We are asked to solve the given equation,
which is considered to be in quadratic form.
But notice how this is not a quadratic equation
because it's degree four, not degree two.
The key to understanding why this is in quadratic form
is to focus on the middle term.
Notice how the middle term has a factor of x squared.
And if we square x squared or square x to the second
we do get the first term of x to the fourth.
Which means we could rewrite the equation as the square
of x squared minus 20 times x squared
plus 64 equals zero.
At this point we will introduce a new variable
so that we can write the equation as a quadratic equation.
We will now let the variable u equal x squared
and then perform a substitution.
So if we let u equal x squared, notice how we can now write
the first term as just u squared minus this would just
be 20u and then we have plus 64 equals zero.
So by performing this substitution we were able to write the
given equation as a quadratic equation which is why the
original equation is considered to be in quadratic form.
And now we will factor this in terms of u.
The left side will factor into two binomial factors.
The first term is u squared which is equal to u times u.
So we have a u in the first position of both binomials.
The second terms of the binomial factors will be the factors
of positive 64 that add to negative 20 because the sum
is negative we must use negative factors of positive 64.
Notice that negative 16 times negative four
is equal to positive 64.
And negative 16 plus negative four is equal to negative 20.
The two factors we need are negative 16 and negative four,
which means one binomial factor
is u minus 16 and the other is u minus four.
Now that we have it factored in terms of u,
we perform another substitution
and substitute x squared for u.
If we substitute x squared for u the first factor becomes
x squared minus 16 and the second factor becomes
x squared minus four.
This is the factored form of the original equation in terms
of x and now because this product is equal to zero,
either the first factor of x squared minus 16
must equal zero or the second factor
of x squared minus four must equal zero.
And now we solve these two equations for x.
To solve here we could factor again,
but let's isolate the x squared
and square root both sides of the equation.
If we had 16 to both sides we have x squared equals 16,
and now we square root both sides of the equation,
but to make sure we get both solutions
we include a plus or minus on the right.
The square root of x squared is equal to x.
So we have x equals plus or minus
the square root of 16 is equal to four.
Here we have x equals plus or minus four.
For this equation we add four to both sides giving us
x squared equals four, and now we square root both sides
of the equation with a plus or minus on the right.
Simplifying we have x equals plus or minus
the square root of four is two.
We have four possible solutions to the original equation;
x equals positive four, negative four,
positive two, or negative two.
To verify all four values are solutions we should substitute
these values back into the original equation
to make sure they satisfy the equation.
And to save some time I've already done this.
Here's the check for x equals four.
Notice how we get zero equals zero, which is true.
Here's the check for x equals negative four, which is true.
Here's the check for x equals two, which is true.
And here's the check for x equals
negative two, which is also true.
And therefore this equation has four solutions.
Again, x equals plus or minus four,
or x equals plus or minus two.
I hope you found this helpful.