Q. \(x^{2}-2x-35=0\)

Q: How do I factor x^2-2x-35=0?

Find numbers with product -35 and sum -2: -7 and 5. So x^2-2x-35=(x-7)(x+5)=0.

Q: What are the solutions using the zero-product rule?

From (x-7)(x+5)=0, set each factor to zero: x-7=0\Rightarrow x=7, and x+5=0\Rightarrow x=-5.

Q: Can I solve it with the quadratic formula?

With a=1,b=-2,c=-35: x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{2\pm\sqrt{4+140}}{2}=\frac{2\pm 12}{2}, giving x=7,-5.

Q: What is the discriminant b^2-4ac?

b^2-4ac=(-2)^2-4(1)(-35)=4+140=144. Since it’s positive, there are two real roots.

Q: Is there a complete-the-square method?

x^2-2x-35=0\Rightarrow (x-1)^2=36. Then x-1=\pm 6, so x=7 or x=-5.

Q: How do I check the solutions quickly?

Substitute x=7: 49-14-35=0. Substitute x=-5: 25+10-35=0. Both satisfy the equation.

Answer

We solve the quadratic equation \(x^2-2x-35=0\).

Factor the expression:

\[
x^2-2x-35=(x-7)(x+5)=0
\]

Set each factor equal to zero:

\[
x-7=0 \Rightarrow x=7
\]
\[
x+5=0 \Rightarrow x=-5
\]

Final answers: \(x=7\) or \(x=-5\).

Detailed Explanation

We want to solve the equation

\[
x^2 – 2x – 35 = 0
\]

Step 1: Factor the quadratic.

We look for two numbers whose product is

\[
-35
\]

and whose sum is

\[
-2.
\]

The numbers are \(5\) and \(-7\), because

\[
5 \cdot (-7) = -35
\]

and

\[
5 + (-7) = -2.
\]

So we can factor the quadratic as

\[
x^2 – 2x – 35 = (x + 5)(x – 7).
\]

Step 2: Set each factor equal to zero (Zero Product Property).

\[
(x + 5)(x – 7) = 0
\]

This means

\[
x + 5 = 0
\]

\[
x – 7 = 0.
\]

Step 3: Solve each simple equation.

First equation:

\[
x + 5 = 0
\]
\[
x = -5.
\]

Second equation:

\[
x – 7 = 0
\]
\[
x = 7.
\]

Final answer:

\[
x = -5 \quad \text{or} \quad x = 7.
\]

See full solution

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Algebra FAQ

How do I factor \(x^2-2x-35=0\)?

Find numbers with product \(-35\) and sum \(-2\): \(-7\) and \(5\). So \(x^2-2x-35=(x-7)(x+5)=0\).

What are the solutions using the zero-product rule?

From \((x-7)(x+5)=0\), set each factor to zero: \(x-7=0\Rightarrow x=7\), and \(x+5=0\Rightarrow x=-5\).

Can I solve it with the quadratic formula?

With \(a=1,b=-2,c=-35\): \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{2\pm\sqrt{4+140}}{2}=\frac{2\pm 12}{2}\), giving \(x=7,-5\).

What is the discriminant \(b^2-4ac\)?

\(b^2-4ac=(-2)^2-4(1)(-35)=4+140=144\). Since it’s positive, there are two real roots.

Is there a complete-the-square method?

\(x^2-2x-35=0\Rightarrow (x-1)^2=36\). Then \(x-1=\pm 6\), so \(x=7\) or \(x=-5\).

How do I check the solutions quickly?

Substitute \(x=7\): \(49-14-35=0\). Substitute \(x=-5\): \(25+10-35=0\). Both satisfy the equation.

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