Answer :
Sure, let's solve each of the given quadratic equations step-by-step by extracting square roots.
### 1. [tex]\( x^2 = 100 \)[/tex]
First, isolate the [tex]\( x^2 \)[/tex] term. The equation is already in the proper form:
[tex]\[ x^2 = 100 \][/tex]
Next, take the square root of both sides:
[tex]\[ x = \pm \sqrt{100} \][/tex]
Since the square root of 100 is 10, we get two solutions:
[tex]\[ x = 10 \quad \text{or} \quad x = -10 \][/tex]
So the solutions for [tex]\( x \)[/tex] are [tex]\( 10 \)[/tex] and [tex]\( -10 \)[/tex].
### 2. [tex]\( 2r^2 - 144 = 0 \)[/tex]
First, isolate the [tex]\( r^2 \)[/tex] term. Add 144 to both sides:
[tex]\[ 2r^2 = 144 \][/tex]
Next, divide both sides by 2:
[tex]\[ r^2 = 72 \][/tex]
Now, take the square root of both sides:
[tex]\[ r = \pm \sqrt{72} \][/tex]
Since [tex]\( \sqrt{72} = 6\sqrt{2} \)[/tex], we get two solutions:
[tex]\[ r = 6\sqrt{2} \quad \text{or} \quad r = -6\sqrt{2} \][/tex]
So the solutions for [tex]\( r \)[/tex] are [tex]\( 6\sqrt{2} \)[/tex] and [tex]\( -6\sqrt{2} \)[/tex].
### 3. [tex]\( 3(k + 7)^2 = 289 \)[/tex]
First, isolate the [tex]\( (k + 7)^2 \)[/tex] term. Divide both sides by 3:
[tex]\[ (k + 7)^2 = \frac{289}{3} \][/tex]
Next, take the square root of both sides:
[tex]\[ k + 7 = \pm \sqrt{\frac{289}{3}} \][/tex]
Since [tex]\( \sqrt{\frac{289}{3}} = \frac{17\sqrt{3}}{3} \)[/tex], we get two solutions:
[tex]\[ k + 7 = \frac{17\sqrt{3}}{3} \quad \text{or} \quad k + 7 = -\frac{17\sqrt{3}}{3} \][/tex]
Subtract 7 from both sides to solve for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{17\sqrt{3}}{3} - 7 \quad \text{or} \quad k = -\frac{17\sqrt{3}}{3} - 7 \][/tex]
So the solutions for [tex]\( k \)[/tex] are [tex]\( -7 + \frac{17\sqrt{3}}{3} \)[/tex] and [tex]\( -7 - \frac{17\sqrt{3}}{3} \)[/tex].
### 4. [tex]\( (s - 4)^2 - 169 = 0 \)[/tex]
First, isolate the [tex]\( (s - 4)^2 \)[/tex] term. Add 169 to both sides:
[tex]\[ (s - 4)^2 = 169 \][/tex]
Next, take the square root of both sides:
[tex]\[ s - 4 = \pm \sqrt{169} \][/tex]
Since the square root of 169 is 13, we get two solutions:
[tex]\[ s - 4 = 13 \quad \text{or} \quad s - 4 = -13 \][/tex]
Add 4 to both sides to solve for [tex]\( s \)[/tex]:
[tex]\[ s = 13 + 4 \quad \text{or} \quad s = -13 + 4 \][/tex]
So the solutions for [tex]\( s \)[/tex] are [tex]\( 17 \)[/tex] and [tex]\( -9 \)[/tex].
### Summary of Solutions
1. [tex]\( x^2 = 100 \)[/tex] : [tex]\( x = 10 \)[/tex] or [tex]\( x = -10 \)[/tex]
2. [tex]\( 2r^2 - 144 = 0 \)[/tex] : [tex]\( r = 6\sqrt{2} \)[/tex] or [tex]\( r = -6\sqrt{2} \)[/tex]
3. [tex]\( 3(k + 7)^2 = 289 \)[/tex] : [tex]\( k = -7 + \frac{17\sqrt{3}}{3} \)[/tex] or [tex]\( k = -7 - \frac{17\sqrt{3}}{3} \)[/tex]
4. [tex]\( (s - 4)^2 - 169 = 0 \)[/tex] : [tex]\( s = 17 \)[/tex] or [tex]\( s = -9 \)[/tex]
These are the solutions for the given quadratic equations.
