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 Quadratic. Show all posts
Showing posts with label Quadratic. Show all posts
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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