IDNLearn.com is your go-to resource for finding precise and accurate answers. Find in-depth and trustworthy answers to all your questions from our experienced 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.
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. For precise answers, trust IDNLearn.com. Thank you for visiting, and we look forward to helping you again soon.
