The quadratic formula part two

Quadratic equations The quadratic formula part two

We have seen that a formula to solve quadratic equations can be derived by completing the square from the general form of a quadratic equation, $ax^2 + bx + c = 0$

x can be found by putting the coefficients into the quadratic formula $x = \frac{-bpmsqrt{{b^2} - 4ac}}{2a}$

Here is an example

$x^2 + 3x - 4 = 0$

In this equation $a = 1$ , $b=3$ and $c = -4$

Putting these numbers into the formula, we get

$x = \frac{-bpmsqrt{{b^2} - 4ac}}{2a}$

$x = \frac{-3pmsqrt{{3^2} - 4\times1\times-4}}{2\times1}$

$x = \frac{-3pmsqrt{9 + 16}}{2}$

$x = \frac{-3pmsqrt{25}}{2}$

$x = \frac{-3pm5}{2}$

$x = \frac{2}{2}$ and $x = \frac{-8}{2}$

$x = 1$ and $x = -4$

We can see from the graph below that when y = 0, x = 1 and -4.

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