Q. \(x^2-3x-4=-3x\)

Answer

Interpret the equation as \(x^2-3x-4=-3x\).

Add \(3x\) to both sides:
\[
x^2-4=0.
\]
Factor and solve:
\[
(x-2)(x+2)=0 \implies x=2 \text{ or } x=-2.
\]

Final answer: \(\boxed{x=2,\ -2}\).

Detailed Explanation

Step-by-step solution

Write the given equation:
\[ x^2 – 3x – 4 = -3x \]
Move all terms to one side by adding \(3x\) to both sides to eliminate the \(-3x\) on the left:
\[ x^2 – 3x – 4 + 3x = -3x + 3x \]
which simplifies to \(x^2 – 4 = 0\).
Solve \(x^2 – 4 = 0\) by recognizing it as a difference of squares:
\[ x^2 – 4 = (x – 2)(x + 2) = 0 \]
Set each factor equal to zero and solve:
\(x – 2 = 0\) gives \(x = 2\).
\(x + 2 = 0\) gives \(x = -2\).
Check both solutions in the original equation:
- For \(x = 2\): left side \(2^2 – 3(2) – 4 = 4 – 6 – 4 = -6\), right side \(-3(2) = -6\) → works.
- For \(x = -2\): left side \((-2)^2 – 3(-2) – 4 = 4 + 6 – 4 = 6\), right side \(-3(-2) = 6\) → works.

Solution: \(x = 2\) or \(x = -2\).

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FAQs

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

-A: -Add \(3x\) to both sides to eliminate the \(-3x\) term: \(x^2-3x-4+3x=0\) gives \(x^2-4=0\).

How do you solve \(x^2-4=0\)?

-A: Factor as \((x-2)(x+2)=0\). So \(x=2\) or \(x=-2\).

Can I use the quadratic formula here?

-A: Yes. For \(x^2-4=0\), \(a=1,b=0,c=-4\). Formula gives \(x=\frac{-0\pm\sqrt{0+16}}{2}= \pm2\).

How can I check the solutions?

-A: Substitute into the original equation \(x^2-3x-4=-3x\). For \(x=2\): \(4-6-4=-6\) equals \(-6\). For \(x=-2\): \(4+6-4=6\) equals \(6\) (right side \(-3x=6\)). Both satisfy.

-Are there any extraneous solutions?

-A: No. The equation is polynomial; operations were reversible, so both \(x=2\) and \(x=-2\) are valid.

What is the graphical interpretation?

-A: Solutions are x-values where the parabola \(y=x^2-3x-4\) meets the line \(y=-3x\). Simplifying gives intersections of \(y=x^2-4\) with \(y=0\), at \(x=\pm2\).

Could I solve by completing the square?

-A: Yes: \(x^2-4=0\) gives \(x^2=4\). Taking square roots yields \(x=\pm2\). (No additional terms needed here.)

What is the domain of this equation?

-A: -All real numbers, since it is a polynomial equation and polynomials are defined for every real \(x\).

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