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1 year ago

Factor complete −2x2 + 10x.

Mathematics

1 answer:

mezya [45]1 year ago

6 0

The factors of the quadratic equation is (x + 1) and (x - 6)

Given,

The quadratic equation; -2x² + 10x + 12

We have to find the factors of this equation using quadratic formula;-

Quadratic formula;- $\frac{-b(+-)\sqrt{b^{2} -4ac} }{2a}$

Here,

a = -2, b = 10 and c = 12

Now,

$\frac{-b(+-)\sqrt{b^{2} -4ac} }{2a}$ = $\frac{-10(+-)\sqrt{10^{2} -4 . -2. 12} }{2.-2}$ = $\frac{-10(+-)\sqrt{100 +96} }{-4}$ = (-10±14) ÷ -4

Solve for,

- 10 + 14 / -4 = -4/4 = -1

That is, (x + 1)

Solve for,

- 10 - 14 / -4 = -24/-4 = 6

That is, (x - 6)

Therefore, the factors for the given quadratic equation is (x + 1) and (x - 6)

Learn more about quadratic formula here;

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3 years ago

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First, differentiate the first time, term by term:

${3y^{2}.\frac{dy}{dx} + 3x^{2}} = 0 \\\\{3y^{2}.\frac{dy}{dx} = -3x^{2}} \\\\\frac{dy}{dx} = \frac{-3x^{2}}{3y^{2}} \\\\\frac{dy}{dx} = \frac{-x^{2}}{y^{2}}$

↑ we'll substitute this later (4th step onwards)

Differentiate the second time:

$3y^{2}.\frac{dy}{dx} + 3x^{2} = 0 \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{dy}{dx})^{2} + 6x = 0 \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{dy}{dx})^{2} = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{-x^{2} }{y^{2} })^{2} = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{x^{4} }{y^{4} }) = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + \frac{6x^{4} }{y^{3} } = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} = - 6x - \frac{6x^{4} }{y^{3} } \\\\$

$3y^{2}.\frac{d^{2} y}{dx^{2}} = - \frac{- 6xy^{3} - 6x^{4} }{y^{3}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{- 6xy^{3} - 6x^{4} }{3y^{2}. y^{3}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{- 2xy^{3} - 2x^{4} }{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x (y^{3} + x^{3})}{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x (1)}{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x}{y^{5}}$

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