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3 years ago

I can't solve this equation pleas help!x^2+2x+8=0

Mathematics

1 answer:

tiny-mole [99]3 years ago

8 0

$x^2+2x+8=0\\
x^2+2x+1+7=0\\
(x+1)^2=-7\\
x+1=-\sqrt{-7} \vee x+1=\sqrt{-7}\\
x=-1-\sqrt7 i \vee x=-1+\sqrt7 i$

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Answer:

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Step-by-step explanation:

To solve this question we're going to use trigonometric identities and good ol' Pythagoras theorem.

a) Firstly, sec(θ)=52. we're gonna convert this to cos(θ) using:

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we can substitute the value of sec(θ) in this equation:

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and solve for for cos(θ)

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side note: just to confirm we can find the value of θ and verify that is indeed an acute angle by $\theta = \arccos{\left(\dfrac{1}{52}\right)} = 88.8^\circ$

b) since right triangle is mentioned in the question. We can use:

$\cos{\theta} = \dfrac{\text{adj}}{\text{hyp}}$

we know the value of cos(θ)=1\52. and by comparing the two. we can say that:

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we can find the third side using the Pythagoras theorem.

$(\text{hyp})^2=(\text{adj})^2+(\text{opp})^2$

$(52)^2=(1)^2+(\text{opp})^2$

$\text{opp}=\sqrt{(52)^2-1}$

$\text{opp}=\sqrt{2703}$

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we can find the sin(θ) using this side:

$\sin{\theta} = \dfrac{\text{opp}}{\text{hyp}}$

$\sin{\theta} = \dfrac{\sqrt{2703}}{52}}$

and since $\csc{\theta} = \dfrac{1}{\sin{\theta}}$

$\csc{\theta} = \dfrac{52}{\sqrt{2703}}$

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