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kenny6666 [7]
3 years ago
7

What are the exact solutions of x^2 = 5x + 2?

Mathematics
1 answer:
aleksandrvk [35]3 years ago
5 0
Move everybody to one side
minus 5x+2 both sides
x^2-5x-2=0
use quadratic formula where
if you have
ax^2+bx+c=0
x=\frac{-b+/- \sqrt{b^{2}-4ac} }{2a}

1x^2-5x-2=0
a=1
b=-5
c=-2

x=\frac{-(-5)+/- \sqrt{(-5)^{2}-4(1)(-2)} }{2(1)}
x=\frac{5+/- \sqrt{25+8} }{2}
x=\frac{5+/- \sqrt{33} }{2}

aprox
x=5.372281323 or -0.3722813233
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Evaluate the following in polar and standard forms. Select all answers that apply. [3(cos(27°) isin(27°))]^5
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Using <span>De Moivre's formula
</span>
if z = a ( cos x + i sin x ) ⇒⇒⇒ ∴ z^n = a^n ( cos nx + i sin nx)

Part (1) 
∴ [ 3 (cos 27° + i sin 27°) ]⁵ = (3⁵) ( cos 5*27° + i sin 5*27°)
= 243 ( cos 135° + i sin 135°)  ⇒⇒⇒⇒ Polar form
=================================
Part (2)

The angle 135° in the second quadrant ⇒ sin is (+)  and cos is (-)
and its reference angle = 180° - 135° = 45°
sin 45° = cos 45°= 1/√2 = (√2)/2
∴ sin 135° = (√2)/2    and  cos 135° = -(√2)/2
∴ 243 ( cos 135° + i sin 135°) = 243 [ -(√2)/2 + i (√2)/2 ]

= - 243(√2)/2 + 243 (√2)/2 i ⇒⇒⇒⇒ standard form
===========================================
The correct answers are options (1) and (3)

option (1) ⇒⇒⇒ 243 ( cos 135° + i sin 135°)  ⇒⇒⇒⇒ Polar form  
option (3) ⇒⇒⇒ - 243(√2)/2 + 243 (√2)/2 i     ⇒⇒⇒⇒ standard form

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