Question 1524
<font face = "courier new" size = 2>Hello I am having problems solving this quadratic equation by using the
quadratic formula, could you please help? x<sup>4</sup>-6x<sup>2</sup>+5=0
`<font color = "blue"><b>
Yours is not a quadratic equation, but a fourth-degree equation.  A quadratic
equation must contain no term in a variable higher than 2.  However if a
trinomial in descending order set equal to zero is such that the variable part
of the first term is the square of the variable part of the middle term, then
we can convert the equation into a quadratic equation by making a
substitution.  
`
x<sup>4</sup> - 6x<sup>2</sup> + 5 = 0 is such an equation because the variable part of the first
term is x<sup>4</sup>, which is the square of the variable part of the second equation, x<sup>2</sup>
so we let some other letter, say u, represent the variable part of the second
term.  That is, we let
`
x<sup>2</sup> = u
`
then we square both sides of that and get
`
x<sup>4</sup> = u<sup>2</sup>  
`
So
`
x<sup>4</sup> - 6x<sup>2</sup> + 5 = 0
`
converts to the quadratic equation: 
`
u<sup>2</sup> - 6u + 5 = 0
`
Then we factor the left side as
`
(u - 5)(u - 1) = 0
`
Then we set each factor = 0,
`
u - 5 = 0, which gives u = 5
`
u - 1 = 0, which gives u = 1
`
But the object is not to find u, but to find x, so we
replace u by x<sup>2</sup> in each case: 
`
u = 5 so
`
x<sup>2</sup> = 5
`
Taking square roots of both sides we get
` ` ` _
x = ±<font face = "symbol">Ö</font>5
`
Also
`
u = 1 so 
`
x<sup>2</sup> = 1
`
Taking square roots of both sides we get
`
x = ±1
` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` _` ` _
So there are four answers: x = <font face = "symbol">Ö</font>5, -<font face = "symbol">Ö</font>5, 1, and -1
`
Edwin