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m)determine the interval for where (f(x)<0) (*hint: you need to factor)…

Question

m)determine the interval for where (f(x)<0) (*hint: you need to factor)

  1. (f(x)=x^{2}-4)
  2. (f(x)= - x^{2}+3x - 10)

n)simplify the complex number

  1. (sqrt{-36})
  2. (sqrt{-144})
  3. (sqrt{-20})
  4. (sqrt{-27})
  5. (sqrt{-72})
  6. (i^{7})
  7. (i^{53})
  8. (i^{34})

o)simplify each expression

  1. ((3 + 5i)+(2 + 7i))
  2. ((-9 + 4i)+(2 + 7i))
  3. ((3 + 5i)-(2 + 7i))
  4. ((3 + 5i)+(2 - 7i))

p) find the product of the complex numbers. give the answer in the form (a + bi)

  1. ((3 + 5i)(7 + 2i))
  2. ((9 + 2i)(9 - 2i))
  3. ((2 + 5i)(2 - 5i))
  4. ((3 + i)(3 - i))
  5. ((2 + 3i)(2 - 7i))
  6. ((5 + 4i)(5 - 4i))

q) find each quotient

  1. (\frac{2+3i}{5 + 4i})
  2. (\frac{5+4i}{3 - 2i})

Explanation:

M1: Factor the first - function

For \(f(x)=x^{2}-4\), use the difference - of - squares formula \(a^{2}-b^{2}=(a + b)(a - b)\). Here \(a = x\) and \(b = 2\), so \(f(x)=(x + 2)(x - 2)\). Set \(f(x)=0\), then \((x + 2)(x - 2)=0\), and the roots are \(x=-2\) and \(x = 2\). We can test the intervals \((-\infty,-2)\), \((-2,2)\) and \((2,\infty)\). For \(x=-3\), \(f(-3)=(-3)^{2}-4=9 - 4=5>0\); for \(x = 0\), \(f(0)=0^{2}-4=-4<0\); for \(x = 3\), \(f(3)=3^{2}-4=9 - 4=5>0\). So the interval where \(f(x)<0\) is \((-2,2)\).

M2: Factor the second - function

For \(f(x)=-x^{2}+3x - 10\), first consider the quadratic formula for \(ax^{2}+bx + c = 0\) (\(a=-1\), \(b = 3\), \(c=-10\)). The discriminant \(\Delta=b^{2}-4ac=3^{2}-4\times(-1)\times(-10)=9 - 40=-31<0\). Since \(a=-1<0\), the graph of \(y =-x^{2}+3x - 10\) is a parabola opening downwards and is always negative. So the interval where \(f(x)<0\) is \((-\infty,\infty)\).

N1: Simplify \(\sqrt{-36}\)

We know that \(\sqrt{-a}=\sqrt{a}i\) for \(a>0\). So \(\sqrt{-36}=\sqrt{36}i = 6i\).

N2: Simplify \(\sqrt{-144}\)

\(\sqrt{-144}=\sqrt{144}i=12i\).

N3: Simplify \(\sqrt{-20}\)

\(\sqrt{-20}=\sqrt{4\times5}i = 2\sqrt{5}i\).

N4: Simplify \(\sqrt{-27}\)

\(\sqrt{-27}=\sqrt{9\times3}i=3\sqrt{3}i\).

N5: Simplify \(\sqrt{-72}\)

\(\sqrt{-72}=\sqrt{36\times2}i = 6\sqrt{2}i\).

N6: Simplify \(i^{7}\)

Since \(i^{1}=i\), \(i^{2}=-1\), \(i^{3}=i^{2}\cdot i=-i\), \(i^{4}=(i^{2})^{2}=1\), and \(7 = 4+3\), then \(i^{7}=i^{4}\cdot i^{3}=1\times(-i)=-i\).

N7: Simplify \(i^{53}\)

Since \(53 = 4\times13+1\), then \(i^{53}=(i^{4})^{13}\cdot i^{1}=1^{13}\cdot i=i\).

