Question

Find the real solutions of the equation
4(x + 1)^2 + 21(x + 1) + 5 = 0

Answer 1

@campbell_st

Answer 2

@campbell_st

Answer 3

ok... so this is an equation that is reducible to a quadratic... make a substitution of

u = (x + 1)

so the equation is

\[4u^2 + 21u + 5 = 0\]

can you solve this..?

Answer 4

I'm not sure. Would I just plug it into the quadratic formula?

Answer 5

you can here here is a cool method to show your teacher...

for a quadratic \[ax^2 + bx + c = 0\]

find \[x \times c =\]

so you need

\[4 \times 5 = 20\]

find the factors of 20 that add to b... or 21 in your question...what would you say..?

Answer 6

20 * 1

Answer 7

For where you put x * c did you mean a * c?

Answer 8

yes... so the factors of 20 are 20 and 1... to they add to given the coefficient of the middle term..

so next is

(au + factor 1)(au + factor 2) (4u + 20)(4u + 1)
------------------------- ---------------
a 4

now remove any common factors from the binomials

4(u + 5)(4u + 1)
---------------
4

cancel the common factor and you get (u + 5)(4u + 1) = 0

so now you can find the values of u...that make the equation true

what are they..?

Answer 9

u = -5
u = -1/4

Answer 10

correct... so now use the substitution again

u = x + 1

so

x + 1 = -5 and x + 1 = -1/4

hope it helps

Answer 11

Thanks :)