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3 years ago

Help please #12 Solve using factoring

Mathematics

2 answers:

Serga [27]3 years ago

5 0

(R-2)x(r+2) that what I think it is

zimovet [89]3 years ago

5 0

You have to use the sum/difference formula

$(a+b)(a-b)=a^2-b^2$

In your case, $r^2-4$ is a difference of squares, so it can be factored with the sum and difference of the roots:

$r^2-4 = (r+2)(r-2)$

So, we have

$r^2-4 = 0 \iff (r+2)(r-2) = 0$

A multiplication is zero if and only if one of the factors is zero, so we have

$(r+2)(r-2) = 0 \iff r+2=0\ \lor\ r-2=0 \iff r=-2\ \lor\ r=2$

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2 years ago

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A company wants to increase the 10% peroxide content of its product by adding pure peroxide (100% peroxide). If x liters of pure

makvit [3.9K]

Answer:

There should be 125 litters of pure peroxide added to produce a new product that is 28% peroxide.

Step-by-step explanation:

The problem provides us with an equation we can use to find the concentration C of the new mixture, which is:

$C=\frac{x+0.1(500)}{x+500}$

which will give us the concentration C as a decimal number.

We are interested in getting a concentration of 28%, which is written as 0.28 in decimal form. This is our C.

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$0.28x+140=x+50$

so now we can subtract an x from both sides so we get:

$0.28-x+140=x-x+50$

which yields:

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Which yields:

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and finally we can divide both sides into -0.72 so we get:

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3 years ago

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7. Use the quadratic formula to find the solution(s). x² + 2x - 8 = 0

morpeh [17]

Hey there!

Use the quadratic formula to find the solution(s). x² + 2x - 8 = 0

Answer :

x = -4 or x = 2 ✅

Explanation :

Quadratic formula : ax² + bx + c = 0 where a ≠ 0

The number of real-number solutions (roots) is determined by the discriminant (b² - 4ac) :

If b² - 4ac > 0 , There are 2 real-number solutions

If b² - 4ac = 0 , There is 1 real-number solution.

If b² - 4ac < 0 , There is no real-number solution.

The roots of the equation are determined by the following calculation:

$x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

Here, we have :

a = 1
b = 2
c = -8

1) Calculate the discriminant :

b² - 4ac ⇔ 2² - 4(1)(-8) ⇔ 4 - (-32) ⇔ 36

b² - 4ac = 36 > 0 ; The equation admits two real-number solutions

2) Calculate the roots of the equation:

▪️ (1)

$x_1 = \frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ x_1 = \frac{ - 2 - \sqrt{36} }{2(1) } \\ \\ x_1 = \frac{ - 2 - 6}{2} \\ \\ x_1 = \frac{ - 8}{2} \\ \\ \blue{\boxed{\red{x_1 = -4}}}$

▪️ (2)

$x_2 = \frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ x_2 = \frac{ - 2 + \sqrt{36} }{2(1)} \\ \\ x_2 = \frac{ - 2 + 6}{2} \\ \\ x_2 = \frac{4}{2} \\ \\ \red{\boxed{\blue{x_2 = 2}}}$

>> Therefore, your answers are x = -4 or x = 2.

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