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

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

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

A It's a quadratic equation (degree 2).

Q How do you simplify the equation first?

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

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

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

Q Can you use the quadratic formula here?

A 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?

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

Q Are the solutions rational or integers?

A Both solutions \(0\) and \(-7\) are integers (hence rational).

Q How do you check the solutions?

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

Q Could any step introduce extraneous solutions?

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

Q What is the axis of symmetry of the parabola \(y = x^2 + 7x + 3\)?

A Axis of symmetry is \(x = -\frac{b}{2a} = -\frac{7}{2}\).

Q Where is the vertex of \(y = x^2 + 7x + 3\)?

A Vertex at \(x = -\frac{7}{2}\). The y-value is \(y = \left(-\frac{7}{2}\right)^2 + 7\left(-\frac{7}{2}\right) + 3 = -\frac{37}{4}\) (i.e., \(-9.25\)), so vertex \(\left(-\frac{7}{2}, -\frac{37}{4}\right)\).

Q Where does the parabola cross the x-axis?

A At the roots: \(x = 0\) and \(x = -7\).

Q How would you complete the square for \(x^2 + 7x + 3\)?

A \(x^2+7x+3 = \left(x+\frac{7}{2}\right)^2 - \frac{49}{4} + 3 = \left(x+\frac{7}{2}\right)^2 - \frac{37}{4}\).

Q If the right side had been 0 instead of 3, would the process change?

A You would solve \(x^2 + 7x + 3 = 0\), which does not factor nicely; use the quadratic formula to get \(x = \frac{-7 \pm \sqrt{13}}{2}\).

Q What common mistakes should I watch for solving this problem?

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