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2 years ago

Solve for all values of p:

Mathematics

1 answer:

Alchen [17]2 years ago

3 0

Answer:

p = - 5, p = - 1

Step-by-step explanation:

Given

$\frac{3p}{p-5}$ - $\frac{2}{p+3}$ = $\frac{p}{p+3}$

Multiply through by the LCM of (p - 5)(p + 3)

3p(p + 3) - 2(p - 5) = p(p - 5) ← distribute parenthesis

3p² + 9p - 2p + 10 = p² - 5p

3p² + 7p + 10 = p² - 5p ( subtract p² - 5p from both sides )

2p² + 12p + 10 = 0 ← divide through by 2

p² + 6p + 5 = 0 ← in standard form

(p + 1)(p + 5) = 0 ← in factored form

Equate each factor to zero and solve for p

p + 1 = 0 ⇒ p = - 1

p + 5 = 0 ⇒ p = - 5

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If (2i/2+i)-(3i/3+i)=a+bi, then a= A. 1/10 B. -10 C. 1/50 D. -1/10

Verizon [17]

Answer:

Option A is correct.

Step-by-step explanation:

We are given:

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$=\frac{2i(3+i)}{(2+i)(3+i)}-\frac{3i(2+i)}{(2+i)(3+i)}\\=\frac{6i+2i^2}{(2+i)(3+i)}-\frac{6i+3i^2}{(2+i)(3+i)}\\=\frac{6i+2i^2-(6i+3i^2)}{(2+i)(3+i)}\\=\frac{6i+2i^2-6i-3i^2)}{5+5i}\\=\frac{-i^2}{5+5i}\\i^2=-1\\=\frac{-(-1)}{5+5i}\\=\frac{1}{5+5i}$

Now rationalize the denominator by multiplying by 5-5i/5-5i

$=\frac{1}{5+5i}*\frac{5-5i}{5-5i} \\=\frac{5-5i}{(5+5i)(5-5i)}\\=\frac{5-5i}{(5+5i)(5-5i)}\\(a+b)(a-b)= a^2-b^2\\=\frac{5(1-i)}{(5)^2-(5i)^2}\\=\frac{5(1-i)}{25+25}\\=\frac{5(1-i)}{50}\\=\frac{1-i}{10}\\=\frac{1}{10}-\frac{i}{10}$

We are given

$\frac{2i}{2+i}-\frac{3i}{3+i} = a+bi$

Now after solving we have:

$\frac{1}{10}-\frac{i}{10}=a+bi$

So value of a = 1/10 and value of b = -1/10

So, Option A is correct.

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Solve for all values of p: ​

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