Q. Solve the equation \(x^2 + 5 = -5x – 1\).

Answer

Start with
\[
x^2+5=-5x-1.
\]
Bring all terms to one side:
\[
x^2+5x+6=0.
\]
Factor:
\[
(x+2)(x+3)=0.
\]
Hence
\[
x=-2\quad\text{or}\quad x=-3.
\]

Detailed Explanation

Solution — step-by-step

Write the original equation:

\(x^{2} + 5 = -5x – 1\)
Move all terms to one side to obtain a standard quadratic form. Add \(5x + 1\) to both sides to get a sum equal to zero:

\(x^{2} + 5 + 5x + 1 = 0\)
Combine like terms (combine the constant terms \(5\) and \(1\), and write the polynomial in descending powers of \(x\)):

\(x^{2} + 5x + 6 = 0\)
Factor the quadratic. Find two integers whose product is \(6\) and whose sum is \(5\). Those integers are \(2\) and \(3\). Thus factorization is:

\((x + 2)(x + 3) = 0\)

(You can check by expanding: \((x + 2)(x + 3) = x^{2} + 3x + 2x + 6 = x^{2} + 5x + 6\).)
Use the zero-product property: if a product of two factors equals zero, then at least one factor is zero. Set each factor equal to zero and solve:

From \(x + 2 = 0\) we get \(x = -2\).

From \(x + 3 = 0\) we get \(x = -3\).
Verify the solutions by substitution into the original equation:

For \(x = -2\): left side \( (-2)^{2} + 5 = 4 + 5 = 9\). Right side \( -5(-2) – 1 = 10 – 1 = 9\). Both sides equal.

For \(x = -3\): left side \( (-3)^{2} + 5 = 9 + 5 = 14\). Right side \( -5(-3) – 1 = 15 – 1 = 14\). Both sides equal.

Solutions: \(x = -2\) and \(x = -3\).

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FAQs

Q What equation should I start from?

A Start with the given equation written clearly: \(x^2 + 5 = -5x - 1\). Bring all terms to one side to get a standard quadratic: \(x^2 + 5x + 6 = 0\).

Q How do I factor \(x^2 + 5x + 6\)?

A Look for two numbers that multiply to 6 and add to 5: 2 and 3. So \(x^2 + 5x + 6 = (x + 2)(x + 3)\).

Q What are the solutions?

A Set each factor to zero: \(x + 2 = 0\) gives \(x = -2\); \(x + 3 = 0\) gives \(x = -3\). Solutions: \(x = -2, -3\).

Q Could I use the quadratic formula instead?

A Yes. With \(a=1,b=5,c=6\): \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{-5 \pm \sqrt{25-24}}{2} = \frac{-5 \pm 1}{2}\), yielding \(-2\) and \(-3\).

Q What is the discriminant and what does it tell me?

A The discriminant is \(b^2-4ac = 25-24 = 1\). Positive and a perfect square, so there are two distinct real rational roots.

Q How can I check my answers quickly?

A Substitute each root back into the original equation: for \(x=-2\), \(4+5 = -5(-2)-1\) gives \(9=9\); for \(x=-3\), \(9+5 = -5(-3)-1\) gives \(14=14\). Both check.

Q Are there any domain issues or extraneous roots?

A No. This is a polynomial (quadratic) equation; domain is all real numbers and no extraneous roots arise from algebraic manipulation.

Q How does the graph of \(y = x^2 + 5x + 6\) look and where are the roots on it?

A It's an upward-opening parabola with roots at \(x=-2\) and \(x=-3\). The vertex is at \(x=-\frac{b}{2a}=-\frac{5}{2}\) with value \(y=-\tfrac{1}{4}\) (approx \(-0.25\)).

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