Q. Solve the equation /3x^2 + 4x + 2 = -x/.

Answer

Interpret the equation as \(3x^2+4x+2=-x\). Move all terms to one side:
\[
3x^2+4x+2+x=0 \implies 3x^2+5x+2=0.
\]
Compute the discriminant: \(\Delta=5^2-4\cdot3\cdot2=25-24=1\). Thus the roots are
\[
x=\frac{-5\pm\sqrt{1}}{2\cdot3}=\frac{-5\pm1}{6}.
\]
So
\[
x=-\tfrac{2}{3}\quad\text{or}\quad x=-1.
\]

Detailed Explanation

Solving the equation step by step

We are given the equation \( 3x^2 + 4x + 2 = -x \).

Move all terms to one side to obtain a standard quadratic form.
Add \( x \) to both sides (so that the right-hand side becomes zero). This gives:
\[ 3x^2 + 4x + 2 + x = 0 \]
Combine like terms (the x-terms):
\[ 3x^2 + 5x + 2 = 0 \]
Factor the quadratic.
We look for two numbers that multiply to the product of the leading coefficient and the constant term, \( 3 \times 2 = 6 \), and add to the middle coefficient \( 5 \). Those numbers are \( 2 \) and \( 3 \) because \( 2 \times 3 = 6 \) and \( 2 + 3 = 5 \).

Rewrite the middle term \( 5x \) as \( 2x + 3x \) to assist factoring by grouping:
\[ 3x^2 + 5x + 2 = 3x^2 + 2x + 3x + 2 \]
Group terms:
\[ (3x^2 + 2x) + (3x + 2) \]
Factor each group:
\[ x(3x + 2) + 1(3x + 2) \]
Factor out the common binomial \( (3x + 2) \):
\[ (3x + 2)(x + 1) \]
So the equation becomes:
\[ (3x + 2)(x + 1) = 0 \]
Use the zero-product property to find the roots.
If a product of factors is zero, at least one factor is zero. Set each factor equal to zero separately:

1) \( 3x + 2 = 0 \) which gives \( 3x = -2 \) and hence \( x = -\frac{2}{3} \).
2) \( x + 1 = 0 \) which gives \( x = -1 \).

Solution: \( x = -1 \) or \( x = -\frac{2}{3} \).

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FAQs

Q What is the first step to solve \(3x^2+4x+2=-x\)?

A Move all terms to one side to form a standard quadratic: \(3x^2+5x+2=0\). Then solve by factoring, quadratic formula, or completing the square.

Q Can this quadratic be factored?

A Yes: \(3x^2+5x+2=(3x+2)(x+1)\). Set each factor to zero to find solutions.

Q What are the solutions?

A From \((3x+2)(x+1)=0\) the roots are \(x=-\tfrac{2}{3}\) and \(x=-1\).

Q What does the discriminant tell us here?

A Discriminant \(\Delta=b^2-4ac=5^2-4\cdot3\cdot2=1\). Positive perfect square means two distinct rational real roots.

Q How would you use the quadratic formula?

A Apply \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) with \(a=3,b=5,c=2\): \(x=\frac{-5\pm1}{6}\), giving \(x=-1\) and \(x=-\tfrac{2}{3}\).

Q What are the parabola’s vertex and direction?

A Since \(a=3>0\), it opens upward. Vertex at \(x=-\tfrac{b}{2a}=-\tfrac{5}{6}\); \(y=f(-\tfrac{5}{6})=-\tfrac{1}{12}\). Vertex: \((-\tfrac{5}{6},-\tfrac{1}{12})\).

Q How can I check my solutions quickly?

A Substitute each root into the original equation \(3x^2+4x+2=-x\). Both \(x=-1\) and \(x=-\tfrac{2}{3}\) satisfy the equation, confirming them.

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