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xxTIMURxx [149]
4 years ago
9

Which two equations would be most appropriately solved by using the zero product property? Select each correct answer. 4x² = 13

0.25x2+0.8x−8=0 −(x−1)(x+9)=0 3x2−6x=0
Mathematics
2 answers:
olga55 [171]4 years ago
5 0
The zero product property tells us that if we multiply two quantities together and get zero, then one or both of these is equal to zero.

In symbols this means that if ab = 0 then a = 0, b = 0 or both.

This means that if we have a product equal to zero one of both of the factors (the things we are multiplying together) is equal to zero.

Of the equations given, the only one that has product (a few terms multiplied together) equal to zero is -(x-1)(x+9)=0 so this is the one that can most appropriately be solved by the zero product property.

We can re-write -(x-1)(x+9) = 0 as (-1)(x-1)(x+9) = 0 which means (x-1) or (x+9) or both equal zero. It should be clear that -1 cannot equal zero.

To solve we set x - 1 = 0 and obtain x = 1.
Next we set x + 9 = 0 and obtain x = -9

The solutions to the equation are x = -9 and x = 1.

We are, however, asked for two equations. Though none of the remaining equations has a product set equal to zero, we can re-write the equation 3 x^{2} -6x=0 by factoring 3x from each term on the left-hand side. Doing so gives us 3x(x-2)=0. Now we have a product equal to zero which means that one or both of the factors on the left-hand side equals zero.

That is, 3x equals 0, or x-2 equals 0 or both do. If 3x = 0 then x = 0. If x-2=0 then x=2. As such the solution to the equation is x = 0, x = 2.
netineya [11]4 years ago
5 0

Answer:

−(x−1)(x+9)=0

And..

3x2−6x=0

Step-by-step explanation:

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Find thd <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D" id="TexFormula1" title="\frac{dy}{dx}" alt="\frac{dy}{dx}" a
NARA [144]

x^3y^2+\sin(x\ln y)+e^{xy}=0

Differentiate both sides, treating y as a function of x. Let's take it one term at a time.

Power, product and chain rules:

\dfrac{\mathrm d(x^3y^2)}{\mathrm dx}=\dfrac{\mathrm d(x^3)}{\mathrm dx}y^2+x^3\dfrac{\mathrm d(y^2)}{\mathrm dx}

=3x^2y^2+x^3(2y)\dfrac{\mathrm dy}{\mathrm dx}

=3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}

Product and chain rules:

\dfrac{\mathrm d(\sin(x\ln y)}{\mathrm dx}=\cos(x\ln y)\dfrac{\mathrm d(x\ln y)}{\mathrm dx}

=\cos(x\ln y)\left(\dfrac{\mathrm d(x)}{\mathrm dx}\ln y+x\dfrac{\mathrm d(\ln y)}{\mathrm dx}\right)

=\cos(x\ln y)\left(\ln y+\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}\right)

=\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}

Product and chain rules:

\dfrac{\mathrm d(e^{xy})}{\mathrm dx}=e^{xy}\dfrac{\mathrm d(xy)}{\mathrm dx}

=e^{xy}\left(\dfrac{\mathrm d(x)}{\mathrm dx}y+x\dfrac{\mathrm d(y)}{\mathrm dx}\right)

=e^{xy}\left(y+x\dfrac{\mathrm dy}{\mathrm dx}\right)

=ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}

The derivative of 0 is, of course, 0. So we have, upon differentiating everything,

3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}+\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}+ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}=0

Isolate the derivative, and solve for it:

\left(6x^3y+\dfrac{\cos(x\ln y)}y+xe^{xy}\right)\dfrac{\mathrm dy}{\mathrm dx}=-\left(3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}\right)

\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}}{6x^3y+\frac{\cos(x\ln y)}y+xe^{xy}}

(See comment below; all the 6s should be 2s)

We can simplify this a bit by multiplying the numerator and denominator by y to get rid of that fraction in the denominator.

\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^3+y\cos(x\ln y)\ln y-y^2e^{xy}}{6x^3y^2+\cos(x\ln y)+xye^{xy}}

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