Question

Solve for the roots in the equation below. x^4+3x^2-4=0

Answer 1

Answer:

$\large\boxed{x=-1\ or\ x=1}$

Step-by-step explanation:

$x^4+3x^2-4=0\\\\x^{(2)(2)}+3x^2-4=0\qquad\text{use}\ (a^n)^m=a^{nm}\\\\(x^2)^2+3x^2-4=0\qquad\text{substitute}\ x^2=t\geq0x=\\\\t^2+3t-4=0\\\\t^2+4t-t-4=0\\\\t(t+4)-1(t+4)=0\\\\(t+4)(t-1)=0\iff t+4=0\ \vee\ t-1=0\\\\t+4=0\qquad\text{subtract 4 from both sides}\\t=-4<0\\\\t-1=0\qquad\text{add 1 to both sides}\\t=1>0\to x^2=1\\\\x^2=1\to x=\pm\sqrt1\to x=-1\ \vee\ x=1$

Answer 2

The root of $x^{4} -3x^{2}$-4=0 are +1, -1, +2, -2.

What is remainder theorem?

If f(x) is a polynomial in x then the remainder on dividing f(x) by x − a is f(a), such that

p(x) = (x-a)·q(x) + r(x), where q(x)= quotient and r(x)= remainder

The given equation is: $x^{4} -3x^{2}$-4=0

Now find the solution by putting x= 0,-1,+1,-2,+2...

When x=1

$x^{4} -3x^{2}$-4=0, then the equation get satisfied.

Now, $x^{4} -3x^{2}$-4=0 ÷ (x-1)

we get, $x^{3} +x^{2} +4x+4$

Now again find the solution for  $x^{3} +x^{2} +4x+4$ by putting x= 0,-1,+1,-2,+2...

So, take x=-1 will give the solution

Now, ($x^{3} +x^{2} +4x+4$ )÷ (x+1) = $x^{2}$ + 4

we get, $x^{4} -3x^{2}$-4=( $x^{3} +x^{2} +4x+4$) * (x-1)   {Using remainder theorem}

  and, $x^{3} +x^{2} +4x+4$=   ( $x^{2}$+ 4) * (x+1)

So, root are:

• x-1=0

x=1

• x+1=0

x=-1

• $x^{2}$ +4=0

x= +2, -2

Hence, the root of $x^{4} -3x^{2}$-4=0 are +1, -1, +2, -2

Learn more about remainder theorem here:

https://brainly.com/question/11658724

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