High School

a) $x^4 - 26x^2 + 25 = 0$;

б) $x^4 + 8x^2 - 9 = 0$;

в) $9x^4 - 32x^2 - 16 = 0$.

Answer :

To solve these equations, we will use a common technique for solving quartic equations by substituting a new variable, say [tex]y = x^2[/tex]. This simplifies the problem to dealing with a quadratic in terms of [tex]y[/tex]. Let's solve each part step-by-step.

a) Solve [tex]x^4 - 26x^2 + 25 = 0[/tex]:

  1. Substitute [tex]y = x^2[/tex]. Then [tex]x^4 = y^2[/tex], so the equation becomes:
    [tex]y^2 - 26y + 25 = 0[/tex]
  2. This is a quadratic equation in [tex]y[/tex]. We can use the quadratic formula [tex]y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex] with [tex]a = 1, b = -26, c = 25[/tex].
  3. Calculate the discriminant:
    [tex](-26)^2 - 4 \times 1 \times 25 = 676 - 100 = 576[/tex]
  4. Solve for [tex]y[/tex]:
    [tex]y = \frac{26 \pm \sqrt{576}}{2} = \frac{26 \pm 24}{2}[/tex]
    • [tex]y_1 = \frac{26 + 24}{2} = 25[/tex]
    • [tex]y_2 = \frac{26 - 24}{2} = 1[/tex]
  5. Recall that [tex]y = x^2[/tex]. So:
    • [tex]x^2 = 25[/tex] gives [tex]x = \pm 5[/tex]
    • [tex]x^2 = 1[/tex] gives [tex]x = \pm 1[/tex]
  6. Therefore, the solutions are [tex]x = 5, -5, 1, -1[/tex].

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

  1. Substitute [tex]y = x^2[/tex], resulting in:
    [tex]y^2 + 8y - 9 = 0[/tex]
  2. Again, apply the quadratic formula:
    [tex]y = \frac{-8 \pm \sqrt{8^2 - 4 \times 1 \times (-9)}}{2} = \frac{-8 \pm \sqrt{64 + 36}}{2}[/tex]
  3. The discriminant is 100, so:
    [tex]y = \frac{-8 \pm 10}{2}[/tex]
    • [tex]y_1 = \frac{-8 + 10}{2} = 1[/tex]
    • [tex]y_2 = \frac{-8 - 10}{2} = -9[/tex]
  4. Since [tex]y = x^2[/tex] and [tex]x^2 = -9[/tex] has no real solutions, we only use [tex]y_1 = 1[/tex]:
    • [tex]x^2 = 1[/tex] gives [tex]x = \pm 1[/tex]
  5. Thus, the real solutions are [tex]x = 1, -1[/tex].

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

  1. Substitute [tex]y = x^2[/tex], resulting in:
    [tex]9y^2 - 32y - 16 = 0[/tex]
  2. Use the quadratic formula:
    [tex]y = \frac{32 \pm \sqrt{(-32)^2 - 4 \times 9 \times (-16)}}{18}[/tex]
  3. Calculate the discriminant:
    [tex]1024 + 576 = 1600[/tex]
  4. Solve for [tex]y[/tex]:
    [tex]y = \frac{32 \pm 40}{18}[/tex]
    • [tex]y_1 = \frac{72}{18} = 4[/tex]
    • [tex]y_2 = \frac{-8}{18} = -\frac{4}{9}[/tex]
  5. Since [tex]y = x^2[/tex] and negative [tex]y[/tex] has no real solutions, we only consider [tex]y_1 = 4[/tex]:
    • [tex]x^2 = 4[/tex] gives [tex]x = \pm 2[/tex]
  6. Therefore, the solutions are [tex]x = 2, -2[/tex].

By breaking down each equation, we simplify solving quartic equations into solving more manageable quadratic equations. Each transformation step is crucial to arrive at the right set of solutions.

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