# Question 0429d

Feb 18, 2017

$x = 5 \pm 3 \sqrt{3}$

#### Explanation:

Really worth committing this bit to memory if you can.

Consider the standardised formula: $\text{ } y = a {x}^{2} + b x + c$

Where: $\text{ } x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

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Given:$\text{ } {x}^{2} - 10 x - 2 = 0$

Thus: $a = 1 \text{; "b=-10"; } c = - 2$ giving:

$x = \frac{- \left(- 10\right) \pm \sqrt{{\left(- 10\right)}^{2} - 4 \left(1\right) \left(- 2\right)}}{2 \left(1\right)}$

$x = \frac{+ 10}{2} \pm \frac{\sqrt{100 + 8}}{2}$

$x = 5 \pm \sqrt{\frac{108}{4}}$

$x = 5 \pm \sqrt{27}$

$x = 5 \pm \sqrt{3 \times {3}^{2}}$

$x = 5 \pm 3 \sqrt{3}$
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This is a sort of cheat method to determine the vertex x-value

Write in the form: $a \left({x}^{2} + \frac{b}{a} x\right) + c$

Then ${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \frac{b}{a}$

In this case $a = 1$ as in ${x}^{2} \to 1 {x}^{2}$ giving:

x_("vertex")=(-1/2)xxb/1" "->(-1/2)xx(-10) = +5#