IDNLearn.com is your trusted platform for finding reliable answers. Discover detailed and accurate answers to your questions from our knowledgeable and dedicated community members.
Sagot :
Sure, let's solve each equation by factoring step-by-step.
### 1. [tex]\( x^2 + 7x = 0 \)[/tex]
Factor out the common term [tex]\( x \)[/tex]:
[tex]\[ x(x + 7) = 0 \][/tex]
Set each factor to 0:
[tex]\[ x = 0 \][/tex]
[tex]\[ x + 7 = 0 \][/tex]
[tex]\[ x = -7 \][/tex]
So the solutions are [tex]\( x = 0 \)[/tex] and [tex]\( x = -7 \)[/tex].
### 2. [tex]\( 6s^2 + 18s = 0 \)[/tex]
Factor out the common term [tex]\( 6s \)[/tex]:
[tex]\[ 6s(s + 3) = 0 \][/tex]
Set each factor to 0:
[tex]\[ 6s = 0 \][/tex]
[tex]\[ s = 0 \][/tex]
[tex]\[ s + 3 = 0 \][/tex]
[tex]\[ s = -3 \][/tex]
So the solutions are [tex]\( s = 0 \)[/tex] and [tex]\( s = -3 \)[/tex].
### 3. [tex]\( t^2 + 8t + 16 = 0 \)[/tex]
This is a perfect square trinomial:
[tex]\[ t^2 + 2 \cdot 4 \cdot t + 4^2 = (t + 4)^2 = 0 \][/tex]
Set the factor to 0:
[tex]\[ (t + 4)^2 = 0 \][/tex]
[tex]\[ t + 4 = 0 \][/tex]
[tex]\[ t = -4 \][/tex]
So the solution is [tex]\( t = -4 \)[/tex].
### 4. [tex]\( x^2 - 10x + 25 = 0 \)[/tex]
This is a perfect square trinomial:
[tex]\[ x^2 - 2 \cdot 5 \cdot x + 5^2 = (x - 5)^2 = 0 \][/tex]
Set the factor to 0:
[tex]\[ (x - 5)^2 = 0 \][/tex]
[tex]\[ x - 5 = 0 \][/tex]
[tex]\[ x = 5 \][/tex]
So the solution is [tex]\( x = 5 \)[/tex].
### 5. [tex]\( h^2 + 6h = 16 \)[/tex]
Rewrite the equation in standard form:
[tex]\[ h^2 + 6h - 16 = 0 \][/tex]
Now, factor the trinomial:
[tex]\[ (h + 8)(h - 2) = 0 \][/tex]
Set each factor to 0:
[tex]\[ h + 8 = 0 \][/tex]
[tex]\[ h = -8 \][/tex]
[tex]\[ h - 2 = 0 \][/tex]
[tex]\[ h = 2 \][/tex]
So the solutions are [tex]\( h = -8 \)[/tex] and [tex]\( h = 2 \)[/tex].
### 6. [tex]\( x^2 - 14 = 5x \)[/tex]
Rewrite the equation in standard form:
[tex]\[ x^2 - 5x - 14 = 0 \][/tex]
Now, factor the trinomial:
[tex]\[ (x - 7)(x + 2) = 0 \][/tex]
Set each factor to 0:
[tex]\[ x - 7 = 0 \][/tex]
[tex]\[ x = 7 \][/tex]
[tex]\[ x + 2 = 0 \][/tex]
[tex]\[ x = -2 \][/tex]
So the solutions are [tex]\( x = 7 \)[/tex] and [tex]\( x = -2 \)[/tex].
