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Lemur [1.5K]
3 years ago
10

Solve by factoring and list only the positive solution: 2x2 - 5x = 88

Mathematics
1 answer:
Semmy [17]3 years ago
8 0
It has to be noted that there are several ways to factor the item. One of these is shown below. 

The general form of a quadratic equation is,
            Ax² + Bx + C = 0

where A and B are numerical coefficients and C is the constant. If we are to express the given equation in this form, 
        2x² - 5x - 88 = 0

The sum of the roots, x₁ and x₂ is -B/A and the product is equal to C/A.

Sum: x₁ + x₂ = -(-5/2) = 5/2
 Product: x₁x₂ = -88/2 = -44

The values of x₁ and x₂ are 8 and -11/2.

The factors are (x - 8) and (x + 11/2)

<em>ANSWERS: 8 and -11/2. </em>
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Answer:

(x-y)^4=x^4-x^3y+6x^2y^2-4xy^3+y^4.

Step-by-step explanation:

We want to find the polynomial that will result from expanding:

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Recall that we can use the Pascal's triangle to obtain the coefficient as:

1     4    6   4   1

Also note how the negative sign is going to alternate.

The <em>power of x </em>will decrease from left to right while the <em>power of y</em> increases from left to right

The expansion then becomes:

(x-y)^4=x^4-x^3y+6x^2y^2-4xy^3+y^4.

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3 years ago
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alexandr402 [8]

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Step-by-step explanation:

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The equations that must be solved for maximum or minimum values of a differentiable function w=​f(x,y,z) subject to two constrai
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L_x=2x+8\lambda x+2\mu=0\implies x(1+4\lambda)+\mu=0

L_y=2y+8\lambda y=0\implies y(1+4\lambda)=0

L_z=2z-2\lambda z+4\mu=0\implies z(1-\lambda)+2\mu=0

L_\lambda=4x^2+4y^2-z^2=0

L_\mu=2x+4z-2=0\implies x+2z=1

Case 1: If y=0, then

4x^2-z^2=0\implies4x^2=z^2\implies2|x|=|z|

Then

x+2z=1\implies x=1-2z\implies2|1-2z|=|z|\implies z=\dfrac25\text{ or }z=\dfrac23

\implies x=\dfrac15\text{ or }x=-\dfrac13

So we have two critical points, \left(\dfrac15,0,\dfrac25\right) and \left(-\dfrac13,0,\dfrac23\right)

Case 2: If \lambda=-\dfrac14, then in the first equation we get

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Then

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but there are no real solutions for y, so this case yields no additional critical points.

So at the two critical points we've found, we get extreme values of

f\left(\dfrac15,0,\dfrac25\right)=\dfrac15 (min)

and

f\left(-\dfrac13,0,\dfrac23\right)=\dfrac59 (max)

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3 years ago
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Answer:

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Step-by-step explanation:

7 0
3 years ago
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