The quadratic formula is a math formula that can be used to solve a quadratic equation that is written in standard form ax^{2} + bx + c = 0
$$ x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a} $$Notice that the general form of a quadratic equation is ax^{2} + bx + c = 0 and it is a second order equation in a single variable. Before using the formula shown above, it is important to check two things:
The plus or minus sign (±) in the formula is there to show that the quadratic equation may have two solutions (x_{1} and x_{2}) generally speaking.
$$ x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a}\ and\ x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a} $$The expression b^{2} - 4ac, called discriminant, shows the nature of the solutions.
It is common to use the symbol Δ (delta from the Greek alphabet) when talking about the discriminant.
Δ = b^{2} - 4ac
If Δ or the discriminant is zero, then it makes no difference whether we choose the plus or the minus sign in the formula.
x_{1} = x_{2} = -b/2a
In this case, we say that there is one repeated real solution.
If Δ or the discriminant is positive, then there will be two real solutions.
If Δ or the discriminant is negative, then we will end up taking the square root of a negative number. In this case, there will be two imaginary-number solutions called complex numbers.
Discriminant
For ax^{2} + bx + c = 0:
Δ = b^{2} - 4ac = 0 ⟶ One real-number solution
Δ = b^{2} - 4ac > 0 ⟶ Two different real-number solutions
Δ = b^{2} - 4ac < 0 ⟶ No real solution, but two different imaginary-number solutions.
Check the lesson about discriminant of the quadratic equation to see what the graph of a quadratic equation looks like when the discriminant is either zero, positive, or negative.
Solve x^{2} - 5x + 4 = 0 using the quadratic formula
a = 1, b = -5, and c = 4
$$ x = \frac{-(-5) ± \sqrt{(-5)^2 - 4(1)(4)}}{2(1)} $$ $$ x = \frac{5 ± \sqrt{25 - 16}}{2} $$ $$ x = \frac{5 ± \sqrt{9}}{2} $$ $$ x = \frac{5 ± 3}{2} $$ $$ x_1 = \frac{5 + 3}{2}\ and\ x_2 = \frac{5 - 3}{2} $$ $$ x_1 = \frac{8}{2}\ and\ x_2 = \frac{2}{2} $$ $$ x_1 = 4\ and\ x_2 = 1 $$
The roots of the equation x^{2} - 5x + 4 = 0 are x_{1} = 4 and x_{2} = 1
Please check the lesson about solve using the quadratic formula to see more examples.
Example #1
Suppose a soccer player shoots a penalty kick with an initial velocity of 28 ft/s. When will the ball reach a height of 30 feet?
Solution
The function h = -16t^{2} + vt + s models the height h in feet of the ball at time t in seconds.
The velocity is v and s is the initial height of the ball.
Since the soccer ball must be on the ground before the soccer player shoots the ball, s is equal to 0.
v = 28 ft/s
h is the height of the ball
30 = -16t^{2} + 28t
Since the standard form of a quadratic equation is ax^{2} + bx + c = 0, you need to put 30 = -16t^{2} + 28t in standard form.
Subtract 30 from both sides of the equation
30 - 30 = -16t^{2} + 28t - 30
0 = -16t^{2} + 28t - 30
-16t^{2} + 28t - 30 = 0
Find the values of a,b, and c and then evaluate the discriminant.
a = -16, b = 28 and c = -30
Δ = b^{2} - 4ac = 28^{2} - 4(-16)(-30)
Δ = b^{2} - 4ac = 784 + 64(-30)
Δ = b^{2} - 4ac = 784 + -1920
Δ = b^{2} - 4ac = -1136
Since the discriminant is negative, the equation 30 = -16t^{2} + 28t has no real solutions.
Therefore, the ball will not reach a height of 30 feet.
Example #2
Find the dimensions of a square that has the same area as a circle whose radius is 10 inches.
Solution
Let x be the length of one side of the square. Then, the area of the square is x^{2}
The area of the circle is pir^{2} = 3.14(10)^{2} = 3.14(100) = 314
x^{2} = 314
x^{2} - 314 = 0
x^{2} - 0x - 314 = 0
a = 1, b = 0, and c = -314
Δ = b^{2} - 4ac = 0^{2} - 4(1)(-314)
Δ = b^{2} - 4ac = 1256
√Δ = √(1256) = 35.44
x_{1} = (-b + 35.44) / 2(1)
x_{1} = (-0 + 35.44) / 2
x_{1} = 35.44 / 2 = 17.72
x_{2} = (-b - 35.44) / 2(1)
x_{2} = (-0 - 35.44) / 2
x_{2} = -35.44 / 2 = -17.72
The length of one side of the square is 17.72 inches
Sep 17, 23 09:46 AM
Jun 09, 23 12:04 PM