are unique.
Recommended:
Polynomial Remainder Theorem
Factor Theorem
Factor representation of polynomials
(x)\big) < deg\big(g(x)\big)$.
Proof:
Existence:
Let $\quad f(x) = a_{n}x^n + a_{n-1}x^{n-1} + \cdots + a_{0}x^0$
and $\quad g(x) = b_{m}x^m + b_{m-1}x^{m-1} + \cdots + b_{0}x^0$
If $\quad n<m$
Then take $q(x) = 0$ and $r(x) = f(x)$
Hence $f(x)=q(x)g(x)+r(x)$ with $deg\big(r(x)\big) = n < deg\big(g(x)\big)$.
If $\quad n\geq m$
Let $\;\;q_{1}(x) = \dfrac{a_{n}}{b_{m}}x^{n-m}$
Then $r_{1}(x) = f(x) - q_{1}(x)g(x)$
Now, if $deg\big(r_{1}(x)\big)< deg\big(g(x)\big)$, then $r(x) = r_{1}(x)$ and $q(x) = q_{1}(x)$. Otherwise:
$\qquad\quad r_{1}(x) = f(x) - q_{1}(x)g(x) = c_{n-1}x^{n-1} + c_{n-2}x^{n-2} + \cdots + c_{0}x^0$
Take $\quad q_{2}(x) = q_{1}(x) + \dfrac{c_{n-1}}{b_{m}}x^{n-m-1}$
This procedure can be continued till degree of $f(x) - q_{k}(x)g(x) < m$. This value of $q_{k}(x) = q(x)$ and corresponding $r_{k}(x) = f(x) - q_{1}(x)g(x) = r(x)$ with $deg\big(r(x)\big) < n = deg\big(g(x)\big)$.
Uniqueness:
If possible, let there exist two sets of polynomials $q_{1}(x)$, $r_{1}(x)$ and $q_{2}(x)$, $r_{2}(x)$ $\left(q_{1}(x)\neq q_{2}(x)\text{ and } r_{1}(x)\neq r_{2}(x)\right)$ such that $f(x)=q_{1}(x)g(x)+r_{1}(x)$ and $f(x)=q_{2}(x)g(x)+r_{2}(x)$ with $deg\big(r_{1}(x)\big)< deg\big(g(x)\big)$ and $deg\big(r_{2}(x)\big)< deg\big(g(x)\big)$.
$\therefore\quad\;\; f(x)=q_{1}(x)g(x)+r_{1}(x) = q_{2}(x)g(x)+r_{2}(x)$
$\Rightarrow\quad\;\; \left(q_{1}(x) - q_{2}(x)\right)g(x) + \left(r_{1}(x)-r_{2}(x)\right) = 0$
$\Rightarrow\quad\;\; \left(q_{1}(x) - q_{2}(x)\right)g(x) = \left(r_{1}(x)-r_{2}(x)\right)$
Since, $deg\big(g(x)\big) = n$, therefore, degree of LHS $>n$. But the degree of RHS$<n$. Hence there is a contradiction. Therefore $q(x)$ and $r(x)$ are unique.
Recommended:
Polynomial Remainder Theorem
Factor Theorem
Factor representation of polynomials
Polynomial Remainder Theorem
Theorem: Remainder of division of a polynomial $f(x)$ by a linear factor $(x-a)$ is $f(a)$
Prerequisites:
Euclidean Polynomials Division (proof)
Proof:
Any polynomial can be written according to the Euclidean polynomial division as:
$\qquad\quad f(x) = q(x)g(x) + r(x)\qquad\text{where }\;\; deg\big(r(x)\big) < deg\big(g(x)\big)\qquad\qquad\ldots (1)$
Here $q(x)$, $g(x)$ and $r(x)$ are the quotient, divisor and remainder respectively and $deg(.)$ represents the degree of the respective polynomial.
Since $g(x) = (x-a)$, therefore $deg\big(g(x)\big) = 1$
Hence, $deg\big(r(x)\big) = 0$
$\Rightarrow \quad\;\; r(x) = r =$ constant
$\therefore \qquad\! r(x)$ is independent of $x$.
Putting $x=a$ and $g(x) = (x-a)$ in equation $(1)$,
$\qquad\quad f(a) = q(a)(a-a) + r\\
\Rightarrow\;\quad\; f(a) = r$
Hence the result
Recommended:
Factor Theorem
Factor representation of polynomials
Prerequisites:
Euclidean Polynomials Division (proof)
Proof:
Any polynomial can be written according to the Euclidean polynomial division as:
$\qquad\quad f(x) = q(x)g(x) + r(x)\qquad\text{where }\;\; deg\big(r(x)\big) < deg\big(g(x)\big)\qquad\qquad\ldots (1)$
Here $q(x)$, $g(x)$ and $r(x)$ are the quotient, divisor and remainder respectively and $deg(.)$ represents the degree of the respective polynomial.
Since $g(x) = (x-a)$, therefore $deg\big(g(x)\big) = 1$
Hence, $deg\big(r(x)\big) = 0$
$\Rightarrow \quad\;\; r(x) = r =$ constant
$\therefore \qquad\! r(x)$ is independent of $x$.
