find the roots of the quadratic equation?

Question

y^2 + 11y + 28 = 0

My problem is that my algebrator gives me a long
complicated sqaure root that I cannot even fathom. What am I doing wrong?

I also have found -4, -7 am I right?

Answer 1

I'm sorry your question hurts my brain

Answer 2

Factor out the 28 = 4 * 7 (since 4y + 7y gives you your 11y).

Hence, the roots are (y + 4) (y + 7) = 0.

Alternatively, you may do the long way using the formula:

x = ( -b plus-minus sqrt(b^2 - 4ac) ) / 2a,
where your a = 2, b = 11, c = 28.

You will still solve the y which is -4 or -7.

Answer 3

perhaps you actually advise: The roots of the quadratic equation 16x^2+7x+4=0 are ?² and ?². Then the equation 16(a million/x)^2+7(a million/x)+4=0 has the roots a million/?² and a million/?². This equation might want to be rearranged into the quadratic 4x^2+7x+16=0.

Answer 4

Just factor the equation...find two multiples of C (28) that add up to B (11). Your answer is correct, you'll get 7 and 4...
So, y + 7 = 0 and y + 4 = 0...Solve for Y

-7, -4

Note: The poster above is also correct. You can use the quadratic formula for any 2nd degree equation.

Answer 5

split 28 into two factors and if sign of constant is +ve then the factors should add up to give the middle term

so 28 is 7*4

so the roots are -4 and -7

Answer 6

This is of the form ax^2+bx+c and the roots are
-b+or-squr.root b^2-4ac/2a
-11+ -sq.rt 11^2-4x1x28/2
x=-4 and -7
There fore -4 and -7 are the roots.
(x+4)(x+7) gives the given equation

Answer 7

Yes, your answer is right.
y^2+11y+28=(y+7)(y+4)=0
y=-7
y=-4

Answer 8

Yes, the answer you got is right! You simply use the quadratic formula.

y= (-b+/-sqr(b^2-4ac))/(2a)

where a= 1, b=11, and c= 28 in your case.

Hope that helped.

Answer 9

yes -4 and -7 are right.


you are smarter than an algabrator

Answer 10

(y+7)(y+4)
y^2+7x+4x+28=0
y(y+7)+4(y+7)=0
(y+7)(y+4)=0