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Concept Version 11
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Other Equations in Quadratic Form

Many equations with no odd-degree terms can be reduced to quadratics and solved with the same methods as quadratics.

Learning Objective

  • Use the quadratic formula to solve any equation in quadratic form


Key Points

    • A biquadratic equation (quartic equation with no terms of odd-degree) has the form 0=ax4+bx2+c0=ax^4+bx^2+c0=ax​4​​+bx​2​​+c. It can be expressed as: 0=ap2+bp+c0=ap^2+bp+c0=ap​2​​+bp+c (where p=x2p=x^2p=x​2​​).
    • The values of ppp can be found by graphing, factoring, completing the square, or using the quadratic formula. Their square roots (positive and negative) are the values of xxx that satisfy the original equation.
    • Higher-order equations can be solved by a similar process that involves reducing their exponents. The requirement is that there are two terms of xxx such that the ratio of the highest exponent of xxx to the lower is 2:12:12:1.

Terms

  • zero

    Also known as a root; an xxx value at which the function of xxx is equal to zero.

  • biquadratic

    When a polynomial involves only the second and fourth powers of a variable.

  • quartic function

    Any polynomial function whose greatest exponent is of power four.


Full Text

Higher degree polynomial equations can be very difficult to solve. In some special situations, however, they can be made more manageable by reducing their exponents via substitution.  If a substitution can be made such that the higher order polynomial takes the form of a quadratic, any method for solving a quadratic equation can be applied.   

For example, if a quartic equation is biquadratic—that is, it includes no terms of an odd-degree— there is a quick way to find the zeroes of the quartic function by reducing it into a quadratic form.  Consider a quadratic function with no odd-degree terms which has the form:

0=ax4+bx2+c0=ax^4+bx^2+c0=ax​4​​+bx​2​​+c

If we let an arbitrary variable ppp equal x2x^2x​2​​, this can be reduced to an equation of a lower degree:

0=ap2+bp+c0=ap^2+bp+c0=ap​2​​+bp+c

With substitution, we were able to reduce a higher order polynomial into a quadratic equation.  It can now be solved with any of a number of methods (via graphing, factoring, completing the square, or by using the quadratic formula).

Once values of ppp are found, each positive value of the temporary variable ppp can be used to find two values of xxx such that:

x=px=\sqrt px=√​p​​​

As with every square root, the root of ppp will have two values, one positive and one negative.  It is important to realize that the same kind of substitution can be done for any equation in quadratic form, not just quartics. 

Example

As an example, consider the equation:

0=x4−12x2+200=x^4-12x^2+200=x​4​​−12x​2​​+20

Quartic graph

Graph of the function f(x)=x4−12x2+20f(x) = x^4-12x^2+20f(x)=x​4​​−12x​2​​+20.

We can substitute the arbitrary variable ppp in place of x2x^2x​2​​:

0=p2−12p+200=p^2-12p+200=p​2​​−12p+20

This equation is now solvable for ppp using the quadratic formula:

p=12±(−12)2−4⋅1⋅202⋅1p=\dfrac {12 \pm \sqrt {(-12)^2-4\cdot 1\cdot 20}}{2\cdot 1}p=​2⋅1​​12±√​(−12)​2​​−4⋅1⋅20​​​​​

Simplifying this, we find ppp equals 2 or 10.

Knowing that p=x2p=x^2p=x​2​​, we can use each value of ppp to solve for two values of xxx:

x=±2x=\pm \sqrt 2x=±√​2​​​ and x=±10x= \pm \sqrt 10x=±√​1​​​0

A similar procedure can be used to solve higher-order equations. The requirement is that there are two terms of xxx such that the ratio of the highest exponent of xxx to the lower is 2:12:12:1.

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