What is the correct way to solve: y² + 6y = 55?

make the equation equal to 0, so take away 55 from both sides:

y^2 + 6y - 55 = 0

Now, this is easily factored:

(y+11)(y-5) = 0

Now set each factor equal to 0 and solve:

y + 11 = 0 y - 5 = 0
y = -11 or y = 5

Hope that helped! =)

i don't know i haven't learned it but i think 11

y(y+6)=55

55=5x11 so y=5

hehe.

set the equation equal to 0

y^2+6y-55 = 0

(y+11)*(y-5) = 0

So y = 5 and y = -11

y^2 + 6y = 55
y^2 + 6y - 55 = 0
y^2 + 11y - 5y - 55 = 0
(y^2 + 11y) - (5y + 55) = 0
y(y + 11) - 5(y + 11) = 0
(y + 11)(y - 5) = 0 (factorization)

y + 11 = 0
y = -11

y - 5 = 0
y = 5

∴ y = -11 , 5

Whilst the method dictated by como works perfectly, it can be at times difficult to intuitively factor a quadratic expression and consequently yield the solution.

First off when presented with a quadractic equation of the form,

ax^2 + bx + c = 0 (as in your case)

One of there situations may arise
(i) There is one solution
(ii) 2 solutions (as with yours)
(ii) no real solutions (complex solutions still exist however I don't believe this is what your after)

There is a simple test for this, and is known as evaluating the discriminant which is defined as
D = sqrt(b^2 - 4ac)
if D > 0 --> (i) ; D = 0 --> (ii) ; D < 0 --> (iii)

In your case D > 0 and as such 2 solutions exist, whilst you can factor the expression (and don't get me wrong when I can identify it easily I do), however there is a full proof method known as the quadratic formula which is used to solve the roots of a quadratic,
it is defined as,

x = (-b + - D)/(2a)

substituting the values you have being a = 1, b = 6 and c = -55 will generate the desired solution,


Hope this helps,

Regards,

David

Hmmm

make the equation equal to 0, by taking the 55 to the other side and changing its sign

y² + 6y - 55 = 0

Now, you need 2 numbers which have the sum of 6, and product of -55

do a little thinking =P you'll get the numbers -5 and 11

put them in brackets with the y in this way << I dunno why but it's this way =P

(y+11)(y-5) = 0

now set each of them so that they'll be equal to 0 and solve...

y + 11 = 0...........y - 5 = 0

take the 11 and -5 to the othe side... you'll get::

y = -11 ......and......y = 5

I'm not good in explaining things =S I'm just 15... but hope that helped =)

ps:: you can also use the formula -b ± √ (b² - 4ac) / 2a but I don't think that'll be correct when factorising is possible =)

good luck

The usual way. Form a quadratic equation, factor and equate each of the factors to 0
y² + 6y = 55
y² + 6y - 55 = 0
(y + 11)(y - 5) = 0
( y + 11) = 0
y = -11
(y - 5) = 0
y = 5
y = {-11, 5}

Rearrange it to be a general quadratic, and then factorise or use the formula.

y² + 6y = 55
⇒ y² + 6y - 55 = 0
⇒ (y + 11)(y - 5) = 0
⇒ y = -11, y = 5

It's important NOT just to divide through by y, as this could (in some cases) lose solutions (when one solution is 0).

The correct way is the way that gets the correct answer. There are many ways.

Here is just one way:
Write the equation like this: y^2 + 6y - 55 = 0
The coefficients are a= 1, b = 6, c = -55
Calculate rad = b^2 - 4ac
Calculate root1 = [ -b + sqrt ( rad ) ] / [2a]
Calculate root2 = [ -b - sqrt ( rad ) ] / [2a]

If 'rad' is less than 0, you need to study about imaginary numbers.

The values of root1 and root2 are the solution for y.