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3 years ago

The quadratic equation 2x^2+4x-30=0 has two solutions. one of the solutions is x=3 the other solution is?

Mathematics

2 answers:

BartSMP [9]3 years ago

6 0

Answer:

2x^2+4x-30=0

Step-by-step explanation:

2x^2+10x-6x-30=0

2x(x+5)-6(x+5)=0

(2x-6)(x+5)=0

2x-6=0 x+5=0

x=6/2 x=-5

x=3

Irina18 [472]3 years ago

5 0

Answer:

Step-by-step explanation:

$2x^2+4x-30=0\\ x^2+2x-15=0$

so we know that the sum of the two solutions is -2 and the multiplication -15

so we are looking for x so that

x+3=-2 and 3x=-15

it gives x = -5

another way to do it is to put (x-3) in factor as we know that 3 is one of the solutions it gives

$2x^2+4x-30 = 2(x-3)(x+5)$

and then the second solution is -5

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Answer:

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Step-by-step explanation:

we would like to figure out the derivative of the following:

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to do so, let,

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use law of exponent which yields:

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take derivative in both sides:

$\displaystyle \frac{dy}{dx} = \frac{d}{dx} (3 - 2 {x}^{ - 1} - { {x}^{ - 2} } )$

use sum derivation rule which yields:

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By constant derivation we acquire:

$\rm\displaystyle \frac{dy}{dx} = 0 - \frac{d}{dx} 2 {x}^{ - 1} - \frac{d}{dx} {x}^{ - 2}$

use exponent rule of derivation which yields:

$\rm\displaystyle \frac{dy}{dx} = 0 - ( - 2 {x}^{ - 1 -1} ) - ( - 2 {x}^{ - 2 - 1} )$

simplify exponent:

$\rm\displaystyle \frac{dy}{dx} = 0 - ( - 2 {x}^{ -2} ) - ( - 2 {x}^{ - 3} )$

two negatives make positive so,

$\displaystyle \frac{dy}{dx} = 2 {x}^{ -2} + 2 {x}^{ - 3}$

<h3>further simplification if needed:</h3>

by law of exponent we acquire:

$\displaystyle \frac{dy}{dx} = \frac{2 }{x^2}+ \frac{2}{x^3}$

simplify addition:

$\displaystyle \frac{dy}{dx} = \frac{2x + 2}{x^3}$

and we are done!

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