VIDEO: Back substitution correctly explained using the example

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Solving biquadratic equations - here's how to proceed

Biquadratic Equations are equations in which the unknown x is to the power of four (x4) and as a square (x2) occurs. Such equations have the general form: ax4 + bx2 + c = 0. The shape is similar to a quadratic equation, only higher Potencies to do.

  1. Such equations can easily be reduced to a quadratic equation by making a substitution: x³ = z, a new unknown which is first calculated.
  2. The result is a quadratic equation of the form az2 + bz + c = 0, which can easily be solved with the abc formula or (after dividing by the coefficient a) with the more familiar pq formula.

Biquadratic equation - a calculated example

As an example, consider the biquadratic equation 16 x4 - 136 x2 + 225 = 0 can be completely calculated.

  1. You substitute, i.e. replace, x² = z and get the quadratic equation:
    2. Resubstitution - Instructions

    If you encounter complicated equations in math, you can solve them by ...

  2. 16 z2 - 136 z + 225 = 0
  3. This equation is to be solved with the pq formula. So you first divide the entire equation by 16 to get the form necessary for this formula:
  4. z2 - 8.5 z + 14.0625 = 0 (If you are using a calculator, you can use Decimal numbers calculate).
  5. The pq formula now provides the two solutions z1 = 6.25 and e.g.2 = 2,25

Back substitution - this is how you calculate "x" in the example

The example is of course not finished yet, because you are supposed to calculate the unknown "x". So far, however, you have only found two solutions for the unknown "z".

  1. The so-called back substitution is due, in which you come back to the unknown "x".
  2. You had set x² = z, you now have to undo this in a certain sense.
  3. In your example, x² = 6.25 and x² = 2.25 apply. In the case of back substitution, you use the solutions you found for z.
  4. These two equations for x are easily solved by taking the root and you get four solutions, namely x1 = 2.5, x2 = -2.5 as well as x3 = 1.5 and x4 = -1,5.

Fourth-degree equations can have a maximum of 4 solutions. In the present example, the biquadratic equation actually has this maximum number of solutions. However, it can also happen that you can only calculate 2 solutions, for example if one of the two solutions for z is negative. If both solutions to z are negative, the biquadratic equation has no solution at all. According to the procedure of substitution and back-substitution, all equations with only (!) Even exponents or also solve equations that only have exponents of the form x6 and x3 Etc. contain x here3 = set z, then take the third root for the back substitution).

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