QuestionSolve for real solutions of the equation using the quadratic formula: $3 x(x+2)=4$ .

Studdy Solution

STEP 1

Assumptions1. We are asked to find the real solutions of the equation, if any. . The equation is quadratic in nature.
3. We will use the quadratic formula to solve the equation.

STEP 2

First, we need to rewrite the equation in the standard form of a quadratic equation, which is $ax^2 + bx + c =0$ .The given equation is $x(x+2) =4$ . We can distribute the $x$ to get $x^2 +6x =4$ .

STEP 3

Subtract from both sides to set the equation equal to zero.
$3x^2 +6x - =0$

STEP 4

Now we can identify the coefficients $a$ , $b$ , and $c$ in the equation $ax^2 + bx + c =0$ .
In this case, $a =3$ , $b =6$ , and $c = -4$ .

STEP 5

The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$

STEP 6

Substitute the values of $a$ , $b$ , and $c$ into the quadratic formula.
$x = \frac{-6 \pm \sqrt{6^2 -4 \cdot3 \cdot -4}}{2 \cdot3}$

STEP 7

implify the expression under the square root.
$x = \frac{-6 \pm \sqrt{36 +48}}{6}$

STEP 8

Further simplify the expression.
$x = \frac{-6 \pm \sqrt{84}}{6}$

STEP 9

implify the square root. $\sqrt{84}$ simplifies to $2\sqrt{21}$ .
$x = \frac{-6 \pm2\sqrt{21}}{6}$

STEP 10

Factor out a2 from the numerator.
$x = \frac{2(-3 \pm \sqrt{21})}{6}$

STEP 11

implify the expression to get the solutions.
$x = -\frac{}{} \pm \frac{\sqrt{21}}{3}$ So the solutions of the equation are $x = -\frac{}{} + \frac{\sqrt{21}}{3}$ and $x = -\frac{}{} - \frac{\sqrt{21}}{3}$ .

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