Thanks.
2006-10-15 16:41:30 · 11 answers · asked by crazylifer 3 in Science & Mathematics ➔ Mathematics
x^2 + 6x + 9 = 0
left-hand side is a perfect square trinomial in x:
x^2 + 6x + 9 = (x + 3)^2 = 0
(x + 3)(x + 3) = 0 ==> x + 3 = 0
x = -3 (i.e. x = -3 is a double/repeated root of the original quadratic equation)
2006-10-15 16:44:45 · answer #1 · answered by JoseABDris 2 · 0⤊ 0⤋
once you amplify (a+b)^2 you get a^2+2ab+b^2 the answer of a quadratic ax^2+bx+c=0 is given by utilising x=(-b+/-sqrt(b^2-4ac))/2a here fixing this in accordance to the above formula we've : x= (-6+/-sqrt(36-36))/2 x=(-6)/2=-3 because of the fact the discriminant (b^2-4ac)=0 it ability that the equation has 2 same roots i.e here x=-3 two times. because of the fact a quadratic could desire to have 2 roots the respond is (x+3)^2=0 or x=-3 The 9 you're asking approximately has been broken up into 3's yet in any different case to envision that's to end the sq. of x^2+6x+9=0 x+a million/2 the different term squared + even if is generated which you do away with here (x+3)^2 yet there isn't something generated so it quite is a suitable sq.
2016-11-23 13:56:04 · answer #2 · answered by alire 4 · 0⤊ 0⤋
x^2 + 6x + 9 = 0
(x + 3)(x + 3) = 0
x = -3
2006-10-15 17:05:22 · answer #3 · answered by CSUFGrad2006 5 · 0⤊ 0⤋
(x+3)(x+3)=0
2006-10-15 16:43:48 · answer #4 · answered by bruinfan 7 · 0⤊ 0⤋
(x+3)(x+3)
2006-10-16 04:08:12 · answer #5 · answered by scheerbarbara 1 · 0⤊ 0⤋
(x+3)(x+3)
2006-10-15 16:43:35 · answer #6 · answered by Sarah B 1 · 0⤊ 0⤋
This is a perfect square
(x+3)^2 = 0
so roots are -3 , -3 or double root-3
2006-10-15 16:52:19 · answer #7 · answered by Mein Hoon Na 7 · 0⤊ 0⤋
(x - ?)(x - 3)
x = 3
2006-10-15 16:52:46 · answer #8 · answered by Anonymous · 0⤊ 0⤋
(x+3)^2=0
so x=-3,-3
2006-10-15 16:45:23 · answer #9 · answered by raj 7 · 0⤊ 0⤋
(x+3 )(x+3 ) or square of (x+3)
2006-10-15 16:44:11 · answer #10 · answered by fancy unicorn 4 · 0⤊ 0⤋