Q. \(x^2+7x+3=3\).

Q: How do you simplify the equation first?

Subtract 3 from both sides to get x^2 + 7x = 0.

Q: How do you solve x^2 + 7x = 0 by factoring?

Factor out x: x(x + 7) = 0, so x = 0 or x = -7.

Q: Can you use the quadratic formula here?

Yes. With a=1, b=7, c=0: x = \frac{-7 \pm \sqrt{49}}{2} = 0, -7.

Q: What is the discriminant and what does it tell us?

Discriminant D = b^2 - 4ac = 49. Since D > 0, there are two distinct real roots.

Q: How do you check the solutions?

Substitute back into the original equation: for x=0: 0^2+7\cdot0+3=3. For x=-7: (-7)^2+7(-7)+3=49-49+3=3.

Answer

Solve \(x^2+7x+3=3\).

Subtract 3 from both sides:
\[
x^2+7x=0.
\]
Factor:
\[
x(x+7)=0.
\]
So \(x=0\) or \(x=-7\).

Final answer: \(x=0,\; x=-7\).

Detailed Explanation

Solution

Write the original equation:
\(x^{2}+7x+3=3\)
Isolate terms by subtracting 3 from both sides:
\(x^{2}+7x+3-3=3-3\) which simplifies to
\(x^{2}+7x=0\)
Factor the left-hand side:
\(x^{2}+7x = x(x+7)\), so the equation becomes
\(x(x+7)=0\)
Use the zero-product property:
If \(x(x+7)=0\), then \(x=0\) or \(x+7=0\). From \(x+7=0\) we get \(x=-7\).
Solution set:
\(\boxed{x=0 \quad \text{or} \quad x=-7}\)
Check:
For \(x=0\): \(0^{2}+7(0)+3=3\) ✓
For \(x=-7\): \((-7)^{2}+7(-7)+3=49-49+3=3\) ✓

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FAQs

What type of equation is \(x^2 + 7x + 3 = 3\)?

It's a quadratic equation (degree 2).

How do you simplify the equation first?

Subtract 3 from both sides to get \(x^2 + 7x = 0\).

How do you solve \(x^2 + 7x = 0\) by factoring?

Factor out \(x\): \(x(x + 7) = 0\), so \(x = 0\) or \(x = -7\).

Can you use the quadratic formula here?

Yes. With \(a=1\), \(b=7\), \(c=0\): \(x = \frac{-7 \pm \sqrt{49}}{2} = 0, -7\).

What is the discriminant and what does it tell us?

Discriminant \(D = b^2 - 4ac = 49\). Since \(D > 0\), there are two distinct real roots.

Are the solutions rational or integers?

How do you check the solutions?

Substitute back into the original equation: for \(x=0\): \(0^2+7\cdot0+3=3\). For \(x=-7\): \((-7)^2+7(-7)+3=49-49+3=3\).

Could any step introduce extraneous solutions?

No, we only used algebraic operations valid for all real numbers (subtracting, factoring), so no extraneous roots arise.

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