Lolaergasheva wrote:
I like more -finding roots and further substitution as you explained first. I cant understand second solving algebraically, because other equations also share x=1 root.
I appreciate your help. Thanks a lot
None of the other equations has either 1 or 5 as root.
The quadratic equations in your example were not factored properly.
1.
x^2+1=0
was factored as
"(x-1)(x+1)", which is not correct
(x-1)(x+1) = x^2 - 1^2 = x^2 - 1 and not {x^2 + 1}
The correct way of factoring it is;
x^2 = -1
Well I see that; it cannot be factored because \(x = \sqrt{-1}\) will result in some imaginary number.
So; x^2+1=0 has zero roots.
2.
x^2-x-2=0
x^2-2x+x-2=0
x(x-2)+1(x-2)=0
(x+1)(x-2)=0
So roots are;
x+1=0; x=-1 (Not 1)
x-2=0; x=2
3.
x^2-10x-5=0
This has two solutions but cannot be factored like other equations;
We will have to use discriminant approach to find roots;
For a quadratic equation; ax^2+bx+c=0
The two roots are;
\(Roots=\frac{-b \pm sqrt{b^2-4*a*c}}{2*a}\)
x^2-10x-5=0
can be written as
1x^2+(-10)x+(-5)=0
a=1
b=-10
c=-5
Plug these values in the formula;
\(Roots = \frac{-(-10) \pm sqrt{(-10)^2-4*1*(-5)}}{2*1}\)
\(Roots = \frac{10 \pm sqrt{100+20}}{2*1}\)
\(Roots = \frac{10 \pm sqrt{120}}{2}\)
\(Roots=5 \pm \frac{sqrt{120}}{2}\)
###\(\frac{sqrt{120}}{2} \approx 5.5\)###
\(Root_1 \approx 5+5.5=10.5\)
\(Root_2 \approx 5-5.5=-0.5\)
Neither of the roots is 1 or 5.
4.
2x^2-2=0
2(x^2-1)=0
x^2-1=0
x^2=1
i.e. x=1(This is the only place where the root is 1) and x=-1
5.
x^2-2x-3=0
x^2-3x+x-3=0
x(x-3)+1(x-3)=0
(x+1)(x-3)=0
x+1=0; x=-1(Not 1)
x-3=0; x=3
Please visit the following link should you need in-depth theory about quadratic equation and factorization;
http://gmatclub.com/forum/algebra-101576.html#p787276