Solved on Mar 07, 2024

Solve for exact roots: $(x-3)^2=4$ , $(x+2)^2=9$ , $(d+1/2)^2=1$ , $(h-3/4)^2=7/16$ , $(s+6)^2=3/4$ , $(x+4)^2=18$ .

STEP 1

Assumptions
1. We are solving a set of quadratic equations of the form $(variable - constant)^2 = value$ or $(variable + constant)^2 = value$ .
2. We will use the square root property, which states that if $x^2 = a$ , then $x = \sqrt{a}$ or $x = -\sqrt{a}$ .
3. The answers should be expressed as exact roots, meaning we will not approximate the square roots.

STEP 2

Solve equation a) $(x-3)^{2}=4$ .
First, take the square root of both sides of the equation.
$x - 3 = \pm\sqrt{4}$

STEP 3

Calculate the square root of 4.
$\sqrt{4} = 2$

STEP 4

Write the two possible solutions for $x$ from equation a).
$x - 3 = 2 \quad \text{or} \quad x - 3 = -2$

STEP 5

Solve for $x$ in both cases.
For the first case: $x = 3 + 2$ $x = 5$
For the second case: $x = 3 - 2$ $x = 1$
The solutions for equation a) are $x = 5$ and $x = 1$ .

STEP 6

Solve equation b) $(x+2)^{2}=9$ .
Take the square root of both sides of the equation.
$x + 2 = \pm\sqrt{9}$

STEP 7

Calculate the square root of 9.
$\sqrt{9} = 3$

STEP 8

Write the two possible solutions for $x$ from equation b).
$x + 2 = 3 \quad \text{or} \quad x + 2 = -3$

STEP 9

Solve for $x$ in both cases.
For the first case: $x = 3 - 2$ $x = 1$
For the second case: $x = -3 - 2$ $x = -5$
The solutions for equation b) are $x = 1$ and $x = -5$ .

STEP 10

Solve equation c) $\left(d+\frac{1}{2}\right)^{2}=1$ .
Take the square root of both sides of the equation.
$d + \frac{1}{2} = \pm\sqrt{1}$

STEP 11

Calculate the square root of 1.
$\sqrt{1} = 1$

STEP 12

Write the two possible solutions for $d$ from equation c).
$d + \frac{1}{2} = 1 \quad \text{or} \quad d + \frac{1}{2} = -1$

STEP 13

Solve for $d$ in both cases.
For the first case: $d = 1 - \frac{1}{2}$ $d = \frac{1}{2}$
For the second case: $d = -1 - \frac{1}{2}$ $d = -\frac{3}{2}$
The solutions for equation c) are $d = \frac{1}{2}$ and $d = -\frac{3}{2}$ .

STEP 14

Solve equation d) $\left(h-\frac{3}{4}\right)^{2}=\frac{7}{16}$ .
Take the square root of both sides of the equation.
$h - \frac{3}{4} = \pm\sqrt{\frac{7}{16}}$

STEP 15

Calculate the square root of $\frac{7}{16}$ .
$\sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$

STEP 16

Write the two possible solutions for $h$ from equation d).
$h - \frac{3}{4} = \frac{\sqrt{7}}{4} \quad \text{or} \quad h - \frac{3}{4} = -\frac{\sqrt{7}}{4}$

STEP 17

Solve for $h$ in both cases.
For the first case: $h = \frac{3}{4} + \frac{\sqrt{7}}{4}$ $h = \frac{3 + \sqrt{7}}{4}$
For the second case: $h = \frac{3}{4} - \frac{\sqrt{7}}{4}$ $h = \frac{3 - \sqrt{7}}{4}$
The solutions for equation d) are $h = \frac{3 + \sqrt{7}}{4}$ and $h = \frac{3 - \sqrt{7}}{4}$ .

STEP 18

Solve equation e) $(s+6)^{2}=\frac{3}{4}$ .
Take the square root of both sides of the equation.
$s + 6 = \pm\sqrt{\frac{3}{4}}$

STEP 19

Calculate the square root of $\frac{3}{4}$ .
$\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$

STEP 20

Write the two possible solutions for $s$ from equation e).
$s + 6 = \frac{\sqrt{3}}{2} \quad \text{or} \quad s + 6 = -\frac{\sqrt{3}}{2}$

STEP 21

Solve for $s$ in both cases.
For the first case: $s = \frac{\sqrt{3}}{2} - 6$ $s = \frac{\sqrt{3} - 12}{2}$
For the second case: $s = -\frac{\sqrt{3}}{2} - 6$ $s = \frac{-\sqrt{3} - 12}{2}$
The solutions for equation e) are $s = \frac{\sqrt{3} - 12}{2}$ and $s = \frac{-\sqrt{3} - 12}{2}$ .

