the roots of the eqn_x3+9x2-52x=0 are.....?

Answer 1

do you mean this:

x^3 +9x^2 - 52x ?

if so, this is the answer

x^3 +9x^2 - 52x = x(x^2 + 9x -52)=x(x-4)(x+13)=0

x=0 , x= -13, x=4

Answer 2

x^3 + 9x^2 - 52x = 0

Factor out an x

x(x^2 + 9x - 52) = 0

So, x = 0 or (x^2 + 9x - 52) = 0

Factor (x^2 + 9x - 52)
(x +___ )(x +___) ---> -52 = -26*2 = 26*-2 = -13*4 = 13*-4
13 and -4 are the only terms that add to make nine, but multiply to make -52

(x + 13)(x - 4) = 0
(x + 13) = 0 OR (x - 4) = 0
So, x = -13, x = 4 or x = 0

Answer 3

x^3 + 9x^2 - 52x = x(x^2 + 9x - 52) = 0

One of the roots is zero. The other two may be determined from the quadratic equation:

x = -9/2 + or - sqrt(81-4*1*(-52))/2 = -9/2 + or - 17/2

The other two roots are thus 4 and -13.

Answer 4

x=0 is the simple one

That leaves x^2 + 9x - 52 = 0
the quadratic equation isn't necessary if you can find 2 numbers that add together to make 9 and multiply together to make -52. Factors of 52 are 26*2 (which can't add up to 9) or 13*4. 13-4 is 9 and 13*-4=52. So they are your factors.

x(x+13)(x-4)=0

so x=0 or x+13=0 or x-4=0

so x=0 or -13 or 4

Answer 5

x^3 +9x^2 - 52x = x(x^2 + 9x -52)=x(x-4)(x+13)=0
so
x=0 , x= -13, x=4

Answer 6

x^3 + 9x^2 - 52x = 0
x ( x^2 + 9x - 52) = 0
x (x + 13) (x - 4) = 0

root of the equation above is x=0 or x=4 or x=-13

Answer 7

x3+9x2-52x=0
x3+9x2-52x=0
x(x2+9x-52)=0
=> x=0 or x2+9x-52=0
if x2+9x-52=0 => x2+13x-4x-52=0
=> x(x+13)-4(x+13)=0
=> (x+13)(x-4)=0
=> x+13=0 or x-4=0
=> x= -13 or x=4
therefore, x = (0)or (-13) or (4).

Answer 8

0, -13 and 4

Answer 9

x³ +9x² -52x =0
x ( x² +9x -52) =0
so X = 0
OR
x² +9x -52 = 0
x = [-b ± √(b²-4ac) ] / 2a
x =( -9 ± √289) /2
x =( -9 ± 17 )/ 2
x = - 13 Or x = 4


So x є { 0 , 4 , -13}

Answer 10

x=0 or x=-9/2+/-sqrt(81+4*52)/2
Answer:
x=0 or x=4 or x=-13