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3 years ago

Which system of equations does not have a real solution?

Mathematics

2 answers:

dolphi86 [110]3 years ago

7 0

Answer:

It's A.

Step-by-step explanation:

Let's look at option A:

From the second equation y = -10 - x. Substituting in the first equation:

-10 - x = x^2 + 3x - 5

x^2 + 4x + 5 = 0

Checking the discriminant b^2 - 4ac we get 16 - 4*1*5 = -4 so there are no real roots. (A negative discriminant means no real roots).

So A has no real solution.

x^2 + 3x - 5 = (20 - 4x)/5 = 4 - 0.8x

x^2 +3.8x - 9 = 0

b^2 - 4ac = (3.8)^2 - 4*1*-9 = 50.44 (positive) so there are real roots.

x^2 + 3x - 5 = -9 - x

x^2 + 4x + 4 = 0

b^2 - 4ac = 4^2 - 4*1*4 = 0 so there are real roots.

Tatiana [17]3 years ago

6 0

Answer:

Correct answer is A!

Step-by-step explanation:

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Find the two square roots of each complex number by creating and solving polynomial equations.

son4ous [18]

Answer:

1) w₁=4 - i w₂= -4 + i

2) w₁= 3 - i w₂= -3 + i

3) w₁= 1 + 2i w₂= - 1 - 2i

4) w₁= 2- 3i w₂= -2 + 3i

5) w₁= 5 - 2i w₂= -5 + 2i

6) w₁= 5 - 3i w₂= -5 + 3i

Step-by-step explanation:

The root of a complex number is given by:

$\sqrt[n]{z}=\sqrt[n]{r}(Cos(\frac{\theta+2k\pi}{n}) + i Sin(\frac{\theta+2k\pi}{n}))$

where:

r: is the module of the complex number

θ: is the angle of the complex number to the positive axis x

n: index of the root

1) z = 15 − 8i ⇒ r=17 θ= -0.4899 rad

w₁= $\sqrt{17}(Cos(\frac{-0.4899}{2}) + i Sin(\frac{-0.4899}{2}))$ =4-i

w₂= $\sqrt{17}(Cos(\frac{-0.4899+2\pi}{2}) + i Sin(\frac{-0.4899+2\pi}{2}))$ =-1+i

2) z = 8 − 6i ⇒ r=10 θ= -0.6435 rad

w₁= $\sqrt{10}(Cos(\frac{ -0.6435}{2}) + i Sin(\frac{ -0.6435}{2}))$ = 3 - i

w₂= $\sqrt{10}(Cos(\frac{ -0.6435+2\pi}{2}) + i Sin(\frac{ -0.6435+2\pi}{2}))$ = -3 + i

3) z = −3 + 4i ⇒ r=5 θ= -0.9316 rad

w₁= $\sqrt{5}(Cos(\frac{-0.9316}{2}) + i Sin(\frac{-0.9316}{2}))$ = 1 + 2i

w₂= $\sqrt{5}(Cos(\frac{-0.9316+2\pi}{2}) + i Sin(\frac{-0.9316+2\pi}{2}))$ = -1 - 2i

4) z = −5 − 12i ⇒ r=13 θ= 0.4426 rad

w₁= $\sqrt{13}(Cos(\frac{0.4426}{2}) + i Sin(\frac{0.4426}{2}))$ = 2- 3i

w₂= $\sqrt{13}(Cos(\frac{0.4426+2\pi}{2}) + i Sin(\frac{0.4426+2\pi}{2}))$ = -2 + 3i

5) z = 21 − 20i ⇒ r=29 θ= -0.8098 rad

w₁= $\sqrt{29}(Cos(\frac{-0.8098}{2}) + i Sin(\frac{-0.8098}{2}))$ = 5 - 2i

w₂= $\sqrt{29}(Cos(\frac{-0.8098+2\pi}{2}) + i Sin(\frac{-0.8098+2\pi}{2}))$ = -5 + 2i

6) z = 16 − 30i ⇒ r=34 θ= -1.0808 rad

w₁= $\sqrt{34}(Cos(\frac{-1.0808}{2}) + i Sin(\frac{-1.0808}{2}))$ = 5 - 3i

w₂= $\sqrt{34}(Cos(\frac{-1.0808+2\pi}{2}) + i Sin(\frac{-1.0808+2\pi}{2}))$ = -5 + 3i

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blsea [12.9K]

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dunno

Step-by-step explanation:

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