### 7. [tex]\( 11r + 15 = -2r^2 \)[/tex]
Rewrite the equation in standard form:
[tex]\[ 2r^2 + 11r + 15 = 0 \][/tex]
Now, factor the trinomial:
[tex]\[ (2r + 5)(r + 3) = 0 \][/tex]
Set each factor to 0:
[tex]\[ 2r + 5 = 0 \][/tex]
[tex]\[ r = -\frac{5}{2} \][/tex]
[tex]\[ r + 3 = 0 \][/tex]
[tex]\[ r = -3 \][/tex]
So the solutions are [tex]\( r = -\frac{5}{2} \)[/tex] and [tex]\( r = -3 \)[/tex].
### 8. [tex]\( x^2 - 25 = 0 \)[/tex]
This is a difference of squares:
[tex]\[ x^2 - 5^2 = (x - 5)(x + 5) = 0 \][/tex]
Set each factor to 0:
[tex]\[ x - 5 = 0 \][/tex]
[tex]\[ x = 5 \][/tex]
[tex]\[ x + 5 = 0 \][/tex]
[tex]\[ x = -5 \][/tex]
So the solutions are [tex]\( x = 5 \)[/tex] and [tex]\( x = -5 \)[/tex].
### 9. [tex]\( 81 - 4x^2 = 0 \)[/tex]
Rewrite the equation:
[tex]\[ 81 = 4x^2 \][/tex]
Divide by 4:
[tex]\[ \frac{81}{4} = x^2 \][/tex]
Take the square root of both sides:
[tex]\[ x = \pm \frac{9}{2} \][/tex]
So the solutions are [tex]\( x = \frac{9}{2} \)[/tex] and [tex]\( x = -\frac{9}{2} \)[/tex].
### 10. [tex]\( 4s^2 + 9 = 12s \)[/tex]
Rewrite the equation in standard form:
[tex]\[ 4s^2 - 12s + 9 = 0 \][/tex]
Now, factor the trinomial:
[tex]\[ (2s - 3)^2 = 0 \][/tex]
Set the factor to 0:
[tex]\[ 2s - 3 = 0 \][/tex]
[tex]\[ s = \frac{3}{2} \][/tex]
So the solution is [tex]\( s = \frac{3}{2} \)[/tex].
By factoring these quadratic equations, we were able to determine the solutions for each.
### 1. [tex]\( x^2 + 7x = 0 \)[/tex]
Factor out the common term [tex]\( x \)[/tex]:
[tex]\[ x(x + 7) = 0 \][/tex]
Set each factor to 0:
[tex]\[ x = 0 \][/tex]
[tex]\[ x + 7 = 0 \][/tex]
[tex]\[ x = -7 \][/tex]
So the solutions are [tex]\( x = 0 \)[/tex] and [tex]\( x = -7 \)[/tex].
### 2. [tex]\( 6s^2 + 18s = 0 \)[/tex]
Factor out the common term [tex]\( 6s \)[/tex]:
[tex]\[ 6s(s + 3) = 0 \][/tex]
Set each factor to 0:
[tex]\[ 6s = 0 \][/tex]
[tex]\[ s = 0 \][/tex]
[tex]\[ s + 3 = 0 \][/tex]
[tex]\[ s = -3 \][/tex]
So the solutions are [tex]\( s = 0 \)[/tex] and [tex]\( s = -3 \)[/tex].
### 3. [tex]\( t^2 + 8t + 16 = 0 \)[/tex]
This is a perfect square trinomial:
[tex]\[ t^2 + 2 \cdot 4 \cdot t + 4^2 = (t + 4)^2 = 0 \][/tex]
Set the factor to 0:
[tex]\[ (t + 4)^2 = 0 \][/tex]
[tex]\[ t + 4 = 0 \][/tex]
[tex]\[ t = -4 \][/tex]
So the solution is [tex]\( t = -4 \)[/tex].
### 4. [tex]\( x^2 - 10x + 25 = 0 \)[/tex]
This is a perfect square trinomial:
[tex]\[ x^2 - 2 \cdot 5 \cdot x + 5^2 = (x - 5)^2 = 0 \][/tex]
Set the factor to 0:
[tex]\[ (x - 5)^2 = 0 \][/tex]
[tex]\[ x - 5 = 0 \][/tex]
[tex]\[ x = 5 \][/tex]
So the solution is [tex]\( x = 5 \)[/tex].
