Solving disguised quadratics

In some cases, we might see a polynomial that has a higher degree than 2 (the degree of a quadratic), but it may still be possible to solve it by treating it like a quadratic equation. These are called disguised quadratics.

What is a disguised quadratic?

A disguised quadratic is a polynomial where we can substitute a variable to transform it into a quadratic form. This usually involves us spotting a pattern in the powers in the expression.

Disguised quadratics where powers are multiples

A common type of disguised quadratic is one where the powers are consecutive multiples of 2. For example, one in the form ax^4+bx^2+c=0. In this case, we can substitute y=x^2, which transforms the equation into ay^2+by+c=0, which is a normal quadratic in terms of y.

Example: Solve the equation x^4 - 5x^2 + 6 = 0

• Substitute y = x^2:
  • y^2 - 5y + 6 = 0
• Factorise the quadratic:
  • (y - 2)(y - 3) = 0
• Find the values of y:
  • y = 2 or y = 3
• Substitute back to find x:
  • If y = 2, then x^2 = 2:
    • x = \pm \sqrt{2}
  • If y = 3, then x^2 = 3:
    • x = \pm \sqrt{3}

Answer: x = \pm \sqrt{2} or x = \pm \sqrt{3}

Disguised quadratics where terms are exponents

Solve the equation 2^{2x} - 5(2^x) + 6 = 0

• Substitute y = 2^x:
  • y^2 - 5y + 6 = 0
• Factorise the quadratic:
  • (y - 2)(y - 3) = 0
• Find the values of y:
  • y = 2 or y = 3
• Substitute back to find x:
  • If y = 2, then 2^x = 2:
    • \log_2{2^x} = \log_2{2}
    • x = 1
  • If y = 3, then 2^x = 3:
    • \log_2{2^x} = \log_2{3}
    • x = \log_2{3} Answer: x = 1 or x = \log_2{3}

Solve the equation 3^{2x} + 5(3^x) - 24 = 0

• Substitute y = 3^x:
  • y^2 + 5y - 24 = 0
• Factorise the quadratic:
  • (y + 8)(y - 3) = 0
• Find the values of y:
  • y = -8 or y = 3
• Substitute back to find x:
  • If y = -8, then 3^x = -8:
    • No solution, since 3^x is always positive.
  • If y = 3, then 3^x = 3:
    • \log_3{3^x} = \log_3{3}
    • x = 1 Answer: x = 1

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