Solving Quadratic Functions with Square Roots: When Twins Really Are Different People

Yesterday, we started solving quadratic equations using square roots, and many problems resulted in solutions that were plus or minus the same number. Today, we're taking it a step further and exploring scenarios where we can arrive at different, distinct answers.

Remember, when solving by square roots, a key skill is simplifying radicals. Let's review this process. Simplifying square roots involves finding the largest perfect square factor of the number under the radical.

For example, let's simplify $\sqrt{84}$. From the attached notes, we can use a factor tree:

84 can be factored into 2 x 2 x 3 x 7, or $2^2 * 3 * 7$. Therefore, $\sqrt{84} = \sqrt{2^2 * 3 * 7} = 2\sqrt{21}$.

Now, let's look at some examples of solving quadratic equations where we get different answers:

Example 1:

Solve $(x+1)^2 - 4 = 0$.

  1. Isolate the squared term: $(x+1)^2 = 4$
  2. Take the square root of both sides: $\sqrt{(x+1)^2} = \pm\sqrt{4}$
  3. Simplify: $x+1 = \pm 2$
  4. Solve for x: $x = -1 \pm 2$
  5. Therefore, $x = -1 + 2 = 1$ or $x = -1 - 2 = -3$

So, the solutions are $x = 1$ and $x = -3$.

Example 2:

Solve $2(x+3)^2 - 18 = 0$

  1. Isolate the squared term: $2(x+3)^2 = 18$
  2. Divide by 2: $(x+3)^2 = 9$
  3. Take the square root of both sides: $\sqrt{(x+3)^2} = \pm\sqrt{9}$
  4. Simplify: $x+3 = \pm 3$
  5. Solve for x: $x = -3 \pm 3$

So, $x = -3 + 3 = 0$ or $x = -3 - 3 = -6$

Example 3:

Solve $4(x+3)^2 - 8 = 0$

  1. Isolate the squared term: $4(x+3)^2 = 8$
  2. Divide by 4: $(x+3)^2 = 2$
  3. Take the square root of both sides: $\sqrt{(x+3)^2} = \pm\sqrt{2}$
  4. Simplify: $x+3 = \pm \sqrt{2}$
  5. Solve for x: $x = -3 \pm \sqrt{2}$

So, $x = -3 + \sqrt{2}$ or $x = -3 - \sqrt{2}$

Class Notes, In-Class Assignment, and Homework:

  • Homework: Practice B Worksheet #19-26 all.

Discussion Question:

When solving by square root, you must be able to factor numbers to simplify the square roots. I have shown you one way to do it by using factor trees...is there another way to do it? Look online for resources on how to simplify square roots and post them here.