Question

value of x by splitting square method.
abx^2 +{ab(a^2 + b^2)+2a^2×b}^2+(a +b)^2=0​

Answer 1

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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