Solve 2x ^ 3 - 3x ^ 2 + 7x = 3 | 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

$$2x^{3}-3x^{2}+7x=3$$

Solve for x (complex solution)

$x=\frac{-\sqrt{11}i+1}{2}\approx 0.5-1.658312395i$
$x=\frac{1}{2}=0.5$
$x=\frac{1+\sqrt{11}i}{2}\approx 0.5+1.658312395i$

Solution Steps

Subtract $3$ from both sides.

$$2x^{3}-3x^{2}+7x-3=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $-3$ and $q$ divides the leading coefficient $2$. List all candidates $\frac{p}{q}$.

$$±\frac{3}{2},±3,±\frac{1}{2},±1$$

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.

$$x=\frac{1}{2}$$

By Factor theorem, $x-k$ is a factor of the polynomial for each root $k$. Divide $2x^{3}-3x^{2}+7x-3$ by $2\left(x-\frac{1}{2}\right)=2x-1$ to get $x^{2}-x+3$. Solve the equation where the result equals to $0$.

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

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 $a$, $-1$ for $b$, and $3$ for $c$ in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\times 3}}{2}$$

Do the calculations.

$$x=\frac{1±\sqrt{-11}}{2}$$

Solve the equation $x^{2}-x+3=0$ when $±$ is plus and when $±$ is minus.

$$x=\frac{-\sqrt{11}i+1}{2}$$ $$x=\frac{1+\sqrt{11}i}{2}$$

List all found solutions.

$$x=\frac{1}{2}$$ $$x=\frac{-\sqrt{11}i+1}{2}$$ $$x=\frac{1+\sqrt{11}i}{2}$$

Solve for x

$x=\frac{1}{2}=0.5$

Solution Steps

Subtract $3$ from both sides.

$$2x^{3}-3x^{2}+7x-3=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $-3$ and $q$ divides the leading coefficient $2$. List all candidates $\frac{p}{q}$.

$$±\frac{3}{2},±3,±\frac{1}{2},±1$$

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.

$$x=\frac{1}{2}$$

By Factor theorem, $x-k$ is a factor of the polynomial for each root $k$. Divide $2x^{3}-3x^{2}+7x-3$ by $2\left(x-\frac{1}{2}\right)=2x-1$ to get $x^{2}-x+3$. Solve the equation where the result equals to $0$.

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

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 $a$, $-1$ for $b$, and $3$ for $c$ in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\times 3}}{2}$$

Do the calculations.

$$x=\frac{1±\sqrt{-11}}{2}$$

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.

$$x=\frac{1}{2}$$