STEP 22

Solve equation f) $(x+4)^{2}=18$ .
Take the square root of both sides of the equation.
$x + 4 = \pm\sqrt{18}$

STEP 23

Calculate the square root of 18.
$\sqrt{18} = \sqrt{9 \cdot 2} = \sqrt{9} \cdot \sqrt{2} = 3\sqrt{2}$

STEP 24

Write the two possible solutions for $x$ from equation f).
$x + 4 = 3\sqrt{2} \quad \text{or} \quad x + 4 = -3\sqrt{2}$

STEP 25

Solve for $x$ in both cases.
For the first case: $x = 3\sqrt{2} - 4$
For the second case: $x = -3\sqrt{2} - 4$
The solutions for equation f) are $x = 3\sqrt{2} - 4$ and $x = -3\sqrt{2} - 4$ .

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Solve for exact roots: (x−3)2=4(x-3)^2=4(x−3)2=4, (x+2)2=9(x+2)^2=9(x+2)2=9, (d+1/2)2=1(d+1/2)^2=1(d+1/2)2=1, (h−3/4)2=7/16(h-3/4)^2=7/16(h−3/4)2=7/16, (s+6)2=3/4(s+6)^2=3/4(s+6)2=3/4, (x+4)2=18(x+4)^2=18(x+4)2=18.

STEP 1

STEP 2

Solve equation a) (x−3)2=4(x-3)^{2}=4(x−3)2=4.First, take the square root of both sides of the equation.x−3=±4x - 3 = \pm\sqrt{4}x−3=±4​

STEP 3

Calculate the square root of 4.4=2\sqrt{4} = 24​=2

STEP 4

Write the two possible solutions for xxx from equation a).x−3=2orx−3=−2x - 3 = 2 \quad \text{or} \quad x - 3 = -2x−3=2orx−3=−2

STEP 5

Solve for xxx in both cases.For the first case: x=3+2x = 3 + 2x=3+2 x=5x = 5x=5For the second case: x=3−2x = 3 - 2x=3−2 x=1x = 1x=1The solutions for equation a) are x=5x = 5x=5 and x=1x = 1x=1.

STEP 6

Solve equation b) (x+2)2=9(x+2)^{2}=9(x+2)2=9.Take the square root of both sides of the equation.x+2=±9x + 2 = \pm\sqrt{9}x+2=±9​

STEP 7

Calculate the square root of 9.9=3\sqrt{9} = 39​=3

STEP 8

Write the two possible solutions for xxx from equation b).x+2=3orx+2=−3x + 2 = 3 \quad \text{or} \quad x + 2 = -3x+2=3orx+2=−3

STEP 9

Solve for xxx in both cases.For the first case: x=3−2x = 3 - 2x=3−2 x=1x = 1x=1For the second case: x=−3−2x = -3 - 2x=−3−2 x=−5x = -5x=−5The solutions for equation b) are x=1x = 1x=1 and x=−5x = -5x=−5.

STEP 10

Solve equation c) (d+12)2=1\left(d+\frac{1}{2}\right)^{2}=1(d+21​)2=1.Take the square root of both sides of the equation.d+12=±1d + \frac{1}{2} = \pm\sqrt{1}d+21​=±1​

STEP 11

Calculate the square root of 1.1=1\sqrt{1} = 11​=1

STEP 12

Write the two possible solutions for ddd from equation c).d+12=1ord+12=−1d + \frac{1}{2} = 1 \quad \text{or} \quad d + \frac{1}{2} = -1d+21​=1ord+21​=−1

STEP 13

STEP 14

Solve equation d) (h−34)2=716\left(h-\frac{3}{4}\right)^{2}=\frac{7}{16}(h−43​)2=167​.Take the square root of both sides of the equation.h−34=±716h - \frac{3}{4} = \pm\sqrt{\frac{7}{16}}h−43​=±167​​

STEP 15

Calculate the square root of 716\frac{7}{16}167​.716=74\sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}167​​=47​​

STEP 16

Write the two possible solutions for hhh from equation d).h−34=74orh−34=−74h - \frac{3}{4} = \frac{\sqrt{7}}{4} \quad \text{or} \quad h - \frac{3}{4} = -\frac{\sqrt{7}}{4}h−43​=47​​orh−43​=−47​​

STEP 17

STEP 18

Solve equation e) (s+6)2=34(s+6)^{2}=\frac{3}{4}(s+6)2=43​.Take the square root of both sides of the equation.s+6=±34s + 6 = \pm\sqrt{\frac{3}{4}}s+6=±43​​

STEP 19

Calculate the square root of 34\frac{3}{4}43​.34=32\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}43​​=23​​

STEP 20

Write the two possible solutions for sss from equation e).s+6=32ors+6=−32s + 6 = \frac{\sqrt{3}}{2} \quad \text{or} \quad s + 6 = -\frac{\sqrt{3}}{2}s+6=23​​ors+6=−23​​

