Question

Find x : (x+1)2 + x2

= 221 show the method also .please​

Answer 1

Left Hand Side

$(x+1)^2+x^2$

$=(x^2+2x+1)+x^2$

$=2x^2+2x+1$

Right Hand Side 221

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$2x^2+2x+1=221$

$2x^2+2x-220=0$

divide by 2

$x^2+x-110=0$

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$(x+11)(x-10)=0$

The solutions are at $x+11=0$ or $x-10=0$.

So the two solutions are $x=-11$ or $x=10$.

Answer 2

Answer:

(x+1)² + x² = 221

(x²+2x+1) + x² = 221

x² + 2x + 1 + x² = 221

2x² + 2x + 1 = 221

2x² + 2x = 221-1

2 (x² + x) = 220

x² + x = 220/2

x² + x = 110

x² + x - 110 = 0

x² + 11x - 10x - 110 = 0

x (x + 11) - 10 (x + 11) = 0

(x + 11) (x - 10) = 0

x + 11 = 0 or x - 10 = 0 [Zero Product Rule]

x = -11 or x = 10

x = -11 or 10

Hope it is helpful...