Here are 10 examples of quadratic equations and their solutions:
Example no 1:
x^2 + 5x + 6 = 0
To solve this equation, we can use the quadratic formula:
x = (-b +/- sqrt(b^2 - 4ac)) / (2a)
Plugging in the values for a, b, and c gives us:
x = (-5 +/- sqrt(5^2 - 416)) / (2*1)
x = (-5 +/- sqrt(25 - 24)) / 2
x = (-5 +/- sqrt(1)) / 2
x = (-5 +/- 1) / 2
x = (-4)/2, (-6)/2
x = -2, -3
So the solutions to this equation are x = -2 and x = -3.
Example no 2:
x^2 - 7x + 12 = 0
Using the quadratic formula, we get:
x = (-7 +/- sqrt(7^2 - 4112)) / (2*1)
x = (-7 +/- sqrt(49 - 48)) / 2
x = (-7 +/- sqrt(1)) / 2
x = (-7 +/- 1) / 2
x = (-6)/2, (-8)/2
x = -3, -4
So the solutions to this equation are x = -3 and x = -4.
Example no 3:
x^2 + 3x - 4 = 0
Using the quadratic formula, we get:
x = (3 +/- sqrt(3^2 - 41(-4))) / (2*1)
x = (3 +/- sqrt(9 + 16)) / 2
x = (3 +/- sqrt(25)) / 2
x = (3 +/- 5) / 2
x = 8/2, -2/2
x = 4, -1
So the solutions to this equation are x = 4 and x = -1.
Example no 4:
x^2 - 5x + 6 = 0
Using the quadratic formula, we get:
x = (-5 +/- sqrt(5^2 - 416)) / (2*1)
x = (-5 +/- sqrt(25 - 24)) / 2
x = (-5 +/- sqrt(1)) / 2
x = (-5 +/- 1) / 2
x = (-4)/2, (-6)/2
x = -2, -3
So the solutions to this equation are x = -2 and x = -3.
Example no 5:
x^2 + x - 6 = 0
Using the quadratic formula, we get:
x = (1 +/- sqrt(1^2 - 41(-6))) / (2*1)
x = (1 +/- sqrt(1 + 24)) / 2
x = (1 +/- sqrt(25)) / 2
x = (1 +/- 5) / 2
x = 6/2, -4/2
x = 3, -2
So the solutions to this equation are x = 3 and x = -2.
Example no 6:
x^2 - 6x + 9 = 0
Using the quadratic formula, we get:
x = (-6 +/- sqrt(6^2 - 419)) / (2*1)
x = (-6 +/- sqrt(36 - 36)) / 2
x = (-6 +/- sqrt(0)) / 2
x = (-6 +/- 0) / 2
x = -6/2, -6/2
x = -3, -3
So the solutions to this equation are x = -3 and x = -3.
Example no 7:
x^2 + 2x + 7 = 0
Using the quadratic formula, we get:
x = (2 +/- sqrt(2^2 - 411)) / (2*1)
x = (2 +/- sqrt(4 - 4)) / 2
x = (2 +/- sqrt(0)) / 2
x = (2 +/- 0) / 2
x = 2/2, 2/2
x = 1, 1
So the solutions to this equation are x = 1 and x = 1.
Example no 8:
x^2 - 4x + 4 = 0
Using the quadratic formula, we get:
x = (-4 +/- sqrt(4^2 - 414)) / (2*1)
x = (-4 +/- sqrt(16 - 16)) / 2
x = (-4 +/- sqrt(0)) / 2
x = (-4 +/- 0) / 2
x = -4/2, -4/2
x = -2, -2
So the solutions to this equation are x = -2 and x = -2.
Example no 9:
x^2 - x - 6 = 0
Using the quadratic formula, we get:
x = (-1 +/- sqrt(1^2 - 41(-6))) / (2*1)
x = (-1 +/- sqrt(1 + 24)) / 2
x = (-1 +/- sqrt(25)) / 2
x = (-1 +/- 5) / 2
x = 4/2, -6/2
x = 2, -3
So the solutions to this equation are x = 2 and x = -3.
Example no 10:
x^2 + x + 1 = 0
Using the quadratic formula, we get:
x = (1 +/- sqrt(1^2 - 411)) / (2*1)
x = (1 +/- sqrt(1 - 4)) / 2
x = (1 +/- sqrt(-3)) / 2
There are no real solutions to this equation because the square root of a negative number is not a real number.
Practice Questions:
Here are 10 quadratic equations for you to practice solving:
x^2 - 3x - 10 = 0
x^2 + 2x - 8 = 0
x^2 - 5x + 6 = 0
x^2 + 7x + 10 = 0
x^2 - x - 1 = 0
x^2 - 4x + 3 = 0
x^2 + x - 2 = 0
x^2 - x - 3 = 0
x^2 + 2x + 1 = 0
x^2 - 6x + 9 = 0
I hope these equations are helpful for your practice! Let me know if you have any questions.
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