Solved on Jan 12, 2024

Solve absolute value equations: 1) $|4x| = x^2 - 5$ , 2) $2x^2 = |5x + 3|$ , 3) $|2(x-4)^2 - 5| = 3$ , 4) $0 = |x^2 - 2x - 3| - 4$ .

STEP 1

Assumptions
1. We are solving absolute value equations.
2. The absolute value of a number is its distance from zero on the number line, regardless of direction.
3. To solve an absolute value equation, we consider both the positive and negative scenarios of the expression inside the absolute value.
We will solve each equation separately.
a) $|4x| = x^2 - 5$

STEP 2

We need to consider two cases for the absolute value equation:
Case 1: $4x = x^2 - 5$
Case 2: $-4x = x^2 - 5$
Let's solve each case.

STEP 3

Solve Case 1:
$4x = x^2 - 5$

STEP 4

Move all terms to one side to set the equation to zero:
$x^2 - 4x - 5 = 0$

STEP 5

Factor the quadratic equation:
$(x - 5)(x + 1) = 0$

STEP 6

Set each factor equal to zero and solve for $x$ :
$x - 5 = 0 \quad \text{or} \quad x + 1 = 0$

STEP 7

Find the solutions for Case 1:
$x = 5 \quad \text{or} \quad x = -1$

STEP 8

Solve Case 2:
$-4x = x^2 - 5$

STEP 9

Move all terms to one side to set the equation to zero:
$x^2 + 4x - 5 = 0$

STEP 10

Factor the quadratic equation:
$(x + 5)(x - 1) = 0$

STEP 11

Set each factor equal to zero and solve for $x$ :
$x + 5 = 0 \quad \text{or} \quad x - 1 = 0$

STEP 12

Find the solutions for Case 2:
$x = -5 \quad \text{or} \quad x = 1$

STEP 13

Combine the solutions from both cases:
$x = 5, -1, -5, 1$
b) $2x^2 = |5x + 3|$

STEP 14

We need to consider two cases for the absolute value equation:
Case 1: $2x^2 = 5x + 3$
Case 2: $2x^2 = -(5x + 3)$
Let's solve each case.

STEP 15

Solve Case 1:
$2x^2 = 5x + 3$

STEP 16

Move all terms to one side to set the equation to zero:
$2x^2 - 5x - 3 = 0$

STEP 17

This quadratic equation does not factor nicely, so we use the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = 2$ , $b = -5$ , and $c = -3$ .

STEP 18

Plug in the values into the quadratic formula:
$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-3)}}{2(2)}$

STEP 19

Simplify and calculate the discriminant:
$x = \frac{5 \pm \sqrt{25 + 24}}{4}$
$x = \frac{5 \pm \sqrt{49}}{4}$

STEP 20

Solve for $x$ :
$x = \frac{5 \pm 7}{4}$

STEP 21

Find the solutions for Case 1:
$x = \frac{5 + 7}{4} = 3 \quad \text{or} \quad x = \frac{5 - 7}{4} = -\frac{1}{2}$

STEP 22

Solve Case 2:
$2x^2 = -(5x + 3)$

STEP 23

Move all terms to one side to set the equation to zero:
$2x^2 + 5x + 3 = 0$

STEP 24

This quadratic equation does not factor nicely either, so we again use the quadratic formula with $a = 2$ , $b = 5$ , and $c = 3$ .

STEP 25

Plug in the values into the quadratic formula:
$x = \frac{-5 \pm \sqrt{5^2 - 4(2)(3)}}{2(2)}$

STEP 26

Simplify and calculate the discriminant:
$x = \frac{-5 \pm \sqrt{25 - 24}}{4}$
$x = \frac{-5 \pm \sqrt{1}}{4}$

STEP 27

Solve for $x$ :
$x = \frac{-5 \pm 1}{4}$

STEP 28

Find the solutions for Case 2:
$x = \frac{-5 + 1}{4} = -1 \quad \text{or} \quad x = \frac{-5 - 1}{4} = -\frac{3}{2}$

STEP 29

Combine the solutions from both cases:
$x = 3, -\frac{1}{2}, -1, -\frac{3}{2}$
c) $|2(x - 4)^2 - 5| = 3$

STEP 30

We need to consider two cases for the absolute value equation:
Case 1: $2(x - 4)^2 - 5 = 3$
Case 2: $2(x - 4)^2 - 5 = -3$
Let's solve each case.

