High School

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a) Solve for \( x \) in the equation [tex] x^4 - 20x^2 + 23 = 0 [/tex].

b) Solve for \( x \) in the equation [tex] x^4 + 8x^2 - 9 = 0 [/tex].

c) Solve for \( x \) in the equation [tex] 9x^4 - 32x^2 - 16 = 0 [/tex].

Answer :

The given problems are polynomial equations that can be solved using substitution and factoring techniques. Let's solve each part sequentially.

a) Solve [tex]x^4 - 20x^2 + 23 = 0[/tex]:


  1. Substitution:
    Let [tex]y = x^2[/tex]. Then [tex]x^4 = y^2[/tex], and the equation becomes:

    [tex]y^2 - 20y + 23 = 0[/tex].


  2. Solve the Quadratic Equation:
    Use the quadratic formula [tex]y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex] where [tex]a = 1[/tex], [tex]b = -20[/tex], and [tex]c = 23[/tex]:

    [



y = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 1 \cdot 23}}{2 \cdot 1} = \frac{20 \pm \sqrt{400 - 92}}{2} = \frac{20 \pm \sqrt{308}}{2}].

Simplify [tex]\sqrt{308} = \sqrt{4 \times 77} = 2\sqrt{77}[/tex]:

[tex]y = \frac{20 \pm 2\sqrt{77}}{2} = 10 \pm \sqrt{77}[/tex].


  1. Find [tex]x[/tex]:
    Since [tex]y = x^2[/tex], we have two cases for each solution:


    • [tex]x^2 = 10 + \sqrt{77}[/tex] gives [tex]x = \pm \sqrt{10 + \sqrt{77}}[/tex].

    • [tex]x^2 = 10 - \sqrt{77}[/tex] gives [tex]x = \pm \sqrt{10 - \sqrt{77}}[/tex].




b) Solve [tex]x^4 + 8x^2 - 9 = 0[/tex]:


  1. Substitution:
    Let [tex]y = x^2[/tex]. Then [tex]x^4 = y^2[/tex], and the equation becomes:

    [tex]y^2 + 8y - 9 = 0[/tex].


  2. Solve the Quadratic Equation:
    Use the quadratic formula [tex]y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex] where [tex]a = 1[/tex], [tex]b = 8[/tex], and [tex]c = -9[/tex]:

    [



y = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot (-9)}}{2 \cdot 1} = \frac{-8 \pm \sqrt{64 + 36}}{2} = \frac{-8 \pm \sqrt{100}}{2}].

Simplifying:

[tex]y = \frac{-8 \pm 10}{2}[/tex].


  • [tex]y = 1[/tex]

  • [tex]y = -9[/tex] (Not possible since [tex]x^2[/tex] cannot be negative)



  1. Find [tex]x[/tex]:

    • [tex]x^2 = 1[/tex] gives [tex]x = \pm 1[/tex].




c) Solve [tex]9x^4 - 32x^2 - 16 = 0[/tex]:


  1. Substitution:
    Let [tex]y = x^2[/tex]. Then [tex]x^4 = y^2[/tex], and the equation becomes:

    [tex]9y^2 - 32y - 16 = 0[/tex].


  2. Solve the Quadratic Equation:
    Use the quadratic formula [tex]y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex] where [tex]a = 9[/tex], [tex]b = -32[/tex], and [tex]c = -16[/tex]:

    [



y = \frac{32 \pm \sqrt{(-32)^2 - 4 \cdot 9 \cdot (-16)}}{2 \cdot 9} = \frac{32 \pm \sqrt{1024 + 576}}{18}].

[tex]y = \frac{32 \pm \sqrt{1600}}{18} = \frac{32 \pm 40}{18}[/tex].


  • [tex]y = 4[/tex]

  • [tex]y = -\frac{8}{9}[/tex] (Not possible since [tex]x^2[/tex] cannot be negative)



  1. Find [tex]x[/tex]:

    • [tex]x^2 = 4[/tex] gives [tex]x = \pm 2[/tex].




In summary, the solutions for each equation are:


  • (a) [tex]x = \pm \sqrt{10 + \sqrt{77}}[/tex], [tex]x = \pm \sqrt{10 - \sqrt{77}}[/tex].

  • (b) [tex]x = \pm 1[/tex].

  • (c) [tex]x = \pm 2[/tex].

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