How do you solve using the completing the square method x^2 + 4x = 21x2+4x=21?

1 Answer
Mar 9, 2016

color(blue)("Vertex"->(x,y) = (-2,-25))Vertex(x,y)=(2,25)

Explanation:

Write as:" "x^2+4x-21=0 x2+4x21=0...................(1)

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color(blue)("Shortcut method")Shortcut method

By sight:

(x+2)^2-4-21=0(x+2)2421=0

color(blue)((x+2)^2-25=0)(x+2)225=0

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color(blue)("The logic behind the shortcut method")The logic behind the shortcut method

color(green)("Consider only the left hand side")Consider only the left hand side

color(green)("Place the two left most terms in brackets")Place the two left most terms in brackets

" "(x^(color(magenta)(2)) +4x) -21 (x2+4x)21

color(green)("Move the index of "color(magenta)(2)" to the outside of the brackets")Move the index of 2 to the outside of the brackets

" "(x+4color(red)(x))^(color(magenta)(2))-21 (x+4x)221

color(green)("Remove the "color(red)(x)" from the "4x" inside the brackets")Remove the x from the 4x inside the brackets

" "(x+color(blue)(4))^2-21 (x+4)221

color(green)("Apply "(1/2)xxcolor(blue)(4)=color(red)(2))Apply (12)×4=2

" "(x+color(red)(2))^2-21 (x+2)221

color(green)("Add the constant "color(red)(k))Add the constant k

" "(x+2)^2+color(red)(k)-21 (x+2)2+k21........................(2)

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color(brown)("Determine the value of "k)Determine the value of k

Set equation (1) = equation (2)

" "x^2+4x-21=(x+2)^2+k-21 x2+4x21=(x+2)2+k21

" "cancel(x^2)+cancel(4x)-cancel(21)=cancel(x^2)+cancel(4x)+4+k-cancel(21)

" "k=-4

So for (x+b)^2+k+c" "-> k=-b^2

" "(x+2)^2-25=0" "........................(2_a)

So color(blue)("Vertex"->(x,y) =((-1)xx2,-25) = (-2,-25))

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Tony B