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Zorluk: OrtaQuadratic Equations and Factoring

If xx is a real number that satisfies the equation x4−13x2+36=0x^4 - 13x^2 + 36 = 0, which of the following values could be the value of xx? Select all such values.

  1. −3-3Cevap
  2. −2-2Cevap
  3. 33Cevap
  4. D
    44
  5. E
    99

Cevap

The correct values are −3-3, −2-2, and 33.
Factoring the equation yields (x2−4)(x2−9)=(x−2)(x+2)(x−3)(x+3)=0(x^2 - 4)(x^2 - 9) = (x - 2)(x + 2)(x - 3)(x + 3) = 0, giving the four distinct real solutions x=−3,−2,2,3x = -3, -2, 2, 3. The options corresponding to −3-3, −2-2, and 33 represent valid values of xx.

Adım Adım Çözüm

1
Substitute a variable u=x2u = x^2 to express the equation in quadratic form.
u2−13u+36=0u^2 - 13u + 36 = 0
This reduces the fourth-degree polynomial into a standard quadratic equation.
2
Factor the quadratic equation to solve for uu.
(u−4)(u−9)=0  ⟹  u=4 or u=9(u - 4)(u - 9) = 0 \implies u = 4 \text{ or } u = 9
Factoring determines the values of x2x^2.
3
Substitute back x2=ux^2 = u and solve for xx by taking both positive and negative square roots.
x2=4  ⟹  x=±2x^2 = 4 \implies x = \pm 2, and x2=9  ⟹  x=±3x^2 = 9 \implies x = \pm 3
Each positive value of uu yields two real solutions for xx.

Anahtar Kavram

Solving quadratic-form equations by factoring and taking positive and negative square roots.
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