Answer:
Therefore,
[tex]x=1\ \ or\ \ x = -2[/tex]
Step-by-step explanation:
Kate is solving by completing the square.
[tex]x^{2} + x-2=0[/tex]
Step 1 : Adding 2 to bothe sides we get
[tex]x^{2} + x=2[/tex]
Step 2 : Add [tex](\dfrac{1}{2}\(coefficient\ of\ x))^{2}=\dfrac{1}{4}[/tex]
[tex]x^{2} +x+\dfrac{1}{4}=2+\dfrac{1}{4}[/tex]
Step 3 : Applying (A+B)² = A² +2AB + B²
[tex](x+\dfrac{1}{2})^{2}=\dfrac{9}{4}[/tex]
Step 4 : Square Rooting
[tex](x+\dfrac{1}{2}=\pm \sqrt{\dfrac{9}{4}}=\pm \dfrac{3}{2}[/tex]
Step 5 : Substituting we get
[tex]x+\dfrac{1}{2}=\dfrac{3}{2}\ \ or\ \ x+\dfrac{1}{2}=-\dfrac{3}{2}\\\\x=\dfrac{2}{2}=1\ \ or\ \ x=\dfrac{-4}{2}=-2[/tex]
Therefore,
[tex]x=1\ \ or\ \ x = -2[/tex]