Putting $x=a$ and $g(x) = (x-a)$ in equation $(1)$,
$\qquad\quad f(a) = q(a)(a-a) + r\\
\Rightarrow\;\quad\; f(a) = r$
Hence the result
Recommended:
Factor Theorem
Factor representation of polynomials
Factor Theorem
Theorem: Polynomial $f(x)$ has a factor $(x-a)$ if and only if $f(a)=0$
Prerequisites:
Euclidean Polynomials Division (proof)
Polynomial Remainder Theorem (proof)
Proof:
Any polynomial can be written according to the Euclidean division as:
$\qquad\quad f(x) = q(x)g(x) + r(x)$
Here $q(x)$, $g(x)$ and $r(x)$ are the quotient, divisor and remainder respectively.
Putting $g(x) = (x-a)$, from the remainder theorem, $r(x) = f(a)$
$\therefore\quad\;\; \begin{equation} f(x) = q(x)(x-a) + f(a)\end{equation}\qquad\qquad\qquad\qquad\qquad\qquad \ldots (1)$
Now if $(x-a)$ is a factor of $f(x)$, then remainder must be $0$.
$\therefore\quad\;\; r(x) = f(a) = 0$
Conversely, if $f(a) = 0$, then from equation $(1)$;
$\qquad\quad f(x) = q(x)(x-a)$
Hence $(x-a)$ is the factor of $f(x)$.
Hence the theorem.
Recommended:
Factor Representation of a Polynomial
Polynomial Remainder Theorem
Prerequisites:
Euclidean Polynomials Division (proof)
Polynomial Remainder Theorem (proof)
Proof:
Any polynomial can be written according to the Euclidean division as:
$\qquad\quad f(x) = q(x)g(x) + r(x)$
Here $q(x)$, $g(x)$ and $r(x)$ are the quotient, divisor and remainder respectively.
Putting $g(x) = (x-a)$, from the remainder theorem, $r(x) = f(a)$
$\therefore\quad\;\; \begin{equation} f(x) = q(x)(x-a) + f(a)\end{equation}\qquad\qquad\qquad\qquad\qquad\qquad \ldots (1)$
Now if $(x-a)$ is a factor of $f(x)$, then remainder must be $0$.
$\therefore\quad\;\; r(x) = f(a) = 0$
Conversely, if $f(a) = 0$, then from equation $(1)$;
$\qquad\quad f(x) = q(x)(x-a)$
Hence $(x-a)$ is the factor of $f(x)$.
Hence the theorem.
Recommended:
Factor Representation of a Polynomial
Polynomial Remainder Theorem
Quadratic Formula
Theorem: Roots of a quadratic equation $ax^2+bx+c=0$ with $a \neq 0$ are given by $\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$.
Prerequisites:
$(a+b)^2=a^2+b^2+2ab$ (proof)
Proof:
Given equation is:
$\qquad \quad ax^2+bx+c=0$
Dividing the whole equation by a:
$\qquad \quad x^2+\dfrac{b}{a}x+\dfrac{c}{a}=0$
Adding and subtracting $\dfrac{b^2}{4a^2}$:
$\qquad \quad \left(x^2+\dfrac{b}{a}x+\frac{b^2}{4a^2}\right)-\dfrac{b^2}{4a^2}+\dfrac{c}{a}=0$
Using $(a+b)^2=a^2+b^2+2ab$, equation can be rewritten as:
$\qquad \quad \left(x + \dfrac{b}{2a}\right)^2 -\dfrac{b^2}{4a^2}+\dfrac{c}{a}=0\\
\Rightarrow\quad\;\;\begin{align}\left(x + \dfrac{b}{2a}\right)^2 &= \dfrac{b^2}{4a^2}-\dfrac{c}{a}\\
&= \dfrac{b^2 - 4ac}{4a^2}\end{align}$
Taking Square root on both sides:
$\Rightarrow\quad \left(x + \dfrac{b}{2a}\right)= \pm\dfrac{\sqrt{b^2 - 4ac}}{2a}\\
\Rightarrow\quad\;\; x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$
Hence the result
Prerequisites:
$(a+b)^2=a^2+b^2+2ab$ (proof)
Proof:
Given equation is:
$\qquad \quad ax^2+bx+c=0$
Dividing the whole equation by a:
$\qquad \quad x^2+\dfrac{b}{a}x+\dfrac{c}{a}=0$
Adding and subtracting $\dfrac{b^2}{4a^2}$:
$\qquad \quad \left(x^2+\dfrac{b}{a}x+\frac{b^2}{4a^2}\right)-\dfrac{b^2}{4a^2}+\dfrac{c}{a}=0$
Using $(a+b)^2=a^2+b^2+2ab$, equation can be rewritten as:
$\qquad \quad \left(x + \dfrac{b}{2a}\right)^2 -\dfrac{b^2}{4a^2}+\dfrac{c}{a}=0\\
\Rightarrow\quad\;\;\begin{align}\left(x + \dfrac{b}{2a}\right)^2 &= \dfrac{b^2}{4a^2}-\dfrac{c}{a}\\
&= \dfrac{b^2 - 4ac}{4a^2}\end{align}$
Taking Square root on both sides:
$\Rightarrow\quad \left(x + \dfrac{b}{2a}\right)= \pm\dfrac{\sqrt{b^2 - 4ac}}{2a}\\
\Rightarrow\quad\;\; x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$
Hence the result
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