What are the approximate solutions of 4x^2 + 3 = -12x to the nearest hundredth

Hello from MrBillDoesMath!

   -.28 and   -2.72

Discussion:

This is a quadratic equation (where a = 4, b +12, and c = 3 using the standard notation involved in the solution)

4x^2 + 3= -12x  =>   (given). Add 12x to both sides

4x^2 + 12x + 3 = 0.

From the quadratic formula:

x =   ( -12 +\- sqrt ( 12^2 - 4(4)(3) )    )/ (2 * 4)

x =   ( -12 +\- sqrt (144 -48))/ (2 * 4)      

x =  ( -12 +\- sqrt (96))  / (2 * 4)    

x = ( -12 +\- sqrt(16*6) ) / (2 * 4)

The "+" solution,    x = -12/8  + ( 4 /8)sqrt(6) =  -3/2 + 1/2 (sqrt(6)) =  -.28

The "-" solution     x = -12/8  -  ( 4 /8)sqrt(6)  =  -3/2 - 1/2 (sqrt(6)) = -2.72

Regards,

MrB

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

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