STEP 31

Solve Case 1:
$2(x - 4)^2 - 5 = 3$

STEP 32

Move all terms to one side to set the equation to zero:
$2(x - 4)^2 - 8 = 0$

STEP 33

Divide by 2 to simplify:
$(x - 4)^2 - 4 = 0$

STEP 34

Factor the quadratic equation:
$(x - 4)^2 = 4$

STEP 35

Take the square root of both sides:
$x - 4 = \pm 2$

STEP 36

Solve for $x$ :
$x = 4 \pm 2$

STEP 37

Find the solutions for Case 1:
$x = 4 + 2 = 6 \quad \text{or} \quad x = 4 - 2 = 2$

STEP 38

Solve Case 2:
$2(x - 4)^2 - 5 = -3$

STEP 39

Move all terms to one side to set the equation to zero:
$2(x - 4)^2 - 2 = 0$

STEP 40

Divide by 2 to simplify:
$(x - 4)^2 - 1 = 0$

STEP 41

Factor the quadratic equation:
$(x - 4 + 1)(x - 4 - 1) = 0$

STEP 42

Simplify the factors:
$(x - 3)(x - 5) = 0$

STEP 43

Set each factor equal to zero and solve for $x$ :
$x - 3 = 0 \quad \text{or} \quad x - 5 = 0$

STEP 44

Find the solutions for Case 2:
$x = 3 \quad \text{or} \quad x = 5$

STEP 45

Combine the solutions from both cases:
$x = 6, 2, 3, 5$
d) $0 = |x^2 - 2x - 3| - 4$

STEP 46

We need to consider two cases for the absolute value equation:
Case 1: $x^2 - 2x - 3 = 4$
Case 2: $x^2 - 2x - 3 = -4$
Let's solve each case.

STEP 47

Solve Case 1:
$x^2 - 2x - 3 = 4$

STEP 48

Move all terms to one side to set the equation to zero:
$x^2 - 2x - 7 = 0$

STEP 49

This quadratic equation does not factor nicely, so we use the quadratic formula with $a = 1$ , $b = -2$ , and $c = -7$ .

STEP 50

Plug in the values into the quadratic formula:
$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-7)}}{2(1)}$

STEP 51

Simplify and calculate the discriminant:
$x = \frac{2 \pm \sqrt{4 + 28}}{2}$
$x = \frac{2 \pm \sqrt{32}}{2}$

STEP 52

Simplify the square root:
$x = \frac{2 \pm 4\sqrt{2}}{2}$

STEP 53

Solve for $x$ :
$x = 1 \pm 2\sqrt{2}$

STEP 54

Find the solutions for Case 1:
$x = 1 + 2\sqrt{2} \quad \text{or} \quad x = 1 - 2\sqrt{2}$

STEP 55

Solve Case 2:
$x^2 - 2x - 3 = -4$

STEP 56

Move all terms to one side to set the equation to zero:
$x^2 - 2x + 1 = 0$

STEP 57

Factor the quadratic equation:
$(x - 1)^2 = 0$

STEP 58

Take the square root of both sides:
$x - 1 = 0$

STEP 59

Solve for $x$ :
$x = 1$

STEP 60

Find the solutions for Case 2:
$x = 1$

STEP 61

Combine the solutions from both cases:
$x = 1 + 2\sqrt{2}, 1 - 2\sqrt{2}, 1$

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Solve absolute value equations: 1) ∣4x∣=x2−5|4x| = x^2 - 5∣4x∣=x2−5, 2) 2x2=∣5x+3∣2x^2 = |5x + 3|2x2=∣5x+3∣, 3) ∣2(x−4)2−5∣=3|2(x-4)^2 - 5| = 3∣2(x−4)2−5∣=3, 4) 0=∣x2−2x−3∣−40 = |x^2 - 2x - 3| - 40=∣x2−2x−3∣−4.

STEP 1

STEP 2

We need to consider two cases for the absolute value equation:Case 1: 4x=x2−54x = x^2 - 54x=x2−5Case 2: −4x=x2−5-4x = x^2 - 5−4x=x2−5Let's solve each case.

