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Solve the following equation by transforming it into a perfect square trinomial.* 3x2 + 12x = 63

Sagot :

Lilithidn
[tex]3x^2 + 12x = 63\\ \\3x^2+12x-63=0\\ \\3(x^2+4x-21)=0\\ \\3[x^2+4x+3x-3x-21] =0 \\ \\3[x^2+7x-3x-21] =0 \\ \\3[x(x +7 )-3(x+7)] =0[/tex]

[tex]3 (x +7 ) (x-3) =0 \\ \\x+7=0 \ \ \ or \ \ \ x-3 =0\\ \\x=-7 \ \ \ or \ \ \ x=3[/tex]
 

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