find the value or values of y in the quadratic equation y^2 + 4y +4=7?

Answer 1

Note: There are exactly two values of y. Whatever the degree is (it is 2 because the first term is y^2), there are that many solutions to it.

y^2 + 4y + 4 = 7

Move the 7 over so that it is an equation equal to zero.

y^2 + 4y - 3 = 0

You need two factors of -3 that will add to give you 4. What are those factors? Well, to save you some time, there are no "integer" answers. So now you must solve by either completing the square or by using the quadratic formula. I will do completing the square because using the quadratic formula is easy enough to figure out.

y^2 + 4y - 3 = 0

Move the 3 over and leave a space for a new "c" you are going to add to both sides. I'll explain...

y^2 + 4y + c = 3

What should c be? To figure this out, take "b" (4) and divide it in half. 2. Then square it. You get 4. 4 is what c should b. Now add 4 to both sides (you have to add to BOTH sides to make the equation still make sense!)

y^2 + 4y + c = 3
y^2 + 4y + 4 = 3 + 4
y^2 + 4y + 4 = 7

Yes, I know. You are right back where you started from. But this has still shown you the FULL process of completing the square. But wait, you aren't done yet.

y^2 + 4y + 4 = 7

Factor the left side. Notice it creates a perfect square (thus the naming "completing the square.")

(y + 2)^2 = 7

Now just try to isolate the variable y. Get rid of that "^2."

√((y + 2)^2) = +-√(7)
y + 2 = +- √7

Now just move over the 2...

y = -2 +- √7

That is your solution! Notice I put the "+-", which means "plus or minus" in front of √7 when I first put in the √. Whenever you introduce a √, you MUST put +-! And also notice that this means you have two solutions: one is -2 + √7 and one is -2 - √7.

I hope you understand it better now! Whenever you can't factor it perfectly, you must do completing the square or use the quadratic formula. You would have gotten the same answer whether you were completing the square or doing the quadratic formula.

P.S. If you ever want to write "√", just type "& radic;" But there wouldn't be a space between the & and the radic;. I just put it there so you could see the actual code.

Answer 2

some historic past: Quadra potential "sq.", so a quadratic equation consistently includes a ^2 term. And, for each ^ , there is an element. on your equation, it incredibly is to the skill 2, so there are 2 aspects. permit's now organize your equation so as that the terrific area is 0. i'm getting y^2+4y+4-7=0, or y^2+4y-3=0 i'm no longer able to locate 2 numbers which will combine to offer -3 as a product and +4 because of the fact the sum. I do, although understand that the quadratic formulation consistently works, so I shall attempt that. a, the coefficient of y^2, =+a million b, the coefficient of y, = +4 c, the consistent, = -3 The quadratic formulation is y=[-b+ or - rt(b^2-4ac)]/2a Substituting, i'm getting y=[-4+ or - rt(sixteen+12)]/2 = [-4+ or - rt28)]/2 =[-4+ or -(rt4)(rt7)]/2 =[-4 + or -2(rt7)]/2 =-2[2+ or - rt7]2 =-2+rt7 or 2-rt7 those are your solutions for y I now see yet in any different case to remedy! y^2+4y+4=7 (y+2)^2 =7 Taking the sq. root of the two facets i'm getting (y+2)= + or - rt7 y=-2+rt7 or -2-rt7 i wish it is of use to you.

Answer 3

This equation can be rewritten as

y^2 + 4y - 3 = 0 which is a quadratic equation of the form ax^2 + bx + c and can be solved.

Answer 4

y^2 + 4y +4=7
=>
y^2 + 4y +4 - 7 =0
=>
y^2 + 4y -3 =0
=>
next use the abc formula to find the solutions.

Answer 5

y^2 + 4y + 4 = 7 this is a perfect square trinomial
(y + 2)^2 = 7 take the square root of both sides
y + 2 = +/- /7 (plus or minus square root of 7)
y = -2 + /7 and y = -2 - /7 solve for y

Answer 6

y = -2 +√7 or -2 -√7.

Answer 7

y^2+4y+4=7
y^2+4y+4-7=0
y^2+4y-3=0

NOW WE SOLVE THIS ACCORDING TO QUADRATIC FORMULA OF:
x = -b + sq.rt.(b^2-4ac) /2a or -b - sq.rt.(b^2 -4ac) / 2a

then,

a=1, b=4 and c= -3,

therefore, we get,

b^2 -4ac= (4)^2 - (4*1* -3)
=16+12
=28

now we continue with the formula as follows,

y = -(4) +sq.rt.(28) / 2*1 OR y= -(4) - sq.rt.(28) / 2*1
= -4 + 2 (rt.7) / 2 OR y= -4 - 2(rt.7) /2
= 2 [-2 + (rt. 7)] / 2 OR y= 2[ -2 - (rt,7)] / 2

THEREFORE WE GET THE FINAL ANSWER THAT,

y = -2+(rt. 7) or -2 - (rt.7)

Answer 8

y^2 + 4y - 3 = 0

using quadratic formula...
a = 1
b = 4
c = -3

y = (1/2a)( -b +- sqrt [b^2 - 4ac])
y = (1/2)(-4 +- sqrt[4^2 - 4*1*-3])
y = (1/2)(-4 +- sqrt (16 + 12))
y = (1/2)(-4 +- sqrt (28))
y = (1/2)(-4 +- 2 sqrt 7)
y = -2 +- sqrt 7 = 0.646 and -4.646

Answer 9

expresion is y^2+4y+4=7
by re arranging, y^2+4y-3=0
it is in the form of ax^2+bx+c=0 where
a=1;
b=4;
c=-3;
the solution for this type of expression is
x=(-b+sqrt(b^2-4ac))/2a or
x=(-b-sqrt(b^2-4ac))/2a

Therefore
y=(-4+sqrt( 16-4(1)(-3) ))/2 or y=(-4-sqrt( 16-4(1)(-3) ))/2
=> y= -2+sqrt(7) or -2-sqrt(7)

Answer 10

(y+2)^2 = 7
y = +/- sqrt(7) -2
that is y = sqrt(7) - 2 or - sqrt(7)-2