### 1. [tex]\( x^2 = 100 \)[/tex]
First, isolate the [tex]\( x^2 \)[/tex] term. The equation is already in the proper form:
[tex]\[ x^2 = 100 \][/tex]
Next, take the square root of both sides:
[tex]\[ x = \pm \sqrt{100} \][/tex]
Since the square root of 100 is 10, we get two solutions:
[tex]\[ x = 10 \quad \text{or} \quad x = -10 \][/tex]
So the solutions for [tex]\( x \)[/tex] are [tex]\( 10 \)[/tex] and [tex]\( -10 \)[/tex].
### 2. [tex]\( 2r^2 - 144 = 0 \)[/tex]
First, isolate the [tex]\( r^2 \)[/tex] term. Add 144 to both sides:
[tex]\[ 2r^2 = 144 \][/tex]
Next, divide both sides by 2:
[tex]\[ r^2 = 72 \][/tex]
Now, take the square root of both sides:
[tex]\[ r = \pm \sqrt{72} \][/tex]
Since [tex]\( \sqrt{72} = 6\sqrt{2} \)[/tex], we get two solutions:
[tex]\[ r = 6\sqrt{2} \quad \text{or} \quad r = -6\sqrt{2} \][/tex]
So the solutions for [tex]\( r \)[/tex] are [tex]\( 6\sqrt{2} \)[/tex] and [tex]\( -6\sqrt{2} \)[/tex].
### 3. [tex]\( 3(k + 7)^2 = 289 \)[/tex]
First, isolate the [tex]\( (k + 7)^2 \)[/tex] term. Divide both sides by 3:
[tex]\[ (k + 7)^2 = \frac{289}{3} \][/tex]
Next, take the square root of both sides:
[tex]\[ k + 7 = \pm \sqrt{\frac{289}{3}} \][/tex]
Since [tex]\( \sqrt{\frac{289}{3}} = \frac{17\sqrt{3}}{3} \)[/tex], we get two solutions:
[tex]\[ k + 7 = \frac{17\sqrt{3}}{3} \quad \text{or} \quad k + 7 = -\frac{17\sqrt{3}}{3} \][/tex]
Subtract 7 from both sides to solve for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{17\sqrt{3}}{3} - 7 \quad \text{or} \quad k = -\frac{17\sqrt{3}}{3} - 7 \][/tex]
So the solutions for [tex]\( k \)[/tex] are [tex]\( -7 + \frac{17\sqrt{3}}{3} \)[/tex] and [tex]\( -7 - \frac{17\sqrt{3}}{3} \)[/tex].
### 4. [tex]\( (s - 4)^2 - 169 = 0 \)[/tex]
First, isolate the [tex]\( (s - 4)^2 \)[/tex] term. Add 169 to both sides:
[tex]\[ (s - 4)^2 = 169 \][/tex]
Next, take the square root of both sides:
[tex]\[ s - 4 = \pm \sqrt{169} \][/tex]
Since the square root of 169 is 13, we get two solutions:
[tex]\[ s - 4 = 13 \quad \text{or} \quad s - 4 = -13 \][/tex]
Add 4 to both sides to solve for [tex]\( s \)[/tex]:
[tex]\[ s = 13 + 4 \quad \text{or} \quad s = -13 + 4 \][/tex]
So the solutions for [tex]\( s \)[/tex] are [tex]\( 17 \)[/tex] and [tex]\( -9 \)[/tex].
### Summary of Solutions
1. [tex]\( x^2 = 100 \)[/tex] : [tex]\( x = 10 \)[/tex] or [tex]\( x = -10 \)[/tex]
2. [tex]\( 2r^2 - 144 = 0 \)[/tex] : [tex]\( r = 6\sqrt{2} \)[/tex] or [tex]\( r = -6\sqrt{2} \)[/tex]
3. [tex]\( 3(k + 7)^2 = 289 \)[/tex] : [tex]\( k = -7 + \frac{17\sqrt{3}}{3} \)[/tex] or [tex]\( k = -7 - \frac{17\sqrt{3}}{3} \)[/tex]
4. [tex]\( (s - 4)^2 - 169 = 0 \)[/tex] : [tex]\( s = 17 \)[/tex] or [tex]\( s = -9 \)[/tex]
These are the solutions for the given quadratic equations.
