Question #4190b

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Answer 1 · 2017-01-27T04:41:06

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

for solution for $-9/7$
$x=-9/7$ , then $(x+9/7)=0$

for solution 3/7,
$x=3/7$ , then $(x-3/7)=0$

Therefore,
$(x+9/7)(x-3/7)=0$

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

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

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

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

you also can use method $x^2-(SOR)x+POR=0$ ,

$x^2-(-9/7+3/7)x+(-9/7*3/7)=0$

$x^2-(-6/7)x-27/49=0$

$x^2+6/7x-27/49=0$

multiply by 49,

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