Question

Solve x4 - 12x2 - 64 = 0.x = +4i and x = 2x = +4 and x = +2x = +41 and x = 21x = +4 and x = 21

Answer 1

We will investigate how to solve " reduceable quadratic " equations.

A reduceable quadratic is a polynomial equation that is characterized by the following form:

`a\cdot x^(2n)\text{ + b}\cdot x^(2n-2)+c\text{ = d}`

Where,

`a\text{ , b , c , d are constants}`

And,

`n\text{ = 2}`

The equation given to us as follows:

`x^4-12x^2\text{ - 64 = 0}`

We will follow a step-wise process in how to solve a reduceable quadratic once recognized.

Step 1: Reduce into a quadratic equation

In this step we will make an appropriate substitution using another variable such the given ( quartic ) equation is transformed into a ( quadractic ) equation as follows:

Substitution:-

`y=x^2`

Substitute the above variable into the given equation as follows:

`\begin{gathered} (y)^2\text{ -12}\cdot y\text{ - 64 = 0} \\ \textcolor{#FF7968}{y^2}\text{\textcolor{#FF7968}{ - 12y - 64 = 0}} \end{gathered}`

Step 2: Solve the reduced equation using quadratic formula

Now we will use the quadratic formula for finding roots. These roots are for the variable ( y ) NOT ( x ) as follows:

`y\text{ = }\frac{-b\pm\sqrt{b^2\text{ - 4ac}}}{2a}`

Where,

`\begin{gathered} ay^2\text{ + by + c = 0} \\ \end{gathered}`

From the reduced quadratic we have:

`a\text{ = 1 , b = -12 , c = -64 }`

Plug in the constants into the quadratic formula and solve for roots for ( y ) as follows:

`\begin{gathered} y\text{ = }\frac{12\pm\sqrt{12^2\text{ - 4}\cdot1\cdot(-64)}}{2\cdot1} \\ \\ y\text{ = }\frac{12\pm\sqrt{144+\text{ 256}}}{2} \\ \\ y\text{ = }\frac{12\pm\text{ 20}}{2}\text{ = 6 }\pm\text{ 10} \\ \\ \textcolor{#FF7968}{y}\text{\textcolor{#FF7968}{ = 16 , -4}} \end{gathered}`

We have solved for the two roots for the substitution variable ( y ) using the quadratic formula.

Step 3: Back substitution

We will back substitute the result of above variable ( y ) and solve for ( x ). Use the susbtitution made in the Step 1 as follows:

`\begin{gathered} y=x^2 \\ 16=x^2\to\text{ x = }\pm√(16)\text{ = }\pm4 \\ -4=x^2\to\text{ x = }\pm2i \\ \textcolor{#FF7968}{x}\text{\textcolor{#FF7968}{ = }}\textcolor{#FF7968}{\pm}\text{\textcolor{#FF7968}{ 4 , }}\textcolor{#FF7968}{\pm2i} \end{gathered}`

Hence, the solutions to the given quartic equation is as follows:

`\textcolor{#FF7968}{x}\text{\textcolor{#FF7968}{ = }}\textcolor{#FF7968}{\pm}\text{\textcolor{#FF7968}{ 4}}\text{ AND}\text{\textcolor{#FF7968}{ x = }}\textcolor{#FF7968}{\pm2i}`