X2 +14x-15=0 in this equation which add LHS as a perfect square 49. how this 49 will come... please tell about 49??? how this calculated

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X2 +14x-15=0 in this equation which add LHS as a perfect square 49. how this 49 will come... please tell about 49??? how this calculated

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Answer 1 · 2018-05-07T03:07:44

x = 1, and x = - 15

Explanation:

$x^{2} + 14 x - 15 = 0$ $x^2 + 14x - 15 = 0$
$D = d^{2} = b^{2} - 4 a c = 196 + 60 = 256$ $D = d^2 = b^2 - 4ac = 196 + 60 = 256$ --> $d = \pm 16$ $d = +- 16$
There are 2 real roots:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{14}{2} \pm \frac{16}{2}$ $x = - b/(2a) +- d/(2a) = - 14/2 +- 16/2$
$x = - 7 \pm 8$ $x = - 7 +- 8$
a. x1 = - 7 + 8 = 1
b. x2 = -7 - 8 = - 15
Note.
Because a + b + c = 0, we use the shortcut.
One real root is x1 = 1, and the other is $x 2 = \frac{c}{a} = - 15$ $x2 = c/a = - 15$ .

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X2 +14x-15=0 in this equation which add LHS as a perfect square 49. how this 49 will come... please tell about 49??? how this calculated

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