value of x by splitting square method.
abx^2 +{ab(a^2 + b^2)+2a^2×b}^2+(a +b)^2=0
Answers
Step-by-step explanation:
The roots are x=\frac{c}{b},\frac{-b}{a}x=bc,a−b
Step-by-step explanation:
Given : The quadratic equation - abx^2 + (b^2 - ac)x - bc= 0abx2+(b2−ac)x−bc=0
To find : Solve for x?
Solution :
Using quadratic formula,
General form - ax^2+bx+c=0ax2+bx+c=0 D=b^2-4acD=b2−4ac
Solution is x=\frac{-b\pm\sqrt{D}}{2a}x=2a−b±D
Equation is abx^2 + (b^2 - ac)x - bc= 0abx2+(b2−ac)x−bc=0
where, a=ab , b=b^2-acb=b2−ac , c=-bc
D=b^2-4acD=b2−4ac
D=(b^2-ac)^2-4(ab)(-bc)D=(b2−ac)2−4(ab)(−bc)
D=b^4+a^2c^2-2b^2ac+4b^2acD=b4+a2c2−2b2ac+4b2ac
D=(b^2+ac)^2D=(b2+ac)2
Solution is x=\frac{-b\pm\sqrt{D}}{2a}x=2a−b±D
x=\frac{-(b^2-ac)\pm\sqrt{(b^2+ac)^2}}{2(ab)}x=2(ab)−(b2−ac)±(b2+ac)2
x=\frac{-b^2+ac\pm(b^2+ac}{2ab}x=2ab−b2+ac±(b2+ac
x=\frac{-b^2+ac+b^2+ac}{2ab},\frac{-b^2+ac-b^2-ac}{2ab}x=2ab−b2+ac+b2+ac,2ab−b2+ac−b2−ac
x=\frac{2ac}{2ab},\frac{-2b^2}{2ab}x=2ab2ac,2ab−2b2
x=\frac{c}{b},\frac{-b}{a}x=bc,a−b
Therefore, The roots are x=\frac{c}{b},\frac{-b}{a}x=bc,a−b
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