Quadratic Equation

Definition

The quadratic formula gives us the solutions, or roots, to a quadratic equation of the form \(ax^2 + bx + c\):

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

Technique

What is the sum of the possible solutions of \( x^2 - 2x - 3 = 0 \)?

Using the quadratic formula:

\[ \begin{align} x &= \frac{-(-2) \pm \sqrt{(2)^2 - 4(1)(-3)}}{2(1)} \\ x &= 1 \pm 2 \\ x &=3 \text{ or } x=-1 \end{align} \]

The the answer is \( 3-1=2 \). \( _\square \)

The quadratic formula is helpful even when the solutions are complex numbers:

 

The roots of \( x^2 - 4x + 5 \) are two complex numbers, \( z_1 \) and \( z_2 \). What is the sum of \( z_1 \) and \( z_2 \)?

\[ \begin{align} x &= \frac{-(-4) \pm \sqrt{(-4)^2-4(1)(5)}}{2(1)} \\
x &= \frac{4 \pm \sqrt{-4}}{2} \\ x &= 2 \pm i \end{align} \]

Thus, the two roots are \( z_1 = 2-i \) and \( z_2 = 2 + i \) and their sum is \( (2-i) + (2 + i ) = 4 \). \( _\square \)

Whether the roots are real or complex depends on the quadratic formula's discriminant, \( b^2 - 4ac\), the expression inside the square root. The roots are real when the discriminant is positive and complex when the discriminant is negative.

Application and Extensions

For what value of \(c\) will \( 2x^2 + 7x + c = 0\) have only a single real root?

The quadratic formula's \(\pm\) tells us that there will always be two roots unless the discriminant is equal to 0. So,

\[ \begin{align} b^2 - 4ac &= 0\\
(7)^2 - 4(2)c &= 0\\ c &= \tfrac{49}{8} \end{align} _\square \]

 

\(x\) is a negative number such that \( x^2+9x -22 = 0 \). What is the sum of all possible values of \(y\) which satisfy the equation \( x = y^2 - 13y + 24 \)?

Since \( x^2+9x -22 = 0 \), we know that

\[ \begin{align} x &= \frac{-9 \pm \sqrt{81-4(-22)}}{2} \\ &= \frac{-9 \pm \sqrt{169}}{2} \\ &= \frac{-9 \pm 13}{2} \\ &= 2 \text{ or } -11 \end{align} \]

Since \( x \) is negative, \( x=-11 \). So now we need to solve \( -11 = y^2 - 13y + 24 \).

This produces the following quadratic equation:

\[ \begin{align} y^2 - 13y + 35 &=0 \\ (y-5)(y-7)&=0 \end{align} \]

Thus \( y=5 \text{ or } 7\), and the answer is 12. \( _\square \)

Note by Arron Kau
4 years, 6 months ago

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