STEP 21

STEP 22

Solve equation f) (x+4)2=18(x+4)^{2}=18(x+4)2=18.Take the square root of both sides of the equation.x+4=±18x + 4 = \pm\sqrt{18}x+4=±18​

STEP 23

Calculate the square root of 18.18=9⋅2=9⋅2=32\sqrt{18} = \sqrt{9 \cdot 2} = \sqrt{9} \cdot \sqrt{2} = 3\sqrt{2}18​=9⋅2​=9​⋅2​=32​

STEP 24

Write the two possible solutions for xxx from equation f).x+4=32orx+4=−32x + 4 = 3\sqrt{2} \quad \text{or} \quad x + 4 = -3\sqrt{2}x+4=32​orx+4=−32​

STEP 25

Solve for xxx in both cases.For the first case: x=32−4x = 3\sqrt{2} - 4x=32​−4For the second case: x=−32−4x = -3\sqrt{2} - 4x=−32​−4The solutions for equation f) are x=32−4x = 3\sqrt{2} - 4x=32​−4 and x=−32−4x = -3\sqrt{2} - 4x=−32​−4.

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Solve for exact roots: $(x-3)^2=4$ , $(x+2)^2=9$ , $(d+1/2)^2=1$ , $(h-3/4)^2=7/16$ , $(s+6)^2=3/4$ , $(x+4)^2=18$ .

Solve equation a) $(x-3)^{2}=4$ .
First, take the square root of both sides of the equation.
$x - 3 = \pm\sqrt{4}$

Calculate the square root of 4.
$\sqrt{4} = 2$

Write the two possible solutions for $x$ from equation a).
$x - 3 = 2 \quad \text{or} \quad x - 3 = -2$

Solve for $x$ in both cases.
For the first case: $x = 3 + 2$ $x = 5$
For the second case: $x = 3 - 2$ $x = 1$
The solutions for equation a) are $x = 5$ and $x = 1$ .

Solve equation b) $(x+2)^{2}=9$ .
Take the square root of both sides of the equation.
$x + 2 = \pm\sqrt{9}$

Calculate the square root of 9.
$\sqrt{9} = 3$

Write the two possible solutions for $x$ from equation b).
$x + 2 = 3 \quad \text{or} \quad x + 2 = -3$

Solve for $x$ in both cases.
For the first case: $x = 3 - 2$ $x = 1$
For the second case: $x = -3 - 2$ $x = -5$
The solutions for equation b) are $x = 1$ and $x = -5$ .

Solve equation c) $\left(d+\frac{1}{2}\right)^{2}=1$ .
Take the square root of both sides of the equation.
$d + \frac{1}{2} = \pm\sqrt{1}$

Calculate the square root of 1.
$\sqrt{1} = 1$

Write the two possible solutions for $d$ from equation c).
$d + \frac{1}{2} = 1 \quad \text{or} \quad d + \frac{1}{2} = -1$

Solve equation d) $\left(h-\frac{3}{4}\right)^{2}=\frac{7}{16}$ .
Take the square root of both sides of the equation.
$h - \frac{3}{4} = \pm\sqrt{\frac{7}{16}}$

Calculate the square root of $\frac{7}{16}$ .
$\sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$

Write the two possible solutions for $h$ from equation d).
$h - \frac{3}{4} = \frac{\sqrt{7}}{4} \quad \text{or} \quad h - \frac{3}{4} = -\frac{\sqrt{7}}{4}$

Solve equation e) $(s+6)^{2}=\frac{3}{4}$ .
Take the square root of both sides of the equation.
$s + 6 = \pm\sqrt{\frac{3}{4}}$

Calculate the square root of $\frac{3}{4}$ .
$\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$

Write the two possible solutions for $s$ from equation e).
$s + 6 = \frac{\sqrt{3}}{2} \quad \text{or} \quad s + 6 = -\frac{\sqrt{3}}{2}$

Solve equation f) $(x+4)^{2}=18$ .
Take the square root of both sides of the equation.
$x + 4 = \pm\sqrt{18}$

Calculate the square root of 18.
$\sqrt{18} = \sqrt{9 \cdot 2} = \sqrt{9} \cdot \sqrt{2} = 3\sqrt{2}$

Write the two possible solutions for $x$ from equation f).
$x + 4 = 3\sqrt{2} \quad \text{or} \quad x + 4 = -3\sqrt{2}$

Solve for $x$ in both cases.
For the first case: $x = 3\sqrt{2} - 4$
For the second case: $x = -3\sqrt{2} - 4$
The solutions for equation f) are $x = 3\sqrt{2} - 4$ and $x = -3\sqrt{2} - 4$ .