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Sagot :
Answer:
The answer:
A. [tex]x= \frac{-5+ \sqrt{-11} }{2}[/tex]
F.[tex]x= \frac{-5- \sqrt{-11} }{2}[/tex]
Step-by-step explanation:
Step 1: Solve with quadratic formula
- [tex]x_{1},2 = \frac{-b+- \sqrt{b^2}- 4ac }{2a}[/tex]
- For [tex]a=1, b=5, c=9\\x_{1},2 = \frac{-5 + \sqrt{5^2 -4 *1 *9} }{2*1}[/tex]
Step 2: Simplify
- [tex]\sqrt{5^2 - 4 *1 *9} : \sqrt{11} i[/tex]
- Multiply the numbers: [tex]4*1*9 = 36[/tex]
- [tex]i\sqrt{ 36 -5^2}[/tex] = [tex]\sqrt{-5^2 + 36} = i\sqrt{11}[/tex]
- [tex]5^2 = 25 = \sqrt{-25 + 36}[/tex]
- Add/subtract the numbers [tex]-25 +36 = 11 = \sqrt{11} = \sqrt{11}i[/tex]
Step 3: Separate the solution
- [tex]x_{1} = \frac{-5 + \sqrt{-11}i}{2} , x_{2} = \frac{-5 - \sqrt{-11}i}{2}[/tex]
Answer:
[tex]\large {\textsf{A and F}}\ \implies \bold{x_1}=\dfrac{-5-\sqrt{-11}}{2},\ \bold{x_2}=\dfrac{-5+\sqrt{-11}}{2}[/tex]
Step-by-step explanation:
Quadratic Formula: [tex]x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Standard Form of a Quadratic Equation: ax² + bx + c = 0, where a ≠ 0.
Given polynomial: x² + 5x + 9
⇒ a = 1, b = 5, c = 9
Step 1: Rewrite to Standard Form.
⇒ x² + 5x + 9 = 0
Step 2: Substitute the values of a, b, and c into the formula.
⇒ a = 1, b = 5, c = 9
[tex]x=\dfrac{-5\pm\sqrt{\bold{5^2}-4\bold{(1)(9)}}}{\bold{2(1)}}\\\\x=\dfrac{-5\pm\sqrt{25\bold{\ - \ 4(9)}}}{2}\\\\x=\dfrac{-5\pm\sqrt{\bold{25-36}}}{2}\\\\x=\dfrac{-5\pm\sqrt{-11}}{2}[/tex]
Step 3: Separate into two possible cases.
[tex]x_1=\dfrac{-5-\sqrt{-11}}{2}\\\\x_2=\dfrac{-5+\sqrt{-11}}{2}[/tex]
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