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grin007 [14]
3 years ago
10

All three please 10 points

Mathematics
1 answer:
GaryK [48]3 years ago
5 0

Answer:

A) x = 3 or -1

B) x = -7

C)x = -7

Step-by-step explanation:

A) x² + 2x + 1 = 2x² - 2

Rearranging, we have;

2x² - x² - 2x - 2 - 1 = 0

x² - 2x - 3 = 0

Using quadratic formula, we have;

x = [-(-2) ± √((-2)² - 4(1 × -3))]/(2 × 1)

x = (2 ± √16)/2

x = (2 + 4)/2 or (2 - 4)/2

x = 6/2 or -2/2

x = 3 or -1

B) ((x + 2)/3) - 2/15 = (x - 2)/5

Multiply through by 15 to get;

5(x + 2) - 2 = 3(x - 2)

5x + 10 - 2 = 3x - 6

5x - 3x = -6 - 10 + 2

2x = -14

x = -14/2

x = -7

C) log(2x + 3) = 2log x

From log derivations, 2 log x is same as log x²

Thus;

log(2x + 3) = logx²

Log will cancel out to give;

2x + 3 = x²

x² - 2x - 3 = 0

Using quadratic formula, we have;

x = [-(-2) ± √((-2)² - 4(1 × -3))]/(2 × 1)

x = (2 ± √16)/2

x = (2 + 4)/2 or (2 - 4)/2

x = 6/2 or -2/2

x = 3 or -1

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When you are multiplying an exponent directly into a number/variable with an exponent, you multiply the exponents together.

For example:

(x^{2} )^{3} = x^6

(x^{3} )^5=x^{15}


When you are multiplying a variable with an exponent by another variable with an exponent, you add the exponents together.

For example:

(x^{2} )(x^{3})=x^{5}

(x^{1} )(x^{2})=x^{3}


(\frac{(x^{-3})(y^{2})}{(x^{4})(y^{6})} )^{3}=\frac{(x^{-9})(y^{6})}{(x^{12})(y^{18})}

You multiply 3 into each exponent in the numerator and the denominator

\frac{(x^{-9})(y^{6})}{(x^{12})(y^{18})}= \frac{y^{6}}{(x^{9})(x^{12})(y^{18})}

When you have a negative exponent, you move it to the other side of the fraction to make the exponent positive.

\frac{y^{6}}{(x^{21})(y^{18})} = \frac{1}{(x^{21})(y^{12})}


When you have something like this:

\frac{x^{2}}{x^5}

You subtract the exponents together, so:

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Your answer is the second option

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