explain the quadratic formula
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The Quadratic Formula: For ax2 + bx + c= 0, the values of x which are the solutions of the equation are given by:
x = \dfrac{-b \pm\sqrt{b^2 - 4ac\,}}{2a}x=2a−b±b2−4ac
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For the Quadratic Formula to work, you must have your equation arranged in the form "(quadratic) = 0". Also, the "2a" in the denominator of the Formula is underneath everything above, not just the square root. And it's a "2a" under there, not just a plain "2". Make sure that you are careful not to drop the square root or the "plus/minus" in the middle of your calculations, or I can guarantee that you will forget to "put them back in" on your test, and you'll mess yourself up. Remember that "b2" means "the square of ALL of b, including its sign", so don't leave b2 being negative, even if b is negative, because the square of a negative is a positive.
In other words, don't be sloppy and don't try to take shortcuts, because it will only hurt you in the long run. Trust me on this!
Here are some examples of how the Quadratic Formula works:
Solve x2 + 3x – 4 = 0
This quadratic happens to factor:
x2 + 3x – 4 = (x + 4)(x – 1) = 0
...so I already know that the solutions are x= –4 and x = 1. How would my solution look in the Quadratic Formula? Using a = 1, b = 3, and c = –4, my solution looks like this:
x = \dfrac{-(3) \pm \sqrt{(3)^2 - 4(1)(-4)\,}}{2(1)}x=2(1)−(3)±(3)2−4(1)(−4)
= \dfrac{-3 \pm \sqrt{9 + 16\,}}{2} = \dfrac{-3 \pm \sqrt{25\,}}{2}=2−3±9+16=2−3±25
= \dfrac{-3 \pm 5}{2} = \dfrac{-3 - 5}{2},\, \dfrac{-3 + 5}{2}=2−3±5=2−3−5,2−3+5
= \dfrac{-8}{2},\, \dfrac{2}{2} = -4,\, 1=2−8,22=−4,1
Then, as expected, the solution is x = –4, x= 1.
x = \dfrac{-b \pm\sqrt{b^2 - 4ac\,}}{2a}x=2a−b±b2−4ac
Affiliate


For the Quadratic Formula to work, you must have your equation arranged in the form "(quadratic) = 0". Also, the "2a" in the denominator of the Formula is underneath everything above, not just the square root. And it's a "2a" under there, not just a plain "2". Make sure that you are careful not to drop the square root or the "plus/minus" in the middle of your calculations, or I can guarantee that you will forget to "put them back in" on your test, and you'll mess yourself up. Remember that "b2" means "the square of ALL of b, including its sign", so don't leave b2 being negative, even if b is negative, because the square of a negative is a positive.
In other words, don't be sloppy and don't try to take shortcuts, because it will only hurt you in the long run. Trust me on this!
Here are some examples of how the Quadratic Formula works:
Solve x2 + 3x – 4 = 0
This quadratic happens to factor:
x2 + 3x – 4 = (x + 4)(x – 1) = 0
...so I already know that the solutions are x= –4 and x = 1. How would my solution look in the Quadratic Formula? Using a = 1, b = 3, and c = –4, my solution looks like this:
x = \dfrac{-(3) \pm \sqrt{(3)^2 - 4(1)(-4)\,}}{2(1)}x=2(1)−(3)±(3)2−4(1)(−4)
= \dfrac{-3 \pm \sqrt{9 + 16\,}}{2} = \dfrac{-3 \pm \sqrt{25\,}}{2}=2−3±9+16=2−3±25
= \dfrac{-3 \pm 5}{2} = \dfrac{-3 - 5}{2},\, \dfrac{-3 + 5}{2}=2−3±5=2−3−5,2−3+5
= \dfrac{-8}{2},\, \dfrac{2}{2} = -4,\, 1=2−8,22=−4,1
Then, as expected, the solution is x = –4, x= 1.
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