Solving Quadratic Equations by Factoring: Complete with ease ... - Free Printable

Here are the solutions for the quadratic equations on your worksheet. I have solved each one step-by-step using factoring.

1. $m^2 - 65m = 0$
* Factor out the common term $m$:
$m(m - 65) = 0$
* Set each part to zero:
$m = 0$ or $m - 65 = 0 \rightarrow m = 65$
* Solutions: $m = 0, 65$

2. $x^2 - 12x + 27 = 0$
* Find two numbers that multiply to $+27$ and add to $-12$. These are $-3$ and $-9$.
* Write the factors:
$(x - 3)(x - 9) = 0$
* Solve for $x$:
$x - 3 = 0 \rightarrow x = 3$
$x - 9 = 0 \rightarrow x = 9$
* Solutions: $x = 3, 9$

3. $k^2 - 12k = 0$
* Factor out the common term $k$:
$k(k - 12) = 0$
* Set each part to zero:
$k = 0$ or $k - 12 = 0 \rightarrow k = 12$
* Solutions: $k = 0, 12$

4. $b^2 + 12b + 35 = 0$
* Find two numbers that multiply to $+35$ and add to $+12$. These are $+5$ and $+7$.
* Write the factors:
$(b + 5)(b + 7) = 0$
* Solve for $b$:
$b + 5 = 0 \rightarrow b = -5$
$b + 7 = 0 \rightarrow b = -7$
* Solutions: $b = -5, -7$

5. $x^2 - 19x = 0$
* Factor out the common term $x$:
$x(x - 19) = 0$
* Set each part to zero:
$x = 0$ or $x - 19 = 0 \rightarrow x = 19$
* Solutions: $x = 0, 19$

6. $x^3 + 3x^2 - 4x = 0$
* First, factor out the common term $x$:
$x(x^2 + 3x - 4) = 0$
* Now factor the part inside the parentheses. Find two numbers that multiply to $-4$ and add to $+3$. These are $+4$ and $-1$.
$x(x + 4)(x - 1) = 0$
* Solve for $x$:
$x = 0$
$x + 4 = 0 \rightarrow x = -4$
$x - 1 = 0 \rightarrow x = 1$
* Solutions: $x = 0, -4, 1$

7. $x^2 + 9x = 0$
* Factor out the common term $x$:
$x(x + 9) = 0$
* Set each part to zero:
$x = 0$ or $x + 9 = 0 \rightarrow x = -9$
* Solutions: $x = 0, -9$

8. $x^2 - 19x + 9 = 0$
* We need two numbers that multiply to $9$ and add to $-19$.
* The factors of $9$ are $(1, 9)$ and $(3, 3)$. Neither pair adds up to $19$ (even if we use negatives like $-1$ and $-9$, they only add to $-10$).
* Since this cannot be factored using simple integers, we use the Quadratic Formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
* $a = 1, b = -19, c = 9$
* $x = \frac{19 \pm \sqrt{(-19)^2 - 4(1)(9)}}{2}$
* $x = \frac{19 \pm \sqrt{361 - 36}}{2}$
* $x = \frac{19 \pm \sqrt{325}}{2}$
* $\sqrt{325}$ simplifies to $5\sqrt{13}$
* Solutions: $x = \frac{19 + 5\sqrt{13}}{2}$ and $x = \frac{19 - 5\sqrt{13}}{2}$
*(Note: If your class only accepts integer answers, there may be a typo in the problem statement, such as the last number being 90 instead of 9. If it were 90, the answers would be 9 and 10. As written, the answer involves square roots.)*

9. $n^2 + 21n = 0$
* Factor out the common term $n$:
$n(n + 21) = 0$
* Set each part to zero:
$n = 0$ or $n + 21 = 0 \rightarrow n = -21$
* Solutions: $n = 0, -21$

10. $y^2 - 10y + 21 = 0$
* Find two numbers that multiply to $+21$ and add to $-10$. These are $-3$ and $-7$.
* Write the factors:
$(y - 3)(y - 7) = 0$
* Solve for $y$:
$y - 3 = 0 \rightarrow y = 3$
$y - 7 = 0 \rightarrow y = 7$
* Solutions: $y = 3, 7$

Final Answer:
1. m = 0, 65
2. x = 3, 9
3. k = 0, 12
4. b = -5, -7
5. x = 0, 19
6. x = 0, -4, 1
7. x = 0, -9
8. x = (19 ± 5√13) / 2
9. n = 0, -21
10. y = 3, 7