3 years ago

Solve: x2 - 20x + 36 = 0

Mathematics

2 answers:

OLEGan [10]3 years ago

5 0

$x^2-20x+36=0\\\\x^2-2x-18x+36=0\\\\x(x-2)-18(x-2)=0\\\\(x-2)(x-18)=0\iff x-2=0\ or\ x-18=0\\\\\boxed{x=2\ or\ x=18}$

natali 33 [55]3 years ago

4 0

$x^2 - 20x + 36 = 0 \\
x^2-2x-18x+36=0\\
x(x-2)-18(x-2)=0\\
(x-18)(x-2)=0\\
x=18 \vee x=2
$

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$\frac{\sin B}{b}=\frac{\sin A}{a} \\ \\ \sin B=\frac{b \sin A}{a} \\ \\ \sin B=\frac{13 \sin 29^{\circ}}{7.2} \\ \\ \boxed{m\angle B =\arcsin \left(\frac{65\sin 29^{\circ}{36} \right), 180^{\circ}-\arcsin \left(\frac{65\sin 29^{\circ}}{36} \right)}$

Considering the diagram. $\angle B$ is shown to be acute, so $\boxed{m\angle B=\arcsin \left(\frac{65\sin 29^{\circ}}{36} \right)}$ .

Angles in a triangle add to 180°, so $m\angle C= 151^{\circ}-\arcsin \left(\frac{65\sin 29^{\circ}}{36} \right)$ .

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$AB=\sqrt{13^2 + 7.2^2-2(13)(7.2)\cos \left(\arcsin \left(\frac{65\sin 29^{\circ}}{36} \right) \right)} \\ \\ \boxed{AB=\sqrt{220.84-187.2\cos \left(\arcsin \left(\frac{65\sin 29^{\circ}}{36} \right) \right)}}$