### 5. [tex]\( h^2 + 6h = 16 \)[/tex]
Rewrite the equation in standard form:
[tex]\[ h^2 + 6h - 16 = 0 \][/tex]
Now, factor the trinomial:
[tex]\[ (h + 8)(h - 2) = 0 \][/tex]
Set each factor to 0:
[tex]\[ h + 8 = 0 \][/tex]
[tex]\[ h = -8 \][/tex]
[tex]\[ h - 2 = 0 \][/tex]
[tex]\[ h = 2 \][/tex]
So the solutions are [tex]\( h = -8 \)[/tex] and [tex]\( h = 2 \)[/tex].
### 6. [tex]\( x^2 - 14 = 5x \)[/tex]
Rewrite the equation in standard form:
[tex]\[ x^2 - 5x - 14 = 0 \][/tex]
Now, factor the trinomial:
[tex]\[ (x - 7)(x + 2) = 0 \][/tex]
Set each factor to 0:
[tex]\[ x - 7 = 0 \][/tex]
[tex]\[ x = 7 \][/tex]
[tex]\[ x + 2 = 0 \][/tex]
[tex]\[ x = -2 \][/tex]
So the solutions are [tex]\( x = 7 \)[/tex] and [tex]\( x = -2 \)[/tex].
### 7. [tex]\( 11r + 15 = -2r^2 \)[/tex]
Rewrite the equation in standard form:
[tex]\[ 2r^2 + 11r + 15 = 0 \][/tex]
Now, factor the trinomial:
[tex]\[ (2r + 5)(r + 3) = 0 \][/tex]
Set each factor to 0:
[tex]\[ 2r + 5 = 0 \][/tex]
[tex]\[ r = -\frac{5}{2} \][/tex]
[tex]\[ r + 3 = 0 \][/tex]
[tex]\[ r = -3 \][/tex]
So the solutions are [tex]\( r = -\frac{5}{2} \)[/tex] and [tex]\( r = -3 \)[/tex].
### 8. [tex]\( x^2 - 25 = 0 \)[/tex]
This is a difference of squares:
[tex]\[ x^2 - 5^2 = (x - 5)(x + 5) = 0 \][/tex]
Set each factor to 0:
[tex]\[ x - 5 = 0 \][/tex]
[tex]\[ x = 5 \][/tex]
[tex]\[ x + 5 = 0 \][/tex]
[tex]\[ x = -5 \][/tex]
So the solutions are [tex]\( x = 5 \)[/tex] and [tex]\( x = -5 \)[/tex].
### 9. [tex]\( 81 - 4x^2 = 0 \)[/tex]
Rewrite the equation:
[tex]\[ 81 = 4x^2 \][/tex]
Divide by 4:
[tex]\[ \frac{81}{4} = x^2 \][/tex]
Take the square root of both sides:
[tex]\[ x = \pm \frac{9}{2} \][/tex]
So the solutions are [tex]\( x = \frac{9}{2} \)[/tex] and [tex]\( x = -\frac{9}{2} \)[/tex].
### 10. [tex]\( 4s^2 + 9 = 12s \)[/tex]
Rewrite the equation in standard form:
[tex]\[ 4s^2 - 12s + 9 = 0 \][/tex]
Now, factor the trinomial:
[tex]\[ (2s - 3)^2 = 0 \][/tex]
Set the factor to 0:
[tex]\[ 2s - 3 = 0 \][/tex]
[tex]\[ s = \frac{3}{2} \][/tex]
So the solution is [tex]\( s = \frac{3}{2} \)[/tex].
By factoring these quadratic equations, we were able to determine the solutions for each.
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. IDNLearn.com has the solutions to your questions. Thanks for stopping by, and come back for more insightful information.
