Quadratic Equations

  • Quadratic Equations

Quadratic Equations


Quadratic equation is an expression or an equation that contains the variable squared, in the form $ax^2+bx+c$, where $a,b,c \in \Re, a \neq 0$. It is also known as second-degree polynomial equation. When $b=0$, then the equation is known as a pure quadratic equation.

Quadratic Formula

The quadratic formula is derived from completing the square of a quadratic equation.

$$ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

The value of x in a quadratic equation can also be solved by factoring the equation if the equation is a perfect square. A perfect square gives a single solution to the quadratic equations.

Example:

  1. Determine the solution set of x in 3m2 - 4m - 2 = 0.
    2. Solution:
    $$ 3m^2 -4m -2 = 0$$ $$\begin{align} x & = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ x_1 & = \frac{-(-4) + \sqrt{(-4)^2-4(3)(-2)}}{2(3)} \\ & = \frac{2}{3} + \frac{\sqrt{10}}{3} \\ x_2 & = \frac{-(-4) - \sqrt{(-4)^2-4(3)(-2)}}{2(3)} \\ & = \frac{2}{3} - \frac{\sqrt{10}}{3} \\ \end{align}$$

  2. Find the roots of n2 - 3n = -2.
    3. Solution:
    $$ n^2 - 3n = -2 $$ $$\begin{align} x & = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ x_1 & = \frac{-(-3) + \sqrt{(-3)^2-4(1)(2)}}{2(1)} \\ & = 2 \\ x_2 & = \frac{-(-3) - \sqrt{(-3)^2-4(1)(2)}}{2(1)} \\ & = 1\\ \end{align}$$

  3. Find the equation whose roots are the reciprocals of the roots of the equation 2x2 - 3x - 5 = 0.
    4. Solution:
    $ 2x^2 - 3x - 5 = 0$ is a perfect square. By factoring,
    $$(2x-5)(x+1) = 0 $$ $$\begin{align} 2x-5 &= 0\\ 2x & =5 \\ x_1 & = 5/2\\ x+1 & = 0\\ x_2 & = -1\\ \end{align}$$ The roots of the required equation must be -1 and 2/5. $$\begin{align} \left( x - \frac{2}{5} \right) \left( x + 1 \right) & = 0 \\ x^2 + x - \frac{2}{5}x - \frac{2}{5} & = 0\\ x^2 + \frac{3}{5}x - \frac{2}{5} & = 0\\ 5x^2 + 3x - 2 & = 0\\ \end{align}$$

  4. Find the value of m that will make 4x2 - 4mx + 4m + 5 a perfect square trinomial.
    5. Solution:
    $$4x^2-4mx+4m+5 = 0$$ Step 1: Transpose all constants to the other side of the equation. Divide each term by $a$ to simplify.
    $$\begin{align} 4x^2 - 4mx & = -4m - 5 \\ \frac{4x^2}{4} - \frac{4mx}{4} & = - \frac{4m+5}{4} \\ x^2 - mx & = -\frac{4m+5}{4} \\ \end{align}$$ Step 2: To make a perfect square, take the coefficient in the first degree term $mx$, multiply by 1/2 and square the result.
    $$\begin{align} \left( m \cdot \frac{1}{2} \right)^2 = \frac{m^2}{4}\\ \end{align}$$ Add the result to both sides.
    $$\begin{align} x^2 - mx + \left( \frac{m^2}{4} \right) & = -\frac{4m+5}{4} + \left( \frac{m^2}{4} \right) \\ \left( x - \frac{m}{2} \right)^2 &= \frac{m^2 - 4m -5}{4} \end{align}$$ Step 3: To get the value of m, let $\left( x - \frac{m}{2} \right)^2 = 0$. Similarly, $$\begin{align} \frac{m^2 - 4m -5}{4} &= 0\\ (m-5)(m+1) &= 0\\ m_1 & = 5\\ m_2 & = -1 \\ \end{align}$$

Discriminant

The quantity $\sqrt{b^2 - 4ac}$ in the quadratic formula is known as the discriminant. It will determine the nature of the roots of the quadratic equation.

$\sqrt{b^2 - 4ac}$ Nature of Roots
0 Only one root, real and equal.
$>0$ Real and unequal
$< 0$ Imaginary and unequal



Sum and Product of Roots

A quadratic equation $ax^2+bx+c=0$ has roots $x_1$ and $x_2$. The sum and product of the roots then are:

Sum of the Roots $$x_1 + x_2 = \frac{-b}{a}$$ Product of the Roots $$ x_1 \times x_2 = \frac{c}{a}$$

Example:

  1. Find the sum of the roots of 5x2 - 10x + 2 = 0.
    2. Solution:
    $$ 5x^2-10x+2=0$$ $$\begin{align} \text{Sum of roots } & = - \frac{b}{a}\\ & = - \frac{(-10)}{5}\\ & = 2 \\ \end{align}$$

  2. If the roots of the equation x2 + Bx + 1 = 0 are the squares of the roots of the equation x2+ bx +1 = 0, then what is the value of B in terms of b?
    3. Solution:
    Let the roots of $x^2+bx+1 = 0$ be $r_1$ and $r_2$.
    Sum of roots of $(x^2+bx+1 = 0)$: $$\begin{align} r_1 + r_2 &= \frac{-b}{1} \\ & = -b\\ \end{align}$$ Product of roots of $(x^2+bx+1 = 0)$: $$\begin{align} r_1 \cdot r_2 &= \frac{1}{1} \\ & = 1\\ \end{align}$$ Square the sum of roots of $(x^2+bx+1 = 0)$, then substitute $(r_1 \cdot r_2)$ $$\begin{align} (r_1 + r_2)^2 & = (-b)^2 \\ r_1^2 + 2 r_1 r_2 + r_2^2 & = b^2 \\ r_1^2 + 2(1) + r_2^2 & = b^2 \\ r_1^2 + r_2^2 & = b^2 - 2 & \text{< Eq. 1>} \\ \end{align}$$ For $(x^2 + Bx + 1 = 0)$, the sum of the roots are equal to the sum of squares of the roots of $(x^2+bx+1 = 0)$, thus, $$\begin{align} r_1^2 + r_2^2 & = \frac{-B}{1} \\ r_1^2 + r_2^2 & = -B & \text{ < Eq. 2>} \\ \end{align}$$ Substitute Eq. 2 to Eq. 1, $$\begin{align} r_1^2 + r_2^2 & = b^2 - 2 \\ -B & = b^2 - 2 \\ B &= 2 - b^2 \\ \end{align}$$