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Math  /  Algebra

QuestionWithout using a calculator, solve each problem in the set of REAL numbers. Show work. Write your Calculus 1 answer as a set of solutions (for equations) or as interval notation (for inequalities). Keep the numbers in their exact simplified form. (Example: x=20=25x=\sqrt{20}=2 \sqrt{5} ). If solving by factoring is easy, utilize factoring.
1. a) x2=8=±22x^{2}=8= \pm 2 \sqrt{2} b) x2<8x^{2}<8 x2−8=0(x−8)(x+8)∣ 2)x+8=0;x=−8=−4×2=−22x−8=0⇒x=8=4×2=22\begin{array}{l} x^{2}-8=0 \\ (x-\sqrt{8})(x+\sqrt{8}) \end{array} \left\lvert\, \begin{array}{l} 2) x+\sqrt{8}=0 ; x=-\sqrt{8} \\ =-\sqrt{4 \times 2}=-2 \sqrt{2} \\ x-\sqrt{8}=0 \Rightarrow x=\sqrt{8}=\sqrt{4 \times 2}=2 \sqrt{2} \end{array}\right.
2. a) x2=−32x^{2}=-32 b) x2>32x^{2}>32
3. a) x2=3x+18x^{2}=3 x+18 b) x2≤3x+18x^{2} \leq 3 x+18
4. a) −5+16x−3x2=0-5+16 x-3 x^{2}=0 b) −5+16x−3x2>0-5+16 x-3 x^{2}>0

Studdy Solution

STEP 1

1. We are solving equations and inequalities involving quadratic expressions.
2. We need to solve for real numbers.
3. The solutions should be presented in exact simplified form.
4. Factoring should be used if it simplifies the solution.

STEP 2

1. Solve x2=8 x^2 = 8 .
2. Solve x2<8 x^2 < 8 .
3. Solve x2=−32 x^2 = -32 .
4. Solve x2>32 x^2 > 32 .
5. Solve x2=3x+18 x^2 = 3x + 18 .
6. Solve x2<3x+18 x^2 < 3x + 18 .
7. Solve −5+16x−3x2=0 -5 + 16x - 3x^2 = 0 .
8. Solve −5+16x−3x2>0 -5 + 16x - 3x^2 > 0 .

STEP 3

Solve x2=8 x^2 = 8 .
x2=8 x^2 = 8
Take the square root of both sides:
x=±8=±22 x = \pm \sqrt{8} = \pm 2\sqrt{2}
Solution: {x=22,x=−22} \{ x = 2\sqrt{2}, x = -2\sqrt{2} \}

STEP 4

Solve x2<8 x^2 < 8 .
x2<8 x^2 < 8
Take the square root of both sides:
−8<x<8 -\sqrt{8} < x < \sqrt{8}
Solution: (−22,22) (-2\sqrt{2}, 2\sqrt{2})

STEP 5

Solve x2=−32 x^2 = -32 .
Since there are no real numbers x x such that x2 x^2 is negative, the solution is:
Solution: ∅ \emptyset

STEP 6

Solve x2>32 x^2 > 32 .
x2>32 x^2 > 32
Take the square root of both sides:
x<−32 or x>32 x < -\sqrt{32} \text{ or } x > \sqrt{32}
Since 32=42 \sqrt{32} = 4\sqrt{2} , the solution is:
Solution: (−∞,−42)∪(42,∞) (-\infty, -4\sqrt{2}) \cup (4\sqrt{2}, \infty)

STEP 7

Solve x2=3x+18 x^2 = 3x + 18 .
Rearrange the equation:
x2−3x−18=0 x^2 - 3x - 18 = 0
Factor the quadratic equation:
(x−6)(x+3)=0 (x - 6)(x + 3) = 0
Solve for x x :
x=6 or x=−3 x = 6 \text{ or } x = -3
Solution: {x=6,x=−3} \{ x = 6, x = -3 \}

STEP 8

Solve x2<3x+18 x^2 < 3x + 18 .
Rearrange the inequality:
x2−3x−18<0 x^2 - 3x - 18 < 0
Solve the corresponding equation x2−3x−18=0 x^2 - 3x - 18 = 0 to find critical points:
(x−6)(x+3)=0 (x - 6)(x + 3) = 0
Critical points are x=6 x = 6 and x=−3 x = -3 .
Test intervals (−∞,−3) (-\infty, -3) , (−3,6) (-3, 6) , and (6,∞) (6, \infty) :
For x2−3x−18<0 x^2 - 3x - 18 < 0 , the solution is:
Solution: (−3,6) (-3, 6)

STEP 9

Solve −5+16x−3x2=0 -5 + 16x - 3x^2 = 0 .
Rewrite the equation in standard form:
−3x2+16x−5=0 -3x^2 + 16x - 5 = 0
Factor the quadratic equation, if possible. If not, use the quadratic formula:
a=−3,b=16,c=−5 a = -3, b = 16, c = -5
Quadratic formula:
x=−b±b2−4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
Plug in the values:
x=−16±162−4(−3)(−5)2(−3) x = \frac{-16 \pm \sqrt{16^2 - 4(-3)(-5)}}{2(-3)}
x=−16±256−60−6 x = \frac{-16 \pm \sqrt{256 - 60}}{-6}
x=−16±196−6 x = \frac{-16 \pm \sqrt{196}}{-6}
x=−16±14−6 x = \frac{-16 \pm 14}{-6}
Solve for x x :
x=−16+14−6=−2−6=13 x = \frac{-16 + 14}{-6} = \frac{-2}{-6} = \frac{1}{3}
x=−16−14−6=−30−6=5 x = \frac{-16 - 14}{-6} = \frac{-30}{-6} = 5
Solution: {x=13,x=5} \{ x = \frac{1}{3}, x = 5 \}

STEP 10

Solve −5+16x−3x2>0 -5 + 16x - 3x^2 > 0 .
Rewrite the inequality in standard form:
−3x2+16x−5>0 -3x^2 + 16x - 5 > 0
Use the roots found in STEP_7, x=13 x = \frac{1}{3} and x=5 x = 5 , to test the intervals (−∞,13) (-\infty, \frac{1}{3}) , (13,5) (\frac{1}{3}, 5) , and (5,∞) (5, \infty) :
For −3x2+16x−5>0 -3x^2 + 16x - 5 > 0 , the solution is:
Solution: (13,5) \left( \frac{1}{3}, 5 \right)

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