Solve the Rational Equation for x (x^2)/(x-3)=(x+2)/(2x-5)

$\frac{x^{2}}{x - 3} = \frac{x + 2}{2 x - 5}$

Step 1

Multiply the numerator of the first fraction by the denominator of the second fraction. Set this equal to the product of the denominator of the first fraction and the numerator of the second fraction.

$x^{2} (2 x - 5) = (x - 3) (x + 2)$

Step 2

Solve the equation for $x$ .

Simplify $x^{2} (2 x - 5)$ .

Start simplifying.

$0 + 0 + x^{2} (2 x - 5) = (x - 3) (x + 2)$

Simplify by multiplying through.

Apply the distributive property.

$x^{2} (2 x) + x^{2} \cdot - 5 = (x - 3) (x + 2)$

Reorder.

Rewrite using the commutative property of multiplication.

$2 x^{2} x + x^{2} \cdot - 5 = (x - 3) (x + 2)$

Move $- 5$ to the left of $x^{2}$ .

$2 x^{2} x - 5 \cdot x^{2} = (x - 3) (x + 2)$

Multiply $x^{2}$ by $x$ by adding the exponents.

Move $x$ .

$2 (x \cdot x^{2}) - 5 \cdot x^{2} = (x - 3) (x + 2)$

Multiply $x$ by $x^{2}$ .

Raise $x$ to the power of $1$ .

$2 (x^{1} x^{2}) - 5 \cdot x^{2} = (x - 3) (x + 2)$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$2 x^{1 + 2} - 5 \cdot x^{2} = (x - 3) (x + 2)$

Add $1$ and $2$ .

$2 x^{3} - 5 \cdot x^{2} = (x - 3) (x + 2)$

$2 x^{3} - 5 x^{2} = (x - 3) (x + 2)$

Simplify $(x - 3) (x + 2)$ .

Expand $(x - 3) (x + 2)$ using the FOIL Method.

Apply the distributive property.

$2 x^{3} - 5 x^{2} = x (x + 2) - 3 (x + 2)$

Apply the distributive property.

$2 x^{3} - 5 x^{2} = x \cdot x + x \cdot 2 - 3 (x + 2)$

Apply the distributive property.

$2 x^{3} - 5 x^{2} = x \cdot x + x \cdot 2 - 3 x - 3 \cdot 2$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$2 x^{3} - 5 x^{2} = x^{2} + x \cdot 2 - 3 x - 3 \cdot 2$

Move $2$ to the left of $x$ .

$2 x^{3} - 5 x^{2} = x^{2} + 2 \cdot x - 3 x - 3 \cdot 2$

Multiply $- 3$ by $2$ .

$2 x^{3} - 5 x^{2} = x^{2} + 2 x - 3 x - 6$

Subtract $3 x$ from $2 x$ .

$2 x^{3} - 5 x^{2} = x^{2} - x - 6$

Move all terms containing $x$ to the left side of the equation.

Subtract $x^{2}$ from both sides of the equation.

$2 x^{3} - 5 x^{2} - x^{2} = - x - 6$

Add $x$ to both sides of the equation.

$2 x^{3} - 5 x^{2} - x^{2} + x = - 6$

Subtract $x^{2}$ from $- 5 x^{2}$ .

$2 x^{3} - 6 x^{2} + x = - 6$

Add $6$ to both sides of the equation.

$2 x^{3} - 6 x^{2} + x + 6 = 0$

Factor $2 x^{3} - 6 x^{2} + x + 6$ using the rational roots test.

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 6, \pm 2, \pm 3 q = \pm 1, \pm 2$

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 0.5, \pm 6, \pm 3, \pm 2, \pm 1.5$

Substitute $2$ and simplify the expression. In this case, the expression is equal to $0$ so $2$ is a root of the polynomial.

Substitute $2$ into the polynomial.

$2 \cdot 2^{3} - 6 \cdot 2^{2} + 2 + 6$

Raise $2$ to the power of $3$ .

$2 \cdot 8 - 6 \cdot 2^{2} + 2 + 6$

Multiply $2$ by $8$ .

$16 - 6 \cdot 2^{2} + 2 + 6$

Raise $2$ to the power of $2$ .

$16 - 6 \cdot 4 + 2 + 6$

Multiply $- 6$ by $4$ .

$16 - 24 + 2 + 6$

Subtract $24$ from $16$ .

$- 8 + 2 + 6$

Add $- 8$ and $2$ .

$- 6 + 6$

Add $- 6$ and $6$ .

$0$

Since $2$ is a known root, divide the polynomial by $x - 2$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{2 x^{3} - 6 x^{2} + x + 6}{x - 2}$

Divide $2 x^{3} - 6 x^{2} + x + 6$ by $x - 2$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$2$

$2 x^{3}$

$6 x^{2}$

$x$

$6$

Divide the highest order term in the dividend $2 x^{3}$ by the highest order term in divisor $x$ .

					$2 x^{2}$
	$x$	-	$2$		$2 x^{3}$	-	$6 x^{2}$	+	$x$	+	$6$

Multiply the new quotient term by the divisor.

				$2 x^{2}$
$x$	-	$2$		$2 x^{3}$	-	$6 x^{2}$	+	$x$	+	$6$
			+	$2 x^{3}$	-	$4 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $2 x^{3} - 4 x^{2}$

				$2 x^{2}$
$x$	-	$2$		$2 x^{3}$	-	$6 x^{2}$	+	$x$	+	$6$
			-	$2 x^{3}$	+	$4 x^{2}$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$2 x^{2}$
$x$	-	$2$		$2 x^{3}$	-	$6 x^{2}$	+	$x$	+	$6$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$2 x^{2}$

Pull the next terms from the original dividend down into the current dividend.

