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babymother [125]
3 years ago
10

(05.08) Solve for x: 2 over quantity x minus 2 plus 7 over quantity x squared minus 4 equals 5 over x. (2 points) x = negative 4

over 3 and x = -5 x = negative 4 over 3 and x = 5 x = 4 over 3 and x = -5 x = 4 over 3 and x = 5
Mathematics
1 answer:
Oxana [17]3 years ago
4 0

Answer:

Step-by-step explanation:

The best way to do this is to use your LCM and eliminate the fractions.  To find the LCM you have to use all the denominators as a multiplier so the denominator in each term cancels out.  We will first factor the x-squared term to simplify and see what 2 factors are hidden there.

x^2-4 factors to (x + 2)(x - 2).  That means that our 3 denominators that make up our LCM are x(x+2)(x-2).  We will mulitply that in to each term in our rational equation, canceling out the denominators where applicable.

x(x+2)(x-2)[\frac{2}{(x-2)}+\frac{7}{(x-2)(x+2)}=\frac{5}{x}]

In the first term, the (x-2) will cancel leaving us with

x(x+2)[2] which simplifies to

x^2+2x[2]

In the second term, the (x+2)(x-2) cancels out leaving us with

x[7].

In the last term, the x cancels out leaving us with

(x+2)(x-2)[5] which simplifies to

x^2-4[5]

Now we will distribute through each cancellation:

2x²+4x;

7x;

5x²-20

Putting them all together we have

2x² + 4x + 7x = 5x² - 20

Combining like terms gives us a quadratic:

3x² - 11x - 20 = 0

Factor that however you find it easiest to factor quadratics and get that

x = 5 and x = -4/3

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How would you solve these 2 maths problems?
sweet-ann [11.9K]

Answer:

see explanation

Step-by-step explanation:

(9)

Expand and simplify right side and compare coefficients of like terms on left side.

(x - 2)(x² + ax + b) + c

= x³ + ax² + bx - 2x² - 2ax - 2b + c

= x³ + x²(a - 2) + x(b - 2a) - 2b + c

compare with

x³ + 2x² - 3x + 4

x² terms

a - 2 = 2 ( add 2 to both sides )

a = 4

x terms

b - 2a = - 3 ← substitute a = 4

b - 8 = - 3 ( add 8 to both sides )

b = 5

constant terms

- 2b + c = 4 ← substitute b = 5

- 10 + c = 4 ( add 10 to both sides )

c = 14

Thus a = 4, b = 5 and c = 14

-------------------------------------------------

(10)

Since both functions cross the x- axis at - 2 then (- 2, 0) satisfies both, that is

f(- 2) = (-2)^{4} + a(- 2)³ + b(- 2)² + 36(- 2) + 144 = 0, that is

- 16 - 8a + 4b - 72 + 144 = 0

- 8a + 4b + 56 = 0

- 8a + 4b = - 56 → (1)

and

g(- 2) = (-2)^{4} + (a + 3)(- 2)³ - 23(- 2)² + (b + 10)(- 2) + 40 = 0, that is

- 16 - 8(a + 3) - 92 - 2(b + 10) + 40 = 0

- 16 - 8a - 24 - 92 - 2b - 20 + 40 = 0

- 8a - 2b - 112 = 0

- 8a - 2b = 112 → (2)

Subtract (1) from (2) term by term

- 6b = 168 ( divide both sides by - 6 )

b = - 28

Substitute b = - 28 into (2)

- 8a + 56 = 112 ( subtract 56 from both sides )

- 8a = 56 ( divide both sides by - 8 )

a = - 7

Thus a = - 7 and b = - 28

5 0
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