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Nonlinear Equations in One Variable and Systems of Equations

SAT Math· difficulty 4/5

Find all real solutions of x4−13x2+36=0x^4 - 13x^2 + 36 = 0.

  • A

    x=±2,±3x = \pm 2, \pm 3

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  • B

    x=2,3x = 2, 3

  • C

    x=±2,±9x = \pm 2, \pm 9

  • D

    x=4,9x = 4, 9

Explanation

Let u=x2u = x^2: u2−13u+36=(u−4)(u−9)=0u^2 - 13u + 36 = (u - 4)(u - 9) = 0, so x2=4x^2 = 4 or 99, giving x=±2,±3x = \pm 2, \pm 3.

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