Solve the equation by completing the square. 8x²=-11x-7?

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Answer 1 · 2018-07-03T19:15:32

$x=-11/16+sqrt103/16*i or x=-11/16-sqrt103/16*i$

Here,

$8x^2=-11x-7$

$=>8x^2+11x+7=0$

$=>8x^2+11x=-7$

Let , $color(violet)(kinRR$ be the $3^(rd) term$ to complet square.

$i.e. 8x^2+11x+color(violet)(k)=color(violet)(k)-7...to(1)$

In the $LHS$ we have ,

$color(blue)(diamond 1^(st)term=8x^2$

$color(blue)(diamond2^(nd)term=11x$

$color(blue)(diamond3^(rd)term)=color(violet)(k$

We have formula for $3^(rd)term :$

$color(red)(3^(rd)term=(2^(nd)term)^2/(4xx1^(st)term))...to(A)$

$=>color(violet)(k)=(11x)^2/(4xx8x^2)=(121x^2)/(32x^2)$

$=>color(violet)(k=121/32$

From $(1)$ ,we get

$8x^2+11x+color(violet)(121/32)=color(violet)(121/32)-7=-103/32$

$=>(2sqrt2x)^2+2(2sqrt2x)(11/(4sqrt2))+(11/(4sqrt2))^2=103/32*i^2$

$=>(2sqrt2x+11/(4sqrt2))^2=(sqrt103/(4sqrt2)*i)^2$

$=>2sqrt2x+11/(4sqrt2)=+-sqrt103/(4sqrt2)*i$

$=>2sqrt2x=-11/(4sqrt2)+-sqrt103/(4sqrt2)*i$

$=>x=-11/16+-sqrt103/16*i$

...................................................................................................
Note :

Formula $(A) :color(red)(3^(rd)term=(2^(nd)term)^2/(4xx1^(st)term))$ can be use

to find THIRD TERM for any eqn. without any doubt .

WHY ??? $to$ Please see below.

$diamond if , a^2+2ab+k=0$ ,then [use $(A)$ ]

$k=(2ab)^2/(4xxa^2)=(4a^2b^2)/(4a^2)=b^2$

$=>a^2+2ab+b^2=(a+b)^2$