Solve x ^ 4 = (4x - 5) ^ 2 . | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$x ^ { 4 } = ( 4 x - 5 ) ^ { 2 }$$

Solve for x (complex solution)

$x=-5$
$x=1$
$x=2-i$
$x=2+i$

Solution Steps

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(4x-5\right)^{2}$.

$$x^{4}=16x^{2}-40x+25$$

Subtract $16x^{2}$ from both sides.

$$x^{4}-16x^{2}=-40x+25$$

Add $40x$ to both sides.

$$x^{4}-16x^{2}+40x=25$$

Subtract $25$ from both sides.

$$x^{4}-16x^{2}+40x-25=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $-25$ and $q$ divides the leading coefficient $1$. List all candidates $\frac{p}{q}$.

$$±25,±5,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=1$$

By Factor theorem, $x-k$ is a factor of the polynomial for each root $k$. Divide $x^{4}-16x^{2}+40x-25$ by $x-1$ to get $x^{3}+x^{2}-15x+25$. Solve the equation where the result equals to $0$.

$$x^{3}+x^{2}-15x+25=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $25$ and $q$ divides the leading coefficient $1$. List all candidates $\frac{p}{q}$.

$$±25,±5,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=-5$$

By Factor theorem, $x-k$ is a factor of the polynomial for each root $k$. Divide $x^{3}+x^{2}-15x+25$ by $x+5$ to get $x^{2}-4x+5$. Solve the equation where the result equals to $0$.

$$x^{2}-4x+5=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. Substitute $1$ for $a$, $-4$ for $b$, and $5$ for $c$ in the quadratic formula.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\times 5}}{2}$$

Do the calculations.

$$x=\frac{4±\sqrt{-4}}{2}$$

Solve the equation $x^{2}-4x+5=0$ when $±$ is plus and when $±$ is minus.

$$x=2-i$$ $$x=2+i$$

List all found solutions.

$$x=1$$ $$x=-5$$ $$x=2-i$$ $$x=2+i$$

Solve for x

$x=-5$
$x=1$

Solution Steps

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(4x-5\right)^{2}$.

$$x^{4}=16x^{2}-40x+25$$

Subtract $16x^{2}$ from both sides.

$$x^{4}-16x^{2}=-40x+25$$

Add $40x$ to both sides.

$$x^{4}-16x^{2}+40x=25$$

Subtract $25$ from both sides.

$$x^{4}-16x^{2}+40x-25=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $-25$ and $q$ divides the leading coefficient $1$. List all candidates $\frac{p}{q}$.

$$±25,±5,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=1$$

By Factor theorem, $x-k$ is a factor of the polynomial for each root $k$. Divide $x^{4}-16x^{2}+40x-25$ by $x-1$ to get $x^{3}+x^{2}-15x+25$. Solve the equation where the result equals to $0$.

$$x^{3}+x^{2}-15x+25=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $25$ and $q$ divides the leading coefficient $1$. List all candidates $\frac{p}{q}$.

$$±25,±5,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=-5$$

By Factor theorem, $x-k$ is a factor of the polynomial for each root $k$. Divide $x^{3}+x^{2}-15x+25$ by $x+5$ to get $x^{2}-4x+5$. Solve the equation where the result equals to $0$.

$$x^{2}-4x+5=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. Substitute $1$ for $a$, $-4$ for $b$, and $5$ for $c$ in the quadratic formula.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\times 5}}{2}$$

Do the calculations.

$$x=\frac{4±\sqrt{-4}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=1$$ $$x=-5$$