# Evaluate x for dy=0 if y=(x^2-ax+b)(x+c).

neela | Student

y = (x^2-ax+b)(x+c)

To evaluate x if y' = 0

We differentiate  y = (ax^2-ax+b)(x+c) and equate to zero and solve for x.

y' = {(x^2-ax+b)(x+c)}' = 0

y' = (x^2-ax+b)'(x+c) +(x^2-ax+c)(x+c)'

y' = (2x-a)(x+c) +(x^2-ax+b)*1

y = 2x^2 +2cx-ax -ac+ x^2-ax+b

y' = 3x^2 + (c-a)x + b-ac.

Equating y' = 0, we get:

x1 = {-(c-a) + sqrt{(c-a)^2 - 4*3(b-ac)}/2*3

x1 = {a-c + sqrt(c^2-2ac +12ac -12b)}/6

X1 = {a-c +sqrt(c^2+10ac-12b}/6

x 2 = {a-c - sqr(c^2+10ac-12b}/6.

Thus x1 and x2 are the values when dy/dx = 0.

william1941 | Student

Here we differentiate the function y.

Now y=(x^2-ax+b)(x+c)

=> dy/dx = [x^3 - ax^2 + bx + cx^2 - acx + bc]'

=> dy/dx = 3x^2 - 2ax + b + 2cx - ac

Now dy =0

=> 3x^2 - 2ax + b + 2cx - ac =0

=> 3x^2 - (2a - 2c)x + b-ac =0

x = [-b + sqrt (b^2-4ac)] / 2a

and x= [-b - sqrt (b^2-4ac)] / 2a

So x = [-(2a-2c) + sqrt ((2a-2c)^2-4*3*(b-ac))] / 2*3

= [-(a-c) + sqrt ((a-c)^2- 3(b-ac))] / 3

and x = [-(2a-2c) - sqrt ((2a-2c)^2-4*3*(b-ac))] / 2*3

= [-(a-c) - sqrt ((a-c)^2- 3(b-ac))] / 3

Therefore x=[-(a-c) + sqrt ((a-c)^2- 3(b-ac))] / 3 and x=[-(a-c) - sqrt ((a-c)^2- 3(b-ac))] / 3

giorgiana1976 | Student

For the beginning, let's differentiate the given function.

dy=(x^2-ax+b)(x+c)dx

Since the function is a product, we'll apply the product rule, when differentiating a product.

(f*g)' = f'*g + f*g'

We'll put f = x^2-ax+b and g = x+c

We'll differentiate, to the right side, with respect to x:

[ (x^2-ax+b)(x+c) ]' = (x^2-ax+b)' * (x+c) + (x^2-ax+b) * (x+c)'

[ (x^2-ax+b)(x+c) ]' = (2x-a) * (x+c) + (x^2-ax+b) * (1)

We'll remove the brackets:

[ (x^2-ax+b)(x+c) ]' = 2x^2 + 2xc - ax - ac + x^2-ax+b

Now, we'll put dy = 0

We'll substitute the expression for dy:

2x^2 + 2xc - ax - ac + x^2-ax+b = 0

We'll combine like terms:

3x^2 + x(2c - 2a) + b - ac = 0

If we'll plug in values for the a,b,c, we'll calculate the values of x for dy = 0.

Since the expression is a quadratic equation, we'll have the following cases:

For the quadratic to have 2 distinct roots, we'll have to impose the constraint: delta > 0

delta = (2c - 2a)^2 - 12( b - ac)

delta = 4c^2 - 8ac + 4a^2 - 12b + 12ac

delta = 4a^2 + 4c^2 + 4ac - 12b > 0

We'll divide by 4:

a^2 + c^2 + ac - 3b > 0

From this expression, we'll conclude that: a,c>b

For the quadratic to have 2 equal roots, we'll have to impose the constraint: delta = 0.

a^2 + c^2 + ac - 3b= 0

For the quadratic not to have any roots, we'll have to impose the constraint: delta < 0.

a^2 + c^2 + ac - 3b<0