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Solving Quadratic Equations Using the Quadratic Formula G9
Introduction:
Quadratic equations can always be solved using the quadratic formula.
It is ax squared plus bx plus c equals 0, where a does not equal 0.
So it is x equals -b plus/minus the square root of (b squared minus 4ac), all over 2a.
\[ ax^2 + bx + c = 0 \]
\[x_{1, 2} = {-b \pm \sqrt{b^2-4ac} \over 2a}\]
Example 1: Solve: 2x squared -3 x – 1 equals 0.
\[ \text{Solve } 2x^2 -3x -1 = 0 \]
Solution 1:
x equals-(-3) plus/minus the square root of (-3) squared -4 times 2 times (-1); all over 2 times 2.
x equals 3 plus/minus the square root (17), all over 4.
\[x = {-(-3) \pm \sqrt{(-3)^2-4(2)(-1)} \over 2(2)}\]
\[x = {3 \pm \sqrt{9-(-8)} \over 4}\]
\[x = {3 \pm \sqrt{17} \over 4} \]
\[ x_1 = {3+\sqrt{17} \over 4} \], and \[ x_2 = {3-\sqrt{17} \over 4} \]
Two roots are: 3 plus the square root 17, all over 4; and 3 minus the square root of 17, over 4.
Example 2: Solve 3x squared -10. Equals 0.
\[ \text{Solve } 3x^2-10 = 0 \]
Solution 2:
a equals 3 b equals 0 c equals -10 x equals 0 plus-square root (0 squared -4(3)(-10)/6
\[x = {-(0) \pm \sqrt{(0)^2-4(3)(-10)} \over 2(3)}\]
\[x = { \pm \sqrt{120} \over 6}\]
Therefore the roots are square root 120/6; and negative square root 120/6.
Conclusion:
The quadratic formula always gives you two roots. If the square root of a negative number comes into play the solution will be a complex number.
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