The equation of the form \(ax^2+bx+c = 0,\) where \(a,b,c \in \mathbb{R},\) and \(a\neq 0\) is called quadratic equation. In this equation \(x\) represents the unknown while \(a,b\) and \(c\) are the coefficients of the equation. The solutions of quadratic
equation are called the roots of quadratic equation. $$ ax^2 + bx+ c = 0 \quad\Rightarrow\quad x_{1,2} = \frac{-b\pm\sqrt{b^2 -4ac}}{2a}$$ The number and the nature of the solution are dependent on discriminant \(D = b^2 -4ac.\)
If discriminant is:
If discriminant is:
- \(D < 0, x_{1, 2} \in \mathbb{C}\) - The solutions are compelx-conjugate numbers,
- \(D = 0, x_{1, 2} \in \mathbb{R}, x_1 = x_2\) - The solutions are real and equal,
- \(D > 0, x_{1, 2} \in \mathbb{R}, x_1 \neq x_2\) - The solutions are real and different.
- Example 1 - Solve the quadratic equation \(3x^2 +4x = 0\)
Solution - This is incomplete quadratic equation since parameter \(c\) is equal to 0. From the equation extract the \(x\). $$ x(3x + 4) =0 \Rightarrow x_1 =0, x_2 = -\frac{4}{3}.$$ - Example 2 - Solve the quadratic equation \(16x^2-4 = 0.\)
Solution - This is incomplete quadratic equation since the parameter \(c\) is equal to 0. $$ 16x^2-4 = 0 \Rightarrow 16x^2 = 4 /\cdot \frac{1}{16} \Rightarrow x^2 =\frac{1}{4}/\sqrt{}$$ $$ x_{1,2} = \pm \frac{1}{2} \Rightarrow x_1 = \frac{1}{2}, x_2 = -\frac{1}{2}$$ - Example 3 - Solve the quadratic equation \(x^2-2x-15 = 0.\)
Solution - In comparison with previous two example in this example the quadratic equation is complete which means that the equations has a,b and c coefficients. The formula for calculating solutions of quadratic equations can be applied. $$ x^2 - 2x - 15 = 0 \Rightarrow a = 1, b=-2, c=-15,$$ $$\begin{eqnarray}x_{1,2} &=& \frac{-b \pm \sqrt{b^2-4ac}}{2a}\\\nonumber x_{1,2}&=& \frac{-(-2) \pm \sqrt{(-2)^2 - 4\cdot 1 \cdot (-15)}}{2\cdot 1} \\ \nonumber x_{1,2} &=& \frac{2 \pm \sqrt{4+60}}{2}\\\nonumber x_{1,2} &=& \frac{2\pm 8}{2} \\ \nonumber x_{1,2} &=& 1\pm 4 \\ \nonumber x_1 &=& 5, x_2 = -3\end{eqnarray}$$ - Example 4 - Solve the equation \(\frac{3x^2 +2}{x^2-1} + \frac{2(x-2)}{x+2} = \frac{5(x^2-x-1)}{x^2 -1}\)
Solution - $$\begin{eqnarray}\frac{3x^2 +2}{x^2-1} + \frac{2(x-2)}{x+2} &=& \frac{5(x^2-x-1)}{x^2 -1}\\ \nonumber \frac{2(x-2)}{x+2} &=&\frac{5x^2-5x-5 -3x^2 -2)}{x^2 -1}\\ \nonumber \frac{2x-4}{x+2} &=& \frac{2x^2-5x-7)}{x^2 -1}/\cdot (x+2)(x^2-1)\\ \nonumber (2x-4)(x^2-1) &=& (x+2)(2x^2-5x-7)\\\nonumber 2x^3 - 2x -4x^2 +4 &=& 2x^3-5x^2-7x + 4x^2 - 10x - 14 \\ \nonumber -3x^2 + 15x +18 &=& 0/: (-3) \\ \nonumber x^2 -5x -6 &=& 0\end{eqnarray}$$ $$a = 1, b = -5, c = -6 $$ $$ \begin{eqnarray}x_{1,2} &=& \frac{-b \pm \sqrt{b^2-4ac}}{2a}\\ \nonumber x_{1,2} &=& \frac{5 \pm \sqrt{25 - 4\cdot 1 \cdot (-6)}}{2}\\ \nonumber x_{1,2} &=& \frac{5\pm\sqrt{25+24}}{2} \\ \nonumber x_{1,2} &=& \frac{5\pm 7}{2} \\ \nonumber x_1 &=& \frac{5+7}{2} = 6\\ \nonumber x_2 &=& \frac{5-7}{2} = -1.\end{eqnarray}$$ - Example 5 - For the given quadratic equation \(2(k-2) = (x+k)^2\) where \(k \in \mathbb{R}\) determine for which values of \(k\) the equation do not have real solutions.
Solution - $$\begin{eqnarray} 2k-4 &=& x^2 + 2xk + k^2\\ \nonumber x^2 +2kx + k^2 -2k + 4 &=& 0 \\ \nonumber a &=& 1, b =2k, c = k^2 -2k +4\end{eqnarray}$$ $$\begin{eqnarray}D & <& 0 \\ \nonumber b^2 -4ac &<& 0 \\ \nonumber 4k^2 -4\cdot 1 \cdot (k^2-2k+4) &<&0 \\ \nonumber 4k^2 -4k^2 +8k - 16 &<& 0 8k &<& 16 \\ \nonumber k &<& 2 \end{eqnarray}$$ - Example 6 - Write the quadratic equation with real coefficients if the solutions of that equation are 4 and -3.
Solution - This example can be solved using Vietes?? formulas i.e. $$x_1 + x_2 = -\frac{b}{a} \Rightarrow 4-3 =\frac{b}{a} \Rightarrow \frac{b}{a} = -1$$ $$x_1\cdot x_2 = \frac{c}{a} \Rightarrow 4\cdot (-3) = \frac{c}{a} \Rightarrow \frac{c}{a} = -12$$ With substitutions of obtained reults into the general formula for equadratic equation $$ x^2 +\frac{b}{a}x + \frac{c}{a} = 0 \Rightarrow x^2 - x -12 = 0.$$ - Example 7 - Write the quadratic equation with real coefficients if the number \(1+\frac{\sqrt{3}}{i}\) is the only solution.
Solution - Rationalize the solution of the quadratic equation. $$ x_1 = 1+\frac{\sqrt{3}}{i}\cdot \frac{-i}{-i} = 1 - i\sqrt{3}$$ $$x_2 = 1-\frac{\sqrt{3}}{i}\cdot \frac{-i}{-i} = 1 + i \sqrt{3}. $$ $$ (1+i\sqrt{3}) + (1-i\sqrt{3})= -\frac{b}{a} \Rightarrow \frac{b}{a} = -2$$ $$ \frac{c}{a} = (1-i\sqrt{3})(1+i\sqrt{3}) = 1 +3 = 4.$$ $$ x^2 + \frac{b}{a}x + \frac{c}{a} = 0 \Rightarrow x^2 -2x + 4 = 0.$$ - Example 8 - Simplify the expression \(\frac{6x^2 +23x +21}{15+7x-2x^2}\)
Solution - The procedure for simplifying this expression can be divided on following steps:- Solve the quadratic equation in numerator,
- Solve the quadratic equation in denomuinator
- Write the solved quadratic expressions in form of product
No comments:
Post a Comment