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Nikitich [7]
3 years ago
7

Solve the quadratic equation 3x² + 2x-4=0 Give your answers to 2 decimal places.

Mathematics
2 answers:
GalinKa [24]3 years ago
6 0

3x² + 2x - 4 = 0

Δ = b² - 4.a.c

Δ = 2² - 4 . 3 . -4

Δ = 4 - 4. 3 . -4

Δ = 52

x = (-b +- √Δ)/2a

 

x' = (-2 + √52)/2.3    

x'' = (-2 - √52)/2.3

x' = 0,8685170918213297    

x'' = -1,5351837584879966

2 Decimal places

x' = 0,87

x'' = -1,54

Kobotan [32]3 years ago
6 0

Answer: .8685

Step-by-step explanation:

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Step-by-step explanation:

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  see attached

Step-by-step explanation:

The Pythagorean theorem can be used to find the hypotenuse associated with each pair of legs. That tells you ...

  c² = a² +b² . . . . . legs a, b; hypotenuse c

__

<h3>alternate form of Pythagorean theorem</h3>

For the purpose of this problem, it is convenient to consider a slightly different form of the equation.

For legs √a and √b, the hypotenuse √c is given by ...

  (√c)² = (√a)² +(√b)²

  c = a +b

That is ...

  legs √a, √b ⇒ hypotenuse √(a+b)

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<h3>application to this problem</h3>

Since the legs are (mostly) given in terms of square roots, the value under the radical for the hypotenuse is simply the sum of those:

legs: √1, √2 ⇒ hypotenuse √(1+2) = √3

legs: √2, √3 ⇒ hypotenuse √(2+3) = √5

legs: √5, √3 ⇒ hypotenuse √(5+3) = √8

legs: √5, √1 ⇒ hypotenuse √(5+1) = √6

_____

<em>Additional comment</em>

You may not see the leg lengths given as square roots very often. This is a rather unusual set of problems for hypotenuse length.

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