STEP 3

Solve Case 1:4x=x2−54x = x^2 - 54x=x2−5

STEP 4

Move all terms to one side to set the equation to zero:x2−4x−5=0x^2 - 4x - 5 = 0x2−4x−5=0

STEP 5

Factor the quadratic equation:(x−5)(x+1)=0(x - 5)(x + 1) = 0(x−5)(x+1)=0

STEP 6

Set each factor equal to zero and solve for xxx:x−5=0orx+1=0x - 5 = 0 \quad \text{or} \quad x + 1 = 0x−5=0orx+1=0

STEP 7

Find the solutions for Case 1:x=5orx=−1x = 5 \quad \text{or} \quad x = -1x=5orx=−1

STEP 8

Solve Case 2:−4x=x2−5-4x = x^2 - 5−4x=x2−5

STEP 9

Move all terms to one side to set the equation to zero:x2+4x−5=0x^2 + 4x - 5 = 0x2+4x−5=0

STEP 10

Factor the quadratic equation:(x+5)(x−1)=0(x + 5)(x - 1) = 0(x+5)(x−1)=0

STEP 11

Set each factor equal to zero and solve for xxx:x+5=0orx−1=0x + 5 = 0 \quad \text{or} \quad x - 1 = 0x+5=0orx−1=0

STEP 12

Find the solutions for Case 2:x=−5orx=1x = -5 \quad \text{or} \quad x = 1x=−5orx=1

STEP 13

Combine the solutions from both cases:x=5,−1,−5,1x = 5, -1, -5, 1x=5,−1,−5,1b) 2x2=∣5x+3∣2x^2 = |5x + 3|2x2=∣5x+3∣

STEP 14

We need to consider two cases for the absolute value equation:Case 1: 2x2=5x+32x^2 = 5x + 32x2=5x+3Case 2: 2x2=−(5x+3)2x^2 = -(5x + 3)2x2=−(5x+3)Let's solve each case.

STEP 15

Solve Case 1:2x2=5x+32x^2 = 5x + 32x2=5x+3

STEP 16

Move all terms to one side to set the equation to zero:2x2−5x−3=02x^2 - 5x - 3 = 02x2−5x−3=0

STEP 17

This quadratic equation does not factor nicely, so we use the quadratic formula:x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}x=2a−b±b2−4ac​​where a=2a = 2a=2, b=−5b = -5b=−5, and c=−3c = -3c=−3.

STEP 18

Plug in the values into the quadratic formula:x=−(−5)±(−5)2−4(2)(−3)2(2)x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-3)}}{2(2)}x=2(2)−(−5)±(−5)2−4(2)(−3)​​

STEP 19

Simplify and calculate the discriminant:x=5±25+244x = \frac{5 \pm \sqrt{25 + 24}}{4}x=45±25+24​​x=5±494x = \frac{5 \pm \sqrt{49}}{4}x=45±49​​

STEP 20

Solve for xxx:x=5±74x = \frac{5 \pm 7}{4}x=45±7​

STEP 21

Find the solutions for Case 1:x=5+74=3orx=5−74=−12x = \frac{5 + 7}{4} = 3 \quad \text{or} \quad x = \frac{5 - 7}{4} = -\frac{1}{2}x=45+7​=3orx=45−7​=−21​

STEP 22

Solve Case 2:2x2=−(5x+3)2x^2 = -(5x + 3)2x2=−(5x+3)

STEP 23

Move all terms to one side to set the equation to zero:2x2+5x+3=02x^2 + 5x + 3 = 02x2+5x+3=0

STEP 24

This quadratic equation does not factor nicely either, so we again use the quadratic formula with a=2a = 2a=2, b=5b = 5b=5, and c=3c = 3c=3.

STEP 25

Plug in the values into the quadratic formula:x=−5±52−4(2)(3)2(2)x = \frac{-5 \pm \sqrt{5^2 - 4(2)(3)}}{2(2)}x=2(2)−5±52−4(2)(3)​​

STEP 26

Simplify and calculate the discriminant:x=−5±25−244x = \frac{-5 \pm \sqrt{25 - 24}}{4}x=4−5±25−24​​x=−5±14x = \frac{-5 \pm \sqrt{1}}{4}x=4−5±1​​

STEP 27

Solve for xxx:x=−5±14x = \frac{-5 \pm 1}{4}x=4−5±1​

STEP 28

Find the solutions for Case 2:x=−5+14=−1orx=−5−14=−32x = \frac{-5 + 1}{4} = -1 \quad \text{or} \quad x = \frac{-5 - 1}{4} = -\frac{3}{2}x=4−5+1​=−1orx=4−5−1​=−23​

