1. Day 146 – Solve 𝑥 2 =𝑘

2. Finding Square Roots
Finding the exact solutions for 0=−16 𝑡 involves finding square roots. Every positive number has a positive and a negative square root. The positive square root of 9 is 3. 9 =3 The negative square root of 9 −3. − 9 =−3 When solving a quadratic equation, you can use the symbol ±, which is read as “plus or minus.”

3. Remember to consider both positive and negative square root of 9.
Finding Square Roots
Therefore, to solve the equation 𝑥 2 =9, take the square root of the expression on each side of the equal sign.
Remember to consider both positive and negative square root of 9.

4. Example 1
Show that 4 9 = Simplify the square root under the radical signs. 4 9 = 2 3 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 2 3 ∙ 2 3 = = 2 3 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 2∙2 3∙3 = 4 9 Since 4 9 = 2 3 𝑎𝑛𝑑 4 9 = 2 3 , 4 9 = 4 9 by subsitution

5. Example 2
a) 𝑥 2 = 4 9 b) 𝑥 2 =1.44 c) 𝑥 2 =10 Answer a)There are two solutions: 4 9 = 2 3 and − 4 9 =− 2 3 b) There are two solutions: 1.44 =1.2 𝑎𝑛𝑑 − 1.44 =−1.2

6. Example 2
c) The solutions are 10 𝑎𝑛𝑑 − 10 . There is no rational number answer. An approximation can be found by using the key on your calculator. The approximate solutions are 3.16 and −3.16. The results of Example 2 lead to a generalization for solving a quadratic equation of the form 𝑥 2 =𝑘. = 10

7. Example 2
SOLVING 𝒙 𝟐 =𝒌 WHEN 𝒌≥𝟎 If 𝑥 2 =𝑘, and 𝑘≥0, then 1. 𝑥=± 𝑘 and 2. The solutions are 𝑘 and − 𝑘 . For example, if 𝑥 2 =16, then 𝑥=± 16 , or ±4. the solutions are 4 and −4.

8. Example 3
Solve each equation. a) 𝑎−2 2 −9=0 b) 𝑥−2 2 =11

9. Answer
a) In this equation, the expression 𝑎−2 plays the role of 𝑥 in the statement 𝑥 2 =𝑘.
The solutions are 2+3, or 5, and 2−3, or −1. Check each solution in the original equation.

10. Answer
b) The approximate solutions are , or 5.32, and 2− 11 , or −1.32. Check each solution in the original equation. The solution to Exampl e 3, Part a) will help you sketch the graph of the function 𝑓 𝑥 = 𝑥−2 2 −9

11. Answer
axis of symmetry
In the form 𝑓 𝑥 = 𝑥−2 2 −9, the vertex is 2, −9 and the axis of symmetry is 𝑥=2. Since the coefficient of the quadratic term is positive, the parabola opens upward and has a minimum value. The solutions to 𝑥−2 2 −9= 0 are 5 and −1. Thus, the graph is a parabola that crosses the x-axis at 5, 0 and (−1, 0).
vertex