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If-x-2-3x-2-y-2-5y-8-Prove-that-x-3-4y-2-20y-33-2-




Question Number 215874 by MathematicalUser2357 last updated on 20/Jan/25
If x^2 +3x+2=y^2 +5y+8,  Prove that x=((−3±(√(4y^2 +20y+33)))/2).
Ifx2+3x+2=y2+5y+8,Provethatx=3±4y2+20y+332.
Answered by MrGaster last updated on 20/Jan/25
x^2 +3x+2=y^2 +5y+8  x^2 +3x−y^2 −5y−6=0  =y^2 +5x+((5/2))^2 +6+((3/2))^2 −((5/2))^2   (x+(3/2))^2 =(y+(5/2))^2 +(9/4)−((25)/( 4))+6  (x+(3/2))^2 =(y+(5/2))^2 +((24−16)/4)  (x+(3/2))^2 =(y+(5/2))^2 +(8/4)  (x+(3/2))^2 =(y+(5/2))^2 +2  x+(3/2)=±(√((y+(5/2))^2 +2))  x=((−3±(√((2y+5)^2 +8)))/2)  x=((−3±(√(4y^2 +20y+25+8)))/2)  x=((−3±(√(4y^2 +20y+33)))/2)  [Q.E.D]
x2+3x+2=y2+5y+8x2+3xy25y6=0=y2+5x+(52)2+6+(32)2(52)2(x+32)2=(y+52)2+94254+6(x+32)2=(y+52)2+24164(x+32)2=(y+52)2+84(x+32)2=(y+52)2+2x+32=±(y+52)2+2x=3±(2y+5)2+82x=3±4y2+20y+25+82x=3±4y2+20y+332[Q.E.D]
Answered by Rasheed.Sindhi last updated on 20/Jan/25
 x^2 +3x+2=y^2 +5y+8,   x^2 +3x−y^2 −5y−6=0  x=((−(3)±(√((3)^2 −4(1)(−y^2 −5y−6))))/(2(1)))  x=((−3±(√(9+4y^2 +20y+24)))/2)  x=((−3±(√(4y^2 +20y+33)))/2)
x2+3x+2=y2+5y+8,x2+3xy25y6=0x=(3)±(3)24(1)(y25y6)2(1)x=3±9+4y2+20y+242x=3±4y2+20y+332
Answered by MATHEMATICSAM last updated on 20/Jan/25
x^2  + 3x + 2 = y^2  + 5y + 8  ⇒ x^2  + 3x + 2 − y^2  − 5y − 8 = 0  ⇒ x^2  + 3x −(y^2  + 5y + 6) = 0  x = ((− 3 ± (√(9 + 4y^2  + 20y + 24)))/2)  ⇒ x = ((−3 ± (√(4y^2  + 20y + 33)))/2) (Proved)
x2+3x+2=y2+5y+8x2+3x+2y25y8=0x2+3x(y2+5y+6)=0x=3±9+4y2+20y+242x=3±4y2+20y+332(Proved)

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