				$2 x^{2}$
$x$	-	$2$		$2 x^{3}$	-	$6 x^{2}$	+	$x$	+	$6$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$2 x^{2}$	+	$x$

Divide the highest order term in the dividend $- 2 x^{2}$ by the highest order term in divisor $x$ .

				$2 x^{2}$	-	$2 x$
$x$	-	$2$		$2 x^{3}$	-	$6 x^{2}$	+	$x$	+	$6$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$2 x^{2}$	+	$x$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	-	$2 x$
$x$	-	$2$		$2 x^{3}$	-	$6 x^{2}$	+	$x$	+	$6$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$2 x^{2}$	+	$x$
					-	$2 x^{2}$	+	$4 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 2 x^{2} + 4 x$

				$2 x^{2}$	-	$2 x$
$x$	-	$2$		$2 x^{3}$	-	$6 x^{2}$	+	$x$	+	$6$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$2 x^{2}$	+	$x$
					+	$2 x^{2}$	-	$4 x$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$2 x^{2}$	-	$2 x$
$x$	-	$2$		$2 x^{3}$	-	$6 x^{2}$	+	$x$	+	$6$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$2 x^{2}$	+	$x$
					+	$2 x^{2}$	-	$4 x$
							-	$3 x$

Pull the next terms from the original dividend down into the current dividend.

				$2 x^{2}$	-	$2 x$
$x$	-	$2$		$2 x^{3}$	-	$6 x^{2}$	+	$x$	+	$6$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$2 x^{2}$	+	$x$
					+	$2 x^{2}$	-	$4 x$
							-	$3 x$	+	$6$

Divide the highest order term in the dividend $- 3 x$ by the highest order term in divisor $x$ .

				$2 x^{2}$	-	$2 x$	-	$3$
$x$	-	$2$		$2 x^{3}$	-	$6 x^{2}$	+	$x$	+	$6$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$2 x^{2}$	+	$x$
					+	$2 x^{2}$	-	$4 x$
							-	$3 x$	+	$6$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	-	$2 x$	-	$3$
$x$	-	$2$		$2 x^{3}$	-	$6 x^{2}$	+	$x$	+	$6$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$2 x^{2}$	+	$x$
					+	$2 x^{2}$	-	$4 x$
							-	$3 x$	+	$6$
							-	$3 x$	+	$6$

The expression needs to be subtracted from the dividend, so change all the signs in $- 3 x + 6$

				$2 x^{2}$	-	$2 x$	-	$3$
$x$	-	$2$		$2 x^{3}$	-	$6 x^{2}$	+	$x$	+	$6$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$2 x^{2}$	+	$x$
					+	$2 x^{2}$	-	$4 x$
							-	$3 x$	+	$6$
							+	$3 x$	-	$6$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$2 x^{2}$	-	$2 x$	-	$3$
$x$	-	$2$		$2 x^{3}$	-	$6 x^{2}$	+	$x$	+	$6$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$2 x^{2}$	+	$x$
					+	$2 x^{2}$	-	$4 x$
							-	$3 x$	+	$6$
							+	$3 x$	-	$6$
										$0$

Since the remander is $0$ , the final answer is the quotient.

$2 x^{2} - 2 x - 3$

Write $2 x^{3} - 6 x^{2} + x + 6$ as a set of factors.

$(x - 2) (2 x^{2} - 2 x - 3) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 2 = 0$

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

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Set $2 x^{2} - 2 x - 3$ equal to $0$ and solve for $x$ .

Set $2 x^{2} - 2 x - 3$ equal to $0$ .

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

Solve $2 x^{2} - 2 x - 3 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 2$ , $b = - 2$ , and $c = - 3$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (2 \cdot - 3)}}{2 \cdot 2}$

Simplify.

Simplify the numerator.

Raise $- 2$ to the power of $2$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 2 \cdot - 3}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 3$ .

Multiply $- 4$ by $2$ .

$x = \frac{2 \pm \sqrt{4 - 8 \cdot - 3}}{2 \cdot 2}$

Multiply $- 8$ by $- 3$ .

$x = \frac{2 \pm \sqrt{4 + 24}}{2 \cdot 2}$

Add $4$ and $24$ .

$x = \frac{2 \pm \sqrt{28}}{2 \cdot 2}$

Rewrite $28$ as $2^{2} \cdot 7$ .

Factor $4$ out of $28$ .

$x = \frac{2 \pm \sqrt{4 (7)}}{2 \cdot 2}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{2 \pm \sqrt{2^{2} \cdot 7}}{2 \cdot 2}$

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{7}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{2 \pm 2 \sqrt{7}}{4}$

Simplify $\frac{2 \pm 2 \sqrt{7}}{4}$ .

$x = \frac{1 \pm \sqrt{7}}{2}$

The final answer is the combination of both solutions.

$x = \frac{1 + \sqrt{7}}{2}, \frac{1 - \sqrt{7}}{2}$

The final solution is all the values that make $(x - 2) (2 x^{2} - 2 x - 3) = 0$ true.

$x = 2, \frac{1 + \sqrt{7}}{2}, \frac{1 - \sqrt{7}}{2}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = 2, \frac{1 + \sqrt{7}}{2}, \frac{1 - \sqrt{7}}{2}$

Decimal Form:

$x = 2, 1.82287565 \dots, - 0.82287565 \dots$

Solve the Rational Equation for x (x^2)/(x-3)=(x+2)/(2x-5)

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