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What are the solutions to the equation (x-1)(x+2)=-2?

Sagot :

Method used:-

Completing the square method.

Steps :-

[tex]( x - 1 ) ( x + 2 ) = - 2\\x^{2}+x-2=-2 \\x^{2}+x=-2+2 \\x^{2}+x=0[/tex]

Now, make the left hand side of the equation by adding the square of ½ to both sides of the equation.

[tex]x^{2}+x+\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{2} \\x^{2}+x+\frac{1}{4}=\frac{1}{4} \\\left(x+\frac{1}{2}\right)^{2}=\frac{1}{4} \\\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{1}{4}} \\[/tex]

Simplify the equation. You'll get a +ve & -ve value.

[tex]x+\frac{1}{2}=\frac{1}{2} \\x+\frac{1}{2}=-\frac{1}{2}[/tex]

Simplify it further.

[tex]\boxed{\sf\:x=0}\\ \boxed{\sf\:x=-1 }[/tex]

The solutions are :-

x = 0 & x = -1

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Hope it helps ⚜