The analytical proof of the quadratic formula used to solve quadratic equations is presented. Examples show how to apply the quadratic formula and discriminant to solve different quadratic equations with detailed explanations.
A quadratic equation in standard form is given by:
\[ a x^2 + b x + c = 0 \]
where \(a \neq 0\) and \(a, b, c\) are constants.
Solve the above equation to find the quadratic formula:
Solve the equation:
\[ 2x^2 - 7x + 4 = 0 \]
Coefficients: \(a = 2, b = -7, c = 4\)
Discriminant: \(\Delta = b^2 - 4ac = (-7)^2 - 4(2)(4) = 17\)
Since \(\Delta > 0\), two real solutions using the quadratic formula:
\[ x_1 = \frac{7 + \sqrt{17}}{4}, \quad x_2 = \frac{7 - \sqrt{17}}{4} \]
Solve the equation:
\[ x^2 - 2x + 5 = 0 \]
Coefficients: \(a = 1, b = -2, c = 5\)
Discriminant: \(\Delta = b^2 - 4ac = (-2)^2 - 4(1)(5) = -16\)
Since \(\Delta < 0\), two complex solutions. Let \(i = \sqrt{-1}\).
\[ x_1 = 1 + 2i, \quad x_2 = 1 - 2i \]
How many solutions does the equation \[ 2x^2 + 3x = -9 \] have?
Standard form: \[ 2x^2 + 3x + 9 = 0 \]
Coefficients: \(a = 2, b = 3, c = 9\)
Discriminant: \(\Delta = 3^2 - 4(2)(9) = -63 < 0\), hence two complex solutions.
Find all values of \(m\) such that \[ x^2 + m x + 4 = 0 \] has two real solutions.
Coefficients: \(a = 1, b = m, c = 4\)
Discriminant: \(\Delta = m^2 - 16 > 0\)
Solve the inequality: \((m-4)(m+4) > 0\)
Solution interval: \(m \in (-\infty, -4) \cup (4, \infty)\)
Find all values of \(k\) such that \[ -x^2 + (k+1)x - 2 = k \] has one solution only.
Standard form: \[ -x^2 + (k+1)x - 2 - k = 0 \]
Coefficients: \(a = -1, b = k+1, c = -2 - k\)
Discriminant: \(\Delta = (k+1)^2 - 4(-1)(-2 - k) = k^2 - 2k - 7\)
Solve \(\Delta = 0\):
\[ k_1 = 1 + 2\sqrt{2}, \quad k_2 = 1 - 2\sqrt{2} \]
Using either value of \(k\) in the original equation yields one solution only.