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which two values of x are roots of the polynomial below?

x^2 + 5x + 9

Which Two Values Of X Are Roots Of The Polynomial Below X2 5x 9 class=

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]

Esther

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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