How do you solve #14 x ^2+ 57x + 28 = 0#?

2 Answers
Oct 9, 2015

Answer:

#x=-7/2# and #x=-4/7# are the two solutions. They can be found by the quadratic formula, by factoring, or by completing the square.

Explanation:

(1) For #14x^2+57x+28=0#, the quadratic formula gives:

#x=(-57pm sqrt(57^2-4*14*28))/(2*14)=(-57pm sqrt(3249-1568))/28#

#=(-57pm sqrt(1681))/28=(-57 pm 41)/28#

Now #(-57+41)/28=-16/28=-4/7# and #(-57-41)/28=-98/28=-7/2#.

(2) Factoring can also be done as follows:

#14x^2+57x+28=0\rightarrow (2x+7)(7x+4)=0#, which leads to the same answers.

(3) Completing the square can be done as follows. Note that #3249/784=(57/28)^2#.

#14x^2+57x+28=0\rightarrow 14(x^2+57/14 x+3249/784)+28=14*3249/784#

#\rightarrow 14(x+57/28)^2=1681/56\rightarrow (x+57/28)^2=1681/784#

#\rightarrow x+57/28=pm sqrt(1681/784)= pm 41/28#

#\rightarrow x=(-57pm 41)/28#

As above, #(-57+41)/28=-16/28=-4/7# and #(-57-41)/28=-98/28=-7/2#.

Oct 9, 2015

Answer:

Solve: #y = 14x^2 + 57x + 28 = 0# (1)

Ans #x= -4/7# and #x = -7/2#

Explanation:

I use the new Transforming Method (Socratic Search)
Transformed equation: #y' = x^2 + 57x + 392 = 0# (2).
Both roots are negative. Factor pairs of (392) --> (-2, -196)(-4, -98)(-8, -49). This sum is -57 = -b. Then the 2 real roots of (2) are: -8 and -49.
Therefor, the 2 real roots of original equation (1) are: #x1 = -8/14 = -4/7# and #x2 = -49/14 = -7/2.#

NOTE . Solving by this Transforming Method is simpler and faster because it avoids the lengthy factoring by grouping, or the boring computation with the formula. In addition, it avoids solving the 2 binomials.