How do you solve $14 x^{2} + 57 x + 28 = 0$ $14 x ^2+ 57x + 28 = 0$ ?

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How do you solve $14 x^{2} + 57 x + 28 = 0$ $14 x ^2+ 57x + 28 = 0$ ?

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Answer 1 · 2015-10-09T02:50:41

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

Explanation:

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

$x = \frac{- 57 \pm \sqrt{57^{2} - 4 \cdot 14 \cdot 28}}{2 \cdot 14} = \frac{- 57 \pm \sqrt{3249 - 1568}}{28}$ $x=(-57pm sqrt(57^2-4*14*28))/(2*14)=(-57pm sqrt(3249-1568))/28$

$= \frac{- 57 \pm \sqrt{1681}}{28} = \frac{- 57 \pm 41}{28}$ $=(-57pm sqrt(1681))/28=(-57 pm 41)/28$

Now $\frac{- 57 + 41}{28} = - \frac{16}{28} = - \frac{4}{7}$ $(-57+41)/28=-16/28=-4/7$ and $\frac{- 57 - 41}{28} = - \frac{98}{28} = - \frac{7}{2}$ $(-57-41)/28=-98/28=-7/2$ .

(2) Factoring can also be done as follows:

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

(3) Completing the square can be done as follows. Note that $\frac{3249}{784} = {(\frac{57}{28})}^{2}$ $3249/784=(57/28)^2$ .

$14 x^{2} + 57 x + 28 = 0 \to 14 (x^{2} + \frac{57}{14} x + \frac{3249}{784}) + 28 = 14 \cdot \frac{3249}{784}$ $14x^2+57x+28=0\rightarrow 14(x^2+57/14 x+3249/784)+28=14*3249/784$

$\to 14 {(x + \frac{57}{28})}^{2} = \frac{1681}{56} \to {(x + \frac{57}{28})}^{2} = \frac{1681}{784}$ $\rightarrow 14(x+57/28)^2=1681/56\rightarrow (x+57/28)^2=1681/784$

$\to x + \frac{57}{28} = \pm \sqrt{\frac{1681}{784}} = \pm \frac{41}{28}$ $\rightarrow x+57/28=pm sqrt(1681/784)= pm 41/28$

$\to x = \frac{- 57 \pm 41}{28}$ $\rightarrow x=(-57pm 41)/28$

As above, $\frac{- 57 + 41}{28} = - \frac{16}{28} = - \frac{4}{7}$ $(-57+41)/28=-16/28=-4/7$ and $\frac{- 57 - 41}{28} = - \frac{98}{28} = - \frac{7}{2}$ $(-57-41)/28=-98/28=-7/2$ .

Answer 2 · 2015-10-09T05:14:33

Solve: $y = 14 x^{2} + 57 x + 28 = 0$ $y = 14x^2 + 57x + 28 = 0$ (1)

Ans $x = - \frac{4}{7}$ $x= -4/7$ and $x = - \frac{7}{2}$ $x = -7/2$

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.

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