Equations of the form z^4+az^2+a_0 are known as biquadratic equations. They are quartic equations. In general they can be solved by reducing them to a quadratic equation where x=z^2 is the variable. Then you can use the quadratic formula or factor. So plugging in x to the biquadratic giives us x^2+ax+a_0.
A quartic is a polynomial of degree 4, meaning the highest exponent is 4. Biquadratic can mean the same thing, but most mathematicians use the term biquadratic to refer to an equation of degree 4 with no odd powers. So for example we cannot have an x3 term. An example of a biquadratic is: x4 +x2 + 22=0
it is not an equation (there no equality in it!)
variable equation solve it test it
Use a variable to represent the unknown. 'Translate' the words to math symbols and write an equation to solve. Solve the equation. Check.
There is no equation to solve, only an expression.
A biquadratic equation is a polynomial equation of the fourth degree, typically expressed in the form ( ax^4 + bx^2 + c = 0 ), where ( a ), ( b ), and ( c ) are constants and ( a \neq 0 ). This type of equation can also be viewed as a quadratic equation in terms of ( y = x^2 ), facilitating easier solutions. The roots of a biquadratic equation can be found by solving for ( y ) and then taking the square roots of the resulting values.
A quartic is a polynomial of degree 4, meaning the highest exponent is 4. Biquadratic can mean the same thing, but most mathematicians use the term biquadratic to refer to an equation of degree 4 with no odd powers. So for example we cannot have an x3 term. An example of a biquadratic is: x4 +x2 + 22=0
A biquadratic is a polynomial which involves only the second and fourth powers of a variable.
Sure. You can always 'solve for' a variable, and if it happens to be the only variable in the equation, than that's how you solve the equation.
you don't answer an equation, you solve an equation
it a biquadratic bezier patch.
Jean-Marc Deshouillers has written: 'On sums of sixteen biquadrates' -- subject(s): Biquadratic Equations, Equations, Biquadratic
If you solve such an equation for "y", you get an equation in the slope-intercept form.
solve it
you can only solve for one in an equation so it can equal something
How do you use division to solve a multiplication equation?Answer this question…
It is not an equation if it does not have an equals sign. You could simplify it but not solve it.