How do you solve #7x + 1= - 4x ^ { 2}#?

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Answer 1 · 2018-03-12T11:16:34

See a solution process below:

First, add #color(red)(4x^2)# to each side of the equation to put the equation into standard quadratic form while keeping the equation balanced:

#color(red)(4x^2) + 7x + 1 = color(red)(4x^2) - 4x^2#

#4x^2 + 7x + 1 = 0#

We can now use the quadratic equation to solve this problem:

For #color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0#, the values of #x# which are the solutions to the equation are given by:

#x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))#

Substituting:

#color(red)(4)# for #color(red)(a)#

#color(blue)(7)# for #color(blue)(b)#

#color(green)(1)# for #color(green)(c)# gives:

#x = (-color(blue)(7) +- sqrt(color(blue)(7)^2 - (4 * color(red)(4) * color(green)(1))))/(2 * color(red)(4))#

#x = (-color(blue)(7) +- sqrt(49 - 16))/8#

#x = (-color(blue)(7) +- sqrt(33))/8#

Answer 2 · 2018-03-12T11:16:39

Solution: # x~~ -1.593, x ~~ -0.157#

# 7x+1=-4x^2 or 4x^2+7x+1=0 #

Comparing with standard quadratic equation #ax^2+bx+c=0#

# a=4 ,b=7 ,c=1# Discriminant # D= b^2-4ac# or

#D=49-16 =33#, discriminant positive, we get two real

solutions. Quadratic formula: #x= (-b+-sqrtD)/(2a) #or

#x= (-7+-sqrt33)/8 :. x= -7/8 + sqrt33/8 ~~ -0.157# and

# x= -7/8 - sqrt33/8 ~~ -1.593 #

Solution: # x~~ -1.593, x ~~ -0.157# [Ans]
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