Q. \(x^2-10x+25=0\)

Q: What are the roots of x^2-10x+25=0?

Factor: x^2-10x+25=(x-5)^2=0, so x=5 (double root).

Q: How do you factor x^2-10x+25?

Compute 25=5\cdot 5 and -10=-5-5. Thus x^2-10x+25=(x-5)(x-5)=(x-5)^2.

Q: What is the discriminant of x^2-10x+25=0?

a=1,b=-10,c=25. So D=b^2-4ac=100-100=0, meaning one real repeated root.

Q: Solve using the quadratic formula.

x=\dfrac{-b\pm\sqrt{D}}{2a}=\dfrac{10\pm 0}{2}=5. Both solutions equal 5.

Q: Can you solve by completing the square?

x^2-10x+25=(x^2-10x+25)=(x-5)^2. Set (x-5)^2=0, giving x=5.

Q: Why is x=5 a double root?

Because the polynomial is a perfect square: (x-5)^2. The parabola touches the x-axis at x=5 without crossing.

Answer

We solve the quadratic \(x^2-10x+25=0\). Notice it factors as \(x^2-10x+25=(x-5)^2\).

So \((x-5)^2=0\), which gives \(x=5\).

Final result: \(x=5\) (double root).

Detailed Explanation

We want to solve the quadratic equation

\[
x^2 – 10x + 25 = 0
\]

Step 1: Recognize the quadratic pattern

The expression \(x^2 – 10x + 25\) looks like a perfect square of the form

\[
\left(x-a\right)^2 = x^2 – 2ax + a^2
\]

Step 2: Match coefficients

Compare

\(x^2 – 10x + 25\) with \(x^2 – 2ax + a^2\).

So we need:

\(-2a = -10\)
\(a^2 = 25\)

From \(-2a = -10\), divide both sides by \(-2\):

\[
a = 5
\]

Check \(a^2\):

\[
5^2 = 25
\]

It matches, so the trinomial is a perfect square.

Step 3: Rewrite as a perfect square equation

Substitute \(a=5\) into \(\left(x-a\right)^2\):

\[
\left(x-5\right)^2 = 0
\]

Step 4: Solve the square equation

If \(\left(x-5\right)^2 = 0\), then the quantity inside the square must be zero:

\[
x – 5 = 0
\]

Step 5: Solve for \(x\)

Add \(5\) to both sides:

\[
x = 5
\]

Final Answer

\[
x = 5
\]

See full solution

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Algebra FAQ

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

Factor: \(x^2-10x+25=(x-5)^2=0\), so \(x=5\) (double root).

How do you factor \(x^2-10x+25\)?

Compute \(25=5\cdot 5\) and \(-10=-5-5\). Thus \(x^2-10x+25=(x-5)(x-5)=(x-5)^2\).

What is the discriminant of \(x^2-10x+25=0\)?

\(a=1,b=-10,c=25\). So \(D=b^2-4ac=100-100=0\), meaning one real repeated root.

Solve using the quadratic formula.

\(x=\dfrac{-b\pm\sqrt{D}}{2a}=\dfrac{10\pm 0}{2}=5\). Both solutions equal \(5\).

Can you solve by completing the square?

\(x^2-10x+25=(x^2-10x+25)=(x-5)^2\). Set \((x-5)^2=0\), giving \(x=5\).

Why is \(x=5\) a double root?

Because the polynomial is a perfect square: \((x-5)^2\). The parabola touches the \(x\)-axis at \(x=5\) without crossing.

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