Q. \(x^2 + x – 2 = 0\)

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

Rewrite as (x+2)(x-1)=0, since 2\cdot(-1)=-2 and 2+(-1)=1. Then x=-2 or x=1.

Q: How do I solve using the quadratic formula?

For ax^2+bx+c=0, x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. Here a=1,b=1,c=-2: x=\frac{-1\pm\sqrt{1+8}}{2}=\frac{-1\pm 3}{2}\Rightarrow x=1,-2.

Q: What is the discriminant and what does it mean?

D=b^2-4ac=1-4(1)(-2)=9. Since D>0, there are two distinct real roots: x=1 and x=-2.

Q: How can I check the solutions in the equation?

Substitute x=1: 1+1-2=0. Substitute x=-2: 4-2-2=0. Both satisfy x^2+x-2=0.

Q: What steps lead to factoring (x+2)(x-1)?

Find numbers with product -2 and sum 1: 2 and -1. Then x^2+x-2=x^2+2x-x-2=(x+2)(x-1).

Q: What if I use completing the square instead?

x^2+x-2=0\Rightarrow x^2+x=\!2. Complete square: x^2+x+\frac{1}{4}=\!2+\frac{1}{4}\Rightarrow (x+\frac{1}{2})^2=\frac{9}{4}. So x+\frac{1}{2}=\pm\frac{3}{2}\Rightarrow x=1,-2.

Answer

We factor the quadratic \(x^2+x-2\). Find two numbers that multiply to \(-2\) and add to \(1\): \(2\) and \(-1\).

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

So \(x+2=0\) or \(x-1=0\).

\[
x=-2 \quad \text{or} \quad x=1
\]

Final result: \(x=-2,\;1\).

Detailed Explanation

We want to solve the equation \(x^{2}+x-2=0\).

Step 1: Factor the quadratic.

For a quadratic \(ax^{2}+bx+c\), we look for two numbers that:

Multiply to \(ac\)
Add to \(b\)

Here \(a=1\), \(b=1\), and \(c=-2\). So \(ac=1\cdot(-2)=-2\).

We need two numbers that multiply to \(-2\) and add to \(1\). The numbers are \(2\) and \(-1\), because \(2\cdot(-1)=-2\) and \(2+(-1)=1\).

So we factor the expression:

\[
x^{2}+x-2=(x+2)(x-1)
\]

Step 2: Set each factor equal to zero.

Because the product is zero, we use the zero-product property:

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

This means either:

\(x+2=0\)
or \(x-1=0\)

Step 3: Solve each simple equation.

First case:

\[
x+2=0
\]

Subtract \(2\) from both sides:

\[
x=-2
\]

Second case:

\[
x-1=0
\]

Add \(1\) to both sides:

\[
x=1
\]

Final Answer.

The solutions to \(x^{2}+x-2=0\) are:

\[
x=-2 \quad \text{or} \quad x=1
\]

See full solution

Graph

Get instant AI help with your homework, try it now!

Homework AI

Algebra FAQ

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

Rewrite as \((x+2)(x-1)=0\), since \(2\cdot(-1)=-2\) and \(2+(-1)=1\). Then \(x=-2\) or \(x=1\).

How do I solve using the quadratic formula?

For \(ax^2+bx+c=0\), \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\). Here \(a=1,b=1,c=-2\): \(x=\frac{-1\pm\sqrt{1+8}}{2}=\frac{-1\pm 3}{2}\Rightarrow x=1,-2\).

What is the discriminant and what does it mean?

\(D=b^2-4ac=1-4(1)(-2)=9\). Since \(D>0\), there are two distinct real roots: \(x=1\) and \(x=-2\).

How can I check the solutions in the equation?

Substitute \(x=1\): \(1+1-2=0\). Substitute \(x=-2\): \(4-2-2=0\). Both satisfy \(x^2+x-2=0\).

What steps lead to factoring \((x+2)(x-1)\)?

Find numbers with product \(-2\) and sum \(1\): \(2\) and \(-1\). Then \(x^2+x-2=x^2+2x-x-2=(x+2)(x-1)\).

What if I use completing the square instead?

\(x^2+x-2=0\Rightarrow x^2+x=\!2\). Complete square: \(x^2+x+\frac{1}{4}=\!2+\frac{1}{4}\Rightarrow (x+\frac{1}{2})^2=\frac{9}{4}\). So \(x+\frac{1}{2}=\pm\frac{3}{2}\Rightarrow x=1,-2\).

Use these tools for help.
Try solving the equation now.

298,376+ active customers
Math, Geometry, Trigonometry, etc.

Math Solver Geometry Solver Trigonometry Solver