STEP 29

Combine the solutions from both cases:x=3,−12,−1,−32x = 3, -\frac{1}{2}, -1, -\frac{3}{2}x=3,−21​,−1,−23​c) ∣2(x−4)2−5∣=3|2(x - 4)^2 - 5| = 3∣2(x−4)2−5∣=3

STEP 30

We need to consider two cases for the absolute value equation:Case 1: 2(x−4)2−5=32(x - 4)^2 - 5 = 32(x−4)2−5=3Case 2: 2(x−4)2−5=−32(x - 4)^2 - 5 = -32(x−4)2−5=−3Let's solve each case.

STEP 31

Solve Case 1:2(x−4)2−5=32(x - 4)^2 - 5 = 32(x−4)2−5=3

STEP 32

Move all terms to one side to set the equation to zero:2(x−4)2−8=02(x - 4)^2 - 8 = 02(x−4)2−8=0

STEP 33

Divide by 2 to simplify:(x−4)2−4=0(x - 4)^2 - 4 = 0(x−4)2−4=0

STEP 34

Factor the quadratic equation:(x−4)2=4(x - 4)^2 = 4(x−4)2=4

STEP 35

Take the square root of both sides:x−4=±2x - 4 = \pm 2x−4=±2

STEP 36

Solve for xxx:x=4±2x = 4 \pm 2x=4±2

STEP 37

Find the solutions for Case 1:x=4+2=6orx=4−2=2x = 4 + 2 = 6 \quad \text{or} \quad x = 4 - 2 = 2x=4+2=6orx=4−2=2

STEP 38

Solve Case 2:2(x−4)2−5=−32(x - 4)^2 - 5 = -32(x−4)2−5=−3

STEP 39

Move all terms to one side to set the equation to zero:2(x−4)2−2=02(x - 4)^2 - 2 = 02(x−4)2−2=0

STEP 40

Divide by 2 to simplify:(x−4)2−1=0(x - 4)^2 - 1 = 0(x−4)2−1=0

Solve absolute value equations: 1) $|4x| = x^2 - 5$ , 2) $2x^2 = |5x + 3|$ , 3) $|2(x-4)^2 - 5| = 3$ , 4) $0 = |x^2 - 2x - 3| - 4$ .

We need to consider two cases for the absolute value equation:
Case 1: $4x = x^2 - 5$
Case 2: $-4x = x^2 - 5$
Let's solve each case.

Solve Case 1:
$4x = x^2 - 5$

Move all terms to one side to set the equation to zero:
$x^2 - 4x - 5 = 0$

Factor the quadratic equation:
$(x - 5)(x + 1) = 0$

Set each factor equal to zero and solve for $x$ :
$x - 5 = 0 \quad \text{or} \quad x + 1 = 0$

Find the solutions for Case 1:
$x = 5 \quad \text{or} \quad x = -1$

Solve Case 2:
$-4x = x^2 - 5$

Move all terms to one side to set the equation to zero:
$x^2 + 4x - 5 = 0$

Factor the quadratic equation:
$(x + 5)(x - 1) = 0$

Set each factor equal to zero and solve for $x$ :
$x + 5 = 0 \quad \text{or} \quad x - 1 = 0$

Find the solutions for Case 2:
$x = -5 \quad \text{or} \quad x = 1$

Combine the solutions from both cases:
$x = 5, -1, -5, 1$
b) $2x^2 = |5x + 3|$

We need to consider two cases for the absolute value equation:
Case 1: $2x^2 = 5x + 3$
Case 2: $2x^2 = -(5x + 3)$
Let's solve each case.

Solve Case 1:
$2x^2 = 5x + 3$

Move all terms to one side to set the equation to zero:
$2x^2 - 5x - 3 = 0$

This quadratic equation does not factor nicely, so we use the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = 2$ , $b = -5$ , and $c = -3$ .

