Q. Which are the roots of \(x^{2} + 10x + 25 = 0\)?

Answer

Note: \(x^2+10x+25=(x+5)^2\), so the root is \(x=-5\) (double root).

We interpret the expression as \( (4x+5)(x^2-2x+5) \). Distribute:

\[ (4x+5)(x^2-2x+5) = 4x(x^2-2x+5) + 5(x^2-2x+5) \]
\[ = 4x^3 – 8x^2 + 20x + 5x^2 – 10x + 25 \]
\[ = 4x^3 – 3x^2 + 10x + 25 \]

Final result: \(\boxed{4x^3-3x^2+10x+25}\)

Detailed Explanation

Problem: Solve the equation \(x^{2} + 10x + 25 = 0\)

Step 1 — Recognize a perfect square trinomial

Observe that the left-hand side factors as the square of a binomial:
\[
x^{2}+10x+25=(x+5)(x+5)=(x+5)^{2}.
\]

Step 2 — Use the factorization to solve

Replace the quadratic by its factorized form and set equal to zero:
\[
(x+5)^{2}=0.
\]
A square is zero only when its base is zero, so
\[
x+5=0,
\]
hence
\[
x=-5.
\]

Step 3 — Root multiplicity

Since the factor \(x+5\) appears twice, the solution \(x=-5\) is a double root (multiplicity 2).

Optional check — Discriminant

For \(ax^{2}+bx+c\) with \(a=1\), \(b=10\), \(c=25\),
\[
\Delta=b^{2}-4ac=10^{2}-4\cdot1\cdot25=100-100=0,
\]
which confirms a single repeated real root. The root from the quadratic formula is
\[
x=\frac{-b}{2a}=\frac{-10}{2}=-5.
\]

Final answer

\[
x=-5\quad\text{(double root, multiplicity 2)}
\]

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FAQs

What are the roots of (x^2 + 10x + 25 = 0)?

Factor as ((x+5)^2=0). The only real root is (x=-5), with multiplicity 2.

How do I factor (x^2 + 10x + 25)?

Recognize it as a perfect square: (x^2+10x+25=(x+5)^2). Check by expanding: ((x+5)(x+5)=x^2+10x+25).

How does the quadratic formula apply here?

Using (x=frac{-bpmsqrt{b^2-4ac}}{2a}) with (a=1,b=10,c=25): discriminant (100-100=0), so (x=-tfrac{10}{2}=-5) (single solution repeated).

What does the discriminant tell me for this equation?

The discriminant is (b^2-4ac=0). That means one real root of multiplicity 2 (a repeated root); the parabola is tangent to the x-axis.

What is the vertex and graph shape of (y=x^2+10x+25)?

It's an upward-opening parabola (a>0). Vertex at ((-5,0)), which lies on the x-axis; the parabola touches the axis at the repeated root.

How can I verify (x=-5) is a solution?

Substitute: ((-5)^2+10(-5)+25=25-50+25=0). Since it yields 0, (x=-5) satisfies the equation.

What are the sum and product of the roots by Vieta's formulas?

For (ax^2+bx+c), sum = (-tfrac{b}{a}) and product = (tfrac{c}{a}). Here sum = (-10) (two roots adding to (-10)), product = (25). With repeated root (-5), (-5+(-5)=-10) and ((-5)(-5)=25).

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