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

Q: How do you solve x^2-25=0 by factoring?

Rewrite as x^2=25. Then x^2-25=(x-5)(x+5)=0. So x=5 or x=-5.

Q: How do you solve x^2-25=0 using square roots?

From x^2=25, take square roots: x=\pm\sqrt{25}=\pm 5. Therefore x=5 or x=-5.

Q: What are the solutions to the equation x^2-25=0 using the quadratic formula?

For x^2+0x-25=0, a=1,b=0,c=-25. Then x=\frac{-0\pm\sqrt{0-4(1)(-25)}}{2}=\frac{\pm\sqrt{100}}{2}=\pm 5.

Q: What is the discriminant of x^2-25=0, and what does it tell you?

Discriminant D=b^2-4ac=0-4(1)(-25)=100. Since D>0, there are two real solutions: x=\pm 5.

Q: How do you verify that x=5 and x=-5 satisfy x^2-25=0?

Plug in: for x=5, 25-25=0. For x=-5, 25-25=0. Both satisfy the equation.

Q: How do you interpret the equation x^2-25=0 geometrically?

It represents where y=x^2-25 crosses the x-axis. Set y=0: x=\pm 5. Thus the parabola intersects the x-axis at -5 and 5.

Answer

We solve \(x^2-25=0\) by factoring as a difference of squares:

\[
x^2-25=(x-5)(x+5)=0
\]

So \(x-5=0\) or \(x+5=0\), giving \(x=5\) or \(x=-5\).

Final result: \(x=-5,\ 5\).

Detailed Explanation

We want to solve the equation

\[
x^2 – 25 = 0
\]

Step 1: Move the constant term to the other side.

Start with

\[
x^2 – 25 = 0
\]

Add \(25\) to both sides to isolate the \(x^2\) term.

\[
x^2 – 25 + 25 = 0 + 25
\]
\[
x^2 = 25
\]

Step 2: Take the square root of both sides.

Since \(x^2 = 25\), we take square roots. Remember that both positive and negative roots are solutions.

\[
x = \pm \sqrt{25}
\]

Step 3: Simplify the square root.

\[
\sqrt{25} = 5
\]

\[
x = \pm 5
\]

Final Answer:

\[
x = 5 \quad \text{or} \quad x = -5
\]

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Algebra FAQ

How do you solve \(x^2-25=0\) by factoring?

Rewrite as \(x^2=25\). Then \(x^2-25=(x-5)(x+5)=0\). So \(x=5\) or \(x=-5\).

How do you solve \(x^2-25=0\) using square roots?

From \(x^2=25\), take square roots: \(x=\pm\sqrt{25}=\pm 5\). Therefore \(x=5\) or \(x=-5\).

What are the solutions to the equation \(x^2-25=0\) using the quadratic formula?

For \(x^2+0x-25=0\), \(a=1,b=0,c=-25\). Then \(x=\frac{-0\pm\sqrt{0-4(1)(-25)}}{2}=\frac{\pm\sqrt{100}}{2}=\pm 5\).

What is the discriminant of \(x^2-25=0\), and what does it tell you?

Discriminant \(D=b^2-4ac=0-4(1)(-25)=100\). Since \(D>0\), there are two real solutions: \(x=\pm 5\).

How do you verify that \(x=5\) and \(x=-5\) satisfy \(x^2-25=0\)?

Plug in: for \(x=5\), \(25-25=0\). For \(x=-5\), \(25-25=0\). Both satisfy the equation.

How do you interpret the equation \(x^2-25=0\) geometrically?

It represents where \(y=x^2-25\) crosses the \(x\)-axis. Set \(y=0\): \(x=\pm 5\). Thus the parabola intersects the \(x\)-axis at \(-5\) and \(5\).

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