Equations of Quadratic Type

35 min
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Equations of Quadratic Type

An equation of quadratic type becomes quadratic after a substitution such as u=x2u=x^2, u=2xu=2^x, or u=x+1/xu=x+1/x. The goal is to expose a familiar quadratic, solve it, and then return to the original variable. State restrictions before substitution, particularly when denominators or even powers are involved.

Worked reasoning

Solve x45x2+4=0x^4-5x^2+4=0. Let u=x2u=x^2: u25u+4=0u^2-5u+4=0, so u=1u=1 or 4. Therefore x=±1x=\pm1 or ±2\pm2.

Exam method

  1. Spot a repeated expression. 2. Substitute one temporary variable. 3. Solve the quadratic and reverse the substitution, checking all roots.

Substitution hygiene

Write the substitution on its own line, then write the reverse substitution before solving. This small habit stops answers being left in the temporary variable. When the repeated expression has restrictions, write them first: for example, u=2xu=2^x must be positive, so a negative u root can be rejected immediately. The resulting quadratic may look ordinary, but the original variable’s domain controls which algebraic answers survive.

One-minute retrieval: Equations of Quadratic Type

Close the worked solution. From memory, state its central rule, name one condition that makes it valid, and reconstruct one check that would catch a typical exam error.

Solve x45x2+4=0x^4-5x^2+4=0. Give the largest root.

Solve x2=49x^2=49. Give the negative root.

What is a useful substitution for 32x5(3x)+6=03^{2x}-5(3^x)+6=0?

After solving u=x2u=x^2 and getting u=9u=9, write both x-values.