Question #0429d

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Answer 1 · 2017-02-18T20:15:14

$x = 5 \pm 3 \sqrt{3}$ $x=5+-3sqrt(3)$

Really worth committing this bit to memory if you can.

Consider the standardised formula: $y = a x^{2} + b x + c$ $" "y=ax^2+bx+c$

Where: $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $" "x=(-b+-sqrt(b^2-4ac))/(2a)$

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

Thus: $a = 1; b = - 10; c = - 2$ $a=1"; "b=-10"; "c=-2$ giving:

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

$x = \frac{+ 10}{2} \pm \frac{\sqrt{100 + 8}}{2}$ $x=(+10)/2+-(sqrt(100+8))/2$

$x = 5 \pm \sqrt{\frac{108}{4}}$ $x=5+-sqrt(108/4)$

$x = 5 \pm \sqrt{27}$ $x=5+-sqrt(27)$

$x = 5 \pm \sqrt{3 \times 3^{2}}$ $x=5+-sqrt(3xx3^2)$

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

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

Then $x_{vertex} = (- \frac{1}{2}) \times \frac{b}{a}$ $x_("vertex")=(-1/2)xxb/a$

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

$x_{vertex} = (- \frac{1}{2}) \times \frac{b}{1} \to (- \frac{1}{2}) \times (- 10) = + 5$ $x_("vertex")=(-1/2)xxb/1" "->(-1/2)xx(-10) = +5$

Tony B