Question #4190b

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Answer 1 · 2017-01-27T04:41:06

$49 x^{2} + 42 x - 27 = 0$ $49x^2+42x-27=0$

for solution for $- \frac{9}{7}$ $-9/7$
$x = - \frac{9}{7}$ $x=-9/7$ , then $(x + \frac{9}{7}) = 0$ $(x+9/7)=0$

for solution 3/7,
$x = \frac{3}{7}$ $x=3/7$ , then $(x - \frac{3}{7}) = 0$ $(x-3/7)=0$

Therefore,
$(x + \frac{9}{7}) (x - \frac{3}{7}) = 0$ $(x+9/7)(x-3/7)=0$

$(\frac{7 x + 9}{7}) (\frac{7 x - 3}{7}) = 0$ $((7x+9)/7)((7x-3)/7)=0$

$(7 x + 9) (7 x - 3) = 0$ $(7x+9)(7x-3)=0$

$49 x^{2} - 21 x + 63 x - 27 = 0$ $49x^2-21x+63x-27=0$

$49 x^{2} + 42 x - 27 = 0$ $49x^2+42x-27=0$

you also can use method $x^{2} - (S O R) x + P O R = 0$ $x^2-(SOR)x+POR=0$ ,

$x^{2} - (- \frac{9}{7} + \frac{3}{7}) x + (- \frac{9}{7} \cdot \frac{3}{7}) = 0$ $x^2-(-9/7+3/7)x+(-9/7*3/7)=0$

$x^{2} - (- \frac{6}{7}) x - \frac{27}{49} = 0$ $x^2-(-6/7)x-27/49=0$

$x^{2} + \frac{6}{7} x - \frac{27}{49} = 0$ $x^2+6/7x-27/49=0$

multiply by 49,

$49 x^{2} + 42 x - 27 = 0$ $49x^2+42x-27=0$