Q. \(x^2 – 10x + 21 = 0\).

Q: Factor x^2-10x+21=0.

Find two numbers with product 21 and sum -10: -7 and -3. So x^2-10x+21=(x-7)(x-3)=0.

Q: Solve using the quadratic formula.

For a=1, b=-10, c=21: \[ x=\frac{10\pm\sqrt{100-84}}{2}=\frac{10\pm 4}{2} \] So x=7 or x=3.

Q: Complete the square for x^2-10x+21=0.

\[ x^2-10x+21=(x-5)^2-25+21=(x-5)^2-4 \] Set to zero: (x-5)^2=4. Then x=5\pm 2, so x=7 or x=3.

Q: What are the roots’ sum and product?

For x^2-10x+21=0: sum =10 and product =21. Indeed 7+3=10 and 7\cdot 3=21.

Q: How do we check the solutions quickly?

Substitute x=7: 49-70+21=0. Substitute x=3: 9-30+21=0. Both satisfy the equation.

Q: What is the discriminant and what does it tell us?

\Delta=b^2-4ac=100-84=16. Since \Delta>0 and perfect square, there are two distinct rational roots. They are 3 and 7.

Answer

Factor the quadratic.

\[
x^2-10x+21=(x-3)(x-7)=0
\]

Set each factor equal to zero.

\[
x-3=0 \Rightarrow x=3
\]
\[
x-7=0 \Rightarrow x=7
\]

Final result: \(x=3\) or \(x=7\).

Detailed Explanation

We want to solve the quadratic equation

\[
x^2 – 10x + 21 = 0
\]

Step 1: Factor the quadratic (when possible).

For a quadratic of the form

\[
x^2 + bx + c = 0,
\]

we look for two numbers whose:

product is \(c\),
sum is \(b\).

Here, \(b = -10\) and \(c = 21\). So we need two numbers that:

multiply to \(21\),
add to \(-10\).

Step 2: Find the numbers.

The factors of \(21\) are \(1 \cdot 21\) and \(3 \cdot 7\).

Now check which pair adds to \(-10\):

\(\,1 + 21 = 22\) (not \(-10\))
\(\,3 + 7 = 10\) (not \(-10\))
\(-3 + (-7) = -10\) (this works)

So the quadratic factors as:

\[
x^2 – 10x + 21 = (x – 3)(x – 7)
\]

Step 3: Set each factor equal to zero (Zero Product Property).

\[
(x – 3)(x – 7) = 0,
\]

then either

\[
x – 3 = 0
\]

\[
x – 7 = 0.
\]

Step 4: Solve each equation.

First equation:

\[
x – 3 = 0
\]
\[
x = 3
\]

Second equation:

\[
x – 7 = 0
\]
\[
x = 7
\]

Final Answer:

\[
x = 3 \text{ or } x = 7
\]

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Algebra FAQ

Factor \(x^2-10x+21=0\).

Find two numbers with product \(21\) and sum \(-10\): \(-7\) and \(-3\). So \(x^2-10x+21=(x-7)(x-3)=0\).

Solve using the quadratic formula.

For \(a=1\), \(b=-10\), \(c=21\): \[ x=\frac{10\pm\sqrt{100-84}}{2}=\frac{10\pm 4}{2} \] So \(x=7\) or \(x=3\).

Complete the square for \(x^2-10x+21=0\).

\[ x^2-10x+21=(x-5)^2-25+21=(x-5)^2-4 \] Set to zero: \((x-5)^2=4\). Then \(x=5\pm 2\), so \(x=7\) or \(x=3\).

What are the roots’ sum and product?

For \(x^2-10x+21=0\): sum \(=10\) and product \(=21\). Indeed \(7+3=10\) and \(7\cdot 3=21\).

How do we check the solutions quickly?

Substitute \(x=7\): \(49-70+21=0\). Substitute \(x=3\): \(9-30+21=0\). Both satisfy the equation.

What is the discriminant and what does it tell us?

\(\Delta=b^2-4ac=100-84=16\). Since \(\Delta>0\) and perfect square, there are two distinct rational roots. They are \(3\) and \(7\).

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