Plug in the values into the quadratic formula:
$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-3)}}{2(2)}$

Simplify and calculate the discriminant:
$x = \frac{5 \pm \sqrt{25 + 24}}{4}$
$x = \frac{5 \pm \sqrt{49}}{4}$

Solve for $x$ :
$x = \frac{5 \pm 7}{4}$

Find the solutions for Case 1:
$x = \frac{5 + 7}{4} = 3 \quad \text{or} \quad x = \frac{5 - 7}{4} = -\frac{1}{2}$

Solve Case 2:
$2x^2 = -(5x + 3)$

Move all terms to one side to set the equation to zero:
$2x^2 + 5x + 3 = 0$

This quadratic equation does not factor nicely either, so we again use the quadratic formula with $a = 2$ , $b = 5$ , and $c = 3$ .

Plug in the values into the quadratic formula:
$x = \frac{-5 \pm \sqrt{5^2 - 4(2)(3)}}{2(2)}$

Simplify and calculate the discriminant:
$x = \frac{-5 \pm \sqrt{25 - 24}}{4}$
$x = \frac{-5 \pm \sqrt{1}}{4}$

Solve for $x$ :
$x = \frac{-5 \pm 1}{4}$

Find the solutions for Case 2:
$x = \frac{-5 + 1}{4} = -1 \quad \text{or} \quad x = \frac{-5 - 1}{4} = -\frac{3}{2}$

Combine the solutions from both cases:
$x = 3, -\frac{1}{2}, -1, -\frac{3}{2}$
c) $|2(x - 4)^2 - 5| = 3$

We need to consider two cases for the absolute value equation:
Case 1: $2(x - 4)^2 - 5 = 3$
Case 2: $2(x - 4)^2 - 5 = -3$
Let's solve each case.

Solve Case 1:
$2(x - 4)^2 - 5 = 3$

Move all terms to one side to set the equation to zero:
$2(x - 4)^2 - 8 = 0$

Divide by 2 to simplify:
$(x - 4)^2 - 4 = 0$

Factor the quadratic equation:
$(x - 4)^2 = 4$

Take the square root of both sides:
$x - 4 = \pm 2$

Solve for $x$ :
$x = 4 \pm 2$

Find the solutions for Case 1:
$x = 4 + 2 = 6 \quad \text{or} \quad x = 4 - 2 = 2$

Solve Case 2:
$2(x - 4)^2 - 5 = -3$

Move all terms to one side to set the equation to zero:
$2(x - 4)^2 - 2 = 0$

Divide by 2 to simplify:
$(x - 4)^2 - 1 = 0$

Factor the quadratic equation:
$(x - 4 + 1)(x - 4 - 1) = 0$

Simplify the factors:
$(x - 3)(x - 5) = 0$

Set each factor equal to zero and solve for $x$ :
$x - 3 = 0 \quad \text{or} \quad x - 5 = 0$

Find the solutions for Case 2:
$x = 3 \quad \text{or} \quad x = 5$

Combine the solutions from both cases:
$x = 6, 2, 3, 5$
d) $0 = |x^2 - 2x - 3| - 4$

We need to consider two cases for the absolute value equation:
Case 1: $x^2 - 2x - 3 = 4$
Case 2: $x^2 - 2x - 3 = -4$
Let's solve each case.

Solve Case 1:
$x^2 - 2x - 3 = 4$

Move all terms to one side to set the equation to zero:
$x^2 - 2x - 7 = 0$

This quadratic equation does not factor nicely, so we use the quadratic formula with $a = 1$ , $b = -2$ , and $c = -7$ .

Plug in the values into the quadratic formula:
$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-7)}}{2(1)}$

Simplify and calculate the discriminant:
$x = \frac{2 \pm \sqrt{4 + 28}}{2}$
$x = \frac{2 \pm \sqrt{32}}{2}$

Simplify the square root:
$x = \frac{2 \pm 4\sqrt{2}}{2}$

Solve for $x$ :
$x = 1 \pm 2\sqrt{2}$

Find the solutions for Case 1:
$x = 1 + 2\sqrt{2} \quad \text{or} \quad x = 1 - 2\sqrt{2}$

Solve Case 2:
$x^2 - 2x - 3 = -4$

Move all terms to one side to set the equation to zero:
$x^2 - 2x + 1 = 0$

Factor the quadratic equation:
$(x - 1)^2 = 0$

Take the square root of both sides:
$x - 1 = 0$

Solve for $x$ :
$x = 1$

Find the solutions for Case 2:
$x = 1$

Combine the solutions from both cases:
$x = 1 + 2\sqrt{2}, 1 - 2\sqrt{2}, 1$