N8: Simplify \(i^{34}\)

Since \(34 = 4\times8+2\), then \(i^{34}=(i^{4})^{8}\cdot i^{2}=1^{8}\cdot(-1)=-1\).

o1: Simplify \((3 + 5i)+(2 + 7i)\)

Combine the real parts and the imaginary parts separately: \((3 + 2)+(5i+7i)=5 + 12i\).

o2: Simplify \((-9 + 4i)+(2 + 7i)\)

\((-9 + 2)+(4i+7i)=-7 + 11i\).

o3: Simplify \((3 + 5i)-(2 + 7i)\)

\((3-2)+(5i - 7i)=1-2i\).

o4: Simplify \((3 + 5i)+(2 - 7i)\)

\((3 + 2)+(5i-7i)=5 - 2i\).

P1: Find the product \((3 + 5i)(7 + 2i)\)

Use the FOIL method: \((3 + 5i)(7 + 2i)=3\times7+3\times2i+5i\times7+5i\times2i=21 + 6i+35i+10i^{2}\). Since \(i^{2}=-1\), we have \(21+6i + 35i-10=11 + 41i\).

P2: Find the product \((9 + 2i)(9 - 2i)\)

Use the difference - of - squares formula \((a + bi)(a - bi)=a^{2}+b^{2}\). Here \(a = 9\) and \(b = 2\), so \((9 + 2i)(9 - 2i)=9^{2}-(2i)^{2}=81+4=85\).

P3: Find the product \((2 + 5i)(2 - 5i)\)

Using the difference - of - squares formula, \((2 + 5i)(2 - 5i)=2^{2}-(5i)^{2}=4 + 25=29\).

P4: Find the product \((3 + i)(3 - i)\)

Using the difference - of - squares formula, \((3 + i)(3 - i)=3^{2}-i^{2}=9 + 1=10\).

P5: Find the product \((2 + 3i)(2 - 7i)\)

Using the FOIL method: \((2 + 3i)(2 - 7i)=2\times2-2\times7i+3i\times2-3i\times7i=4-14i + 6i-21i^{2}=4-14i + 6i + 21=25-8i\).

P6: Find the product \((5 + 4i)(5 - 4i)\)

Using the difference - of - squares formula, \((5 + 4i)(5 - 4i)=5^{2}-(4i)^{2}=25 + 16=41\).

Q1: Find the quotient \(\frac{2 + 3i}{5+4i}\)

Multiply the numerator and denominator by the conjugate of the denominator \(5 - 4i\): \(\frac{(2 + 3i)(5 - 4i)}{(5 + 4i)(5 - 4i)}=\frac{2\times5-2\times4i+3i\times5-3i\times4i}{5^{2}-(4i)^{2}}=\frac{10-8i + 15i-12i^{2}}{25 + 16}=\frac{10-8i + 15i + 12}{41}=\frac{22 + 7i}{41}=\frac{22}{41}+\frac{7}{41}i\).

Q2: Find the quotient \(\frac{5 + 4i}{3-2i}\)

Multiply the numerator and denominator by the conjugate of the denominator \(3 + 2i\): \(\frac{(5 + 4i)(3 + 2i)}{(3-2i)(3 + 2i)}=\frac{5\times3…

Answer:

M1. \((-2,2)\)
M2. \((-\infty,\infty)\)
N1. \(6i\)
N2. \(12i\)
N3. \(2\sqrt{5}i\)
N4. \(3\sqrt{3}i\)
N5. \(6\sqrt{2}i\)
N6. \(-i\)
N7. \(i\)
N8. \(-1\)
o1. \(5 + 12i\)
o2. \(-7 + 11i\)
o3. \(1-2i\)
o4. \(5 - 2i\)
P1. \(11 + 41i\)
P2. \(85\)
P3. \(29\)
P4. \(10\)
P5. \(25-8i\)
P6. \(41\)
Q1. \(\frac{22}{41}+\frac{7}{41}i\)
Q2. \(\frac{7}{13}+\frac{22}{13}i\)