Solve using the quadratic formula
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This lesson shows how to solve using the quadratic formula. To use the quadratic formula, you need to identify a, b, and c using the standard form of a quadratic equation ax2 + bx + c = 0.
Quadratic formula
$$ x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a} $$To find x or the solution of a quadratic equation, the quadratic formula tells you to take the negative of b. Then, add or subtract negative b to the square root of b squared minus 4ac. Finally divide everything you got so far by 2a.
The plus or minus (±) in the formula means that you will end up with two answers.
You can also write the quadratic formula as x = [-b ± √(b2 - 4ac)]/2aFirst, study carefully the example in the image below. If you do not quite understand it do not worry. Keep reading to see more examples.
How to identify a, b, and c
It is important to avoid substituting the wrong values for a, b, and c so you do not end up with incorrect solutions. As a math teacher, I have seen students making this mistake quite often. Therefore, study the three examples below carefully so you know how to identify a, b, and c.
The standard form is ax2 + bx + c = 0.
1) for 6x2 + 8x + 7 = 0, we get a = 6, b = 8, and c = 7
2) for x2 + 8x - 7 = 0, we get a = 1, b = 8, and c = -7
3) for -x2 - 8x + 7 = 0, we get a = -1, b = -8, and c = 7
The value of the expression under the radical sign or b2 - 4ac has a special name. It is called discriminant! Its value will tell you the number of solutions and whether the solutions are real solutions (real roots) or complex solutions (complex roots).
More examples showing how to solve using the quadratic formula
Example #1:Solve using the quadratic formula x2 + 8x + 7 = 0
a = 1, b = 8, and c = 7
x = [-b ± √(b2 - 4ac))] / 2a
x = [-8 ± √(82 - 4 × 1 × 7)] / 2 × 1
x = [-8 ± √(64 - 4 × 1 × 7)] / 2
x = [-8 ± √(64 - 4 × 7)] / 2
x = [-8 ± √(64 - 28)] / 2
x = [-8 ± √(36)] / 2
x = (-8 ± 6 ) / 2
x1 = (-8 + 6 ) / 2
x1 = (-2 ) / 2
x1 = -1
x2 = (-8 - 6 ) / 2
x2 = (-14 ) / 2
x2 = -7
Example #2:
Solve using the quadratic formula 4x2 - 11x - 3 = 0
a = 4, b = -11, and c = -3
x = [-b ± √(b2 - 4ac)] / 2a
x = [- -11 ± √( (-11)2 - 4 × 4 × -3)] / 2 × 4
x = [11 ± √(121 - 4 × 4 × -3)] / 8
x = [11 ± √(121 - 4 × -12)] / 8
x = [11 ± √(121 + 48)] / 8
x = [11 ± √(169)] / 8
x = (11 ± 13 ) / 8
x1 = (11 + 13 ) / 8
x1 = (24 ) / 8
x1 = 3
x2 = (11 - 13 ) / 8
x2 = (-2 ) / 8
x2 = -1/4
Example #3:
Solve using the quadratic formula x2 + x - 2 = 0
a = 1, b = 1, and c = -2
x = [-b ± √(b2 - 4ac)] / 2a
x = [-1 ± √((1)2 - 4 × 1 × -2)] / 2 × 1
x = [-1 ± √(1 - 4 × 1 × -2)] / 2
x = [-1 ± √(1 - 4 × -2)] / 2
x = [-1 ± √(1 + 8)] / 2
x = [-1 ± √(9)] / 2
x = (-1 ± 3 ) / 2
x1 = (-1 + 3 ) / 2
x1 = (2) / 2
x1 = 1
x2 = (-1 - 3 ) / 2
x2 = (-4 ) / 2
x2 = -2
Notice that in all three examples above, the discriminant is positive and the number of solutions of each quadratic equation is 2. For example, in example #3, the discriminant shown in blue is 9 and the solutions are 1 and -2.
In general, when the discriminant is positive, you will get two answers.
Example #4:
Solve using the quadratic formula 4x2 + 12x + 9 = 0
a = 4, b = 12, and c = 9
x = [-b ± √(b2 - 4ac)] / 2a
x = [-12 ± √((12)2 - 4 × 4 × 9)] / 2 × 4
x = [-12 ± √(144 - 4 × 4 × 9)] / 8
x = [-12 ± √(144 - 4 × 36)] / 8
x = [-12 ± √(144 - 144)] / 8
x = [-12 ± √(0)] / 8
x = (-12 ± 0) / 8
x1 = (-12 + 0) / 8
x1 = (-12) / 8
x1 = -1.5
x2 = (-12 - 0) / 8
x2 = (-12) / 8
x2 = -1.5
Notice that in example #4, the discriminant shown in blue is zero and there is only one solution.
In general, when the discriminant is zero, you will get only one answer.
Example #5:
Solve using the quadratic formula x2 + 8x + 25 = 0
a = 1, b = 8, and c = 25
x = [-b ± √(b2 - 4ac)] / 2a
x = [-8 ± √((-8)2 - 4 × 1 × 25)] / 2 × 1
x = [-8 ± √(64 - 4 × 25)] / 2
x = [-8 ± √(64 - 100)] / 2
x = [-8 ± √(-36)] / 2
x = [-8 ± √(36(-1))] / 2
x = [-8 ± √(36(i2))] / 2
x = (-8 ± 6i) / 2
x1 = (-8 + 6i) / 2
x1 = -8/2 + 6i/2
x1 = -4 + 3i
x2 = (-8 - 6i) / 2
x2 = -8/2 - 6i/2
x2 = -4 - 3i
Notice that in example #5, the discriminant shown in blue is negative and you ended up with two complex numbers. Since it is not possible to take the square root of a negative number, mathematicians came up with a solution by letting i2 equal to -1 and i is called imaginary number.
In general, when the discriminant is negative, you will get complex numbers as solutions.