Q. \(x^2-6x+5=0\)

Q: Solve the equation x^2-6x+5=0.

Factor: (x-1)(x-5)=0. So x=1 or x=5.

Q: How do you find the roots using the quadratic formula?

For a=1,b=-6,c=5: x=\frac{-(-6)\pm\sqrt{(-6)^2-4\cdot1\cdot5}}{2\cdot1}=\frac{6\pm\sqrt{36-20}}{2}=\frac{6\pm4}{2}. Hence x=1,5.

Q: What are the values of the discriminant \Delta and what does it mean?

\Delta=b^2-4ac=(-6)^2-4(1)(5)=36-20=16. Since \Delta>0, there are two distinct real solutions.

Q: Can you complete the square to solve it?

x^2-6x+5=(x^2-6x+9)-4=(x-3)^2-4=0. Then (x-3)^2=4, so x-3=\pm2, giving x=1,5.

Q: How do you factor x^2-6x+5 efficiently?

Find two numbers with sum -6 and product 5: -1 and -5. Thus x^2-6x+5=(x-1)(x-5).

Answer

We solve the quadratic \(x^2-6x+5=0\) by factoring.

\[
x^2-6x+5=(x-1)(x-5)=0
\]

So \(x=1\) or \(x=5\).

\(\boxed{x=1,\ 5}\)

Detailed Explanation

We want to solve the quadratic equation \(x^2 – 6x + 5 = 0\). The goal is to find all values of \(x\) that make the left-hand side equal to zero.

Step 1: Factor the quadratic

To factor \(x^2 – 6x + 5\), we look for two numbers \(a\) and \(b\) such that:

\(a \cdot b = 5\) (the constant term), and \(a + b = -6\) (the coefficient of \(x\)).

The numbers that multiply to \(5\) are \(1\) and \(5\), or \(-1\) and \(-5\). Since we need the sum to be \(-6\), we choose \(-1\) and \(-5\).

So we can rewrite the quadratic as:

\[
x^2 – 6x + 5 = (x – 1)(x – 5)
\]

This is because:

\((x – 1)(x – 5) = x^2 – 5x – x + 5 = x^2 – 6x + 5\).

Step 2: Set each factor equal to zero

Now the equation becomes:

\[
(x – 1)(x – 5) = 0
\]

By the zero product property, if a product is zero, then at least one factor must be zero:

\[
x – 1 = 0 \quad \text{or} \quad x – 5 = 0
\]

Step 3: Solve each simple equation

First equation:

\[
x – 1 = 0
\]

Add \(1\) to both sides:

\[
x = 1
\]

Second equation:

\[
x – 5 = 0
\]

Add \(5\) to both sides:

\[
x = 5
\]

Final Answer

The solutions are:

\[
x = 1 \quad \text{and} \quad x = 5
\]

See full solution

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Algebra FAQ

Solve the equation \(x^2-6x+5=0\).

Factor: \((x-1)(x-5)=0\). So \(x=1\) or \(x=5\).

How do you find the roots using the quadratic formula?

For \(a=1,b=-6,c=5\): \(x=\frac{-(-6)\pm\sqrt{(-6)^2-4\cdot1\cdot5}}{2\cdot1}=\frac{6\pm\sqrt{36-20}}{2}=\frac{6\pm4}{2}\). Hence \(x=1,5\).

What are the values of the discriminant \(\Delta\) and what does it mean?

\(\Delta=b^2-4ac=(-6)^2-4(1)(5)=36-20=16\). Since \(\Delta>0\), there are two distinct real solutions.

Can you complete the square to solve it?

\(x^2-6x+5=(x^2-6x+9)-4=(x-3)^2-4=0\). Then \((x-3)^2=4\), so \(x-3=\pm2\), giving \(x=1,5\).

How do you factor \(x^2-6x+5\) efficiently?

Find two numbers with sum \(-6\) and product \(5\): \(-1\) and \(-5\). Thus \(x^2-6x+5=(x-1)(x-5)\).

Use this to solve the quadratic.
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