Find the Roots (Zeros) (3x-1)(x+3)=x(1+6x)

$(3 x - 1) (x + 3) = x (1 + 6 x)$

Step 1

Simplify $(3 x - 1) (x + 3)$ .

Start simplifying.

$0 + 0 + (3 x - 1) (x + 3) = x (1 + 6 x)$

Simplify by adding zeros.

$(3 x - 1) (x + 3) = x (1 + 6 x)$

Expand $(3 x - 1) (x + 3)$ using the FOIL Method.

Apply the distributive property.

$3 x (x + 3) - 1 (x + 3) = x (1 + 6 x)$

Apply the distributive property.

$3 x \cdot x + 3 x \cdot 3 - 1 (x + 3) = x (1 + 6 x)$

Apply the distributive property.

$3 x \cdot x + 3 x \cdot 3 - 1 x - 1 \cdot 3 = x (1 + 6 x)$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$3 (x \cdot x) + 3 x \cdot 3 - 1 x - 1 \cdot 3 = x (1 + 6 x)$

Multiply $x$ by $x$ .

$3 x^{2} + 3 x \cdot 3 - 1 x - 1 \cdot 3 = x (1 + 6 x)$

Multiply $3$ by $3$ .

$3 x^{2} + 9 x - 1 x - 1 \cdot 3 = x (1 + 6 x)$

Rewrite $- 1 x$ as $- x$ .

$3 x^{2} + 9 x - x - 1 \cdot 3 = x (1 + 6 x)$

Multiply $- 1$ by $3$ .

$3 x^{2} + 9 x - x - 3 = x (1 + 6 x)$

Subtract $x$ from $9 x$ .

$3 x^{2} + 8 x - 3 = x (1 + 6 x)$

Step 2

Simplify $x (1 + 6 x)$ .

Apply the distributive property.

$3 x^{2} + 8 x - 3 = x \cdot 1 + x (6 x)$

Simplify the expression.

Multiply $x$ by $1$ .

$3 x^{2} + 8 x - 3 = x + x (6 x)$

Rewrite using the commutative property of multiplication.

$3 x^{2} + 8 x - 3 = x + 6 x \cdot x$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$3 x^{2} + 8 x - 3 = x + 6 (x \cdot x)$

Multiply $x$ by $x$ .

$3 x^{2} + 8 x - 3 = x + 6 x^{2}$

Step 3

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

Subtract $x$ from both sides of the equation.

$3 x^{2} + 8 x - 3 - x = 6 x^{2}$

Subtract $6 x^{2}$ from both sides of the equation.

$3 x^{2} + 8 x - 3 - x - 6 x^{2} = 0$

Subtract $6 x^{2}$ from $3 x^{2}$ .

$- 3 x^{2} + 8 x - 3 - x = 0$

Subtract $x$ from $8 x$ .

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

Step 4

Use the quadratic formula to find the solutions.

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

Step 5

Substitute the values $a = - 3$ , $b = 7$ , and $c = - 3$ into the quadratic formula and solve for $x$ .

$\frac{- 7 \pm \sqrt{7^{2} - 4 \cdot (- 3 \cdot - 3)}}{2 \cdot - 3}$

Step 6

Simplify.

Simplify the numerator.

Raise $7$ to the power of $2$ .

$x = \frac{- 7 \pm \sqrt{49 - 4 \cdot - 3 \cdot - 3}}{2 \cdot - 3}$

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

Multiply $- 4$ by $- 3$ .

$x = \frac{- 7 \pm \sqrt{49 + 12 \cdot - 3}}{2 \cdot - 3}$

Multiply $12$ by $- 3$ .

$x = \frac{- 7 \pm \sqrt{49 - 36}}{2 \cdot - 3}$

Subtract $36$ from $49$ .

$x = \frac{- 7 \pm \sqrt{13}}{2 \cdot - 3}$

Multiply $2$ by $- 3$ .

$x = \frac{- 7 \pm \sqrt{13}}{- 6}$

Simplify $\frac{- 7 \pm \sqrt{13}}{- 6}$ .

$x = \frac{7 \pm \sqrt{13}}{6}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = \frac{7 \pm \sqrt{13}}{6}$

Decimal Form:

$x = 1.76759187 \dots, 0.56574145 \dots$

Step 8

Find the Roots (Zeros) (3x-1)(x+3)=x(1+6x)

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