Tyrol460
contestada

Write x2 - 2x - 3 = 0 in the form (x - a)2 = b, where a and b are integers. (x - 4)2 = 3 (x - 3)2 = 2 (x - 2)2 = 1 (x - 1)2 = 4

Respuesta :

Benjamin16
[tex]f(x)=x^{2}-2x-3 \\ \\ \Delta=(-2)^{2}-4*1*(-3)=4+12=16 \\ \\ a= \frac{2}{2}=1 \\ \\ b= \frac{-16}{4}=-4 \\ \\ x^{2}-2x-3=(x-1)^{2}-4=(x-1)^{2}=4 [/tex]

\\\ Ben

Answer:

The correct option is 4.

Step-by-step explanation:

The given equation is

[tex]x^2-2x-3=0[/tex]

If an expression is [tex]x^2+bx[/tex], then to make it perfect sqaure add [tex](\frac{b}{2})^2[/tex].

[tex](x^2-2x)-3=0[/tex]

Here b=-2,to make the perfect square we need to add [tex](\frac{-2}{2})^2[/tex],i.e, 1.

[tex](x^2-2x+1)-1-3=0[/tex]

[tex](x-1)^2-4=0[/tex]                [tex][\because (a-b)^2=a^2-2ab+b^2][/tex]

Add 4 on both the sides.

[tex](x-1)^2-4+4=0+4[/tex]

[tex](x-1)^2=4[/tex]

Therefore the correct option is 4.