Gladys S. answered • 01/27/21
ASVAB, Algebra, Geometry, and PreCalculus Tutor
Completing the square for a quadratic in standard form, x2 + bx +c requires writing it in vertex form
( x-h )2 + k.
The choice of h and k depends on factoring the original trinomial as a perfect square.
1.The first step is to set the expression to zero
x2+6x+4=0
2.The second step is to add constants to both sides to make the expression a perfect square trinomial.
In this case, adding 5 to both sides makes it a perfect square because:
x2+2(3x) + ( 3)2 =0 + 5 or x2+ 2(3x) + 4 + 5 = 0+ 5
3.We have now used the mathematical identity: ( x+ a) 2 = x2 + 2ax + a2 to simplify our problem
4.The problem to solve becomes : x2+2(3x) + ( 3)2 = (x + 3) 2= 5
5.Taking the square root of both sides: sqrt((x+3)2 ) = +/- sqrt(5) or x+3 = +/- sqrt(5)
6.Solving for x: x = -3 - sqrt(5) and x = -3 + sqrt(5)
Gladys S.
so in vertex form the equation is (x--3)^2 +5. Is a upward parabola centered at x = -3 and is shifted vertically by 5 .01/27/21
Gladys S.
h =-3; k = 501/27/21
Gladys S.
standard form is (x-h)^2 + k in this case : h = -3 and k = 501/27/21
Gladys S.
standard form is ( x -- 3)^2 + -5 . h = -3 and k = -5 .upward parabola centered at 0. passing thru x = -3 - sqrt(5) and -3+sqrt(5). Minimum at x = 0. where y = -5.01/27/21
Madina A.
h = to what ?? k =501/27/21