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
_________________

~fluke

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