Which of the following equations has a root in common with x^2 - 6x +

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;

https://gmatclub.com/forum/algebra-101576.html#p787276

Thank you for the link and for correcting my mistakes.

Roots of x^2-6x+5=0 are 1 and 5
After scanning the answer choices, I chose easy equation 2x^2-2=0 which gives x = 1 or -1 So D

ok here is how I approached the problem:

I solved the equation to get the x value
x=5, x=1

and I solved each of the equations to see if any of the equations has a common factor with the equation in the stem quation.
non of the had anything in common except for C, which has x=1,x=-1.

hope that helps

The given equation can be simplified to (x-1)(x-5)=0
Roots are 1 and 5.

You don't need to find roots of each equation give as options
Just simplify each option

After quick inspection only E can have 5 as a probable root. So only check whether or not 1 is a root.

A. No real root
B. (x-2)(x+1)=0 [x=1 does not make the left hand side zero]
C. 2(x^2-1)=0 [x=1 does make the left hand side zero]
D. (x-3)(x+1)=0 [x=1 does not make the left hand side zero]
E. Don't care... by observation neither 1 nor 5 can be a root

Hence C

Which of the following has "A" root common. Only C has one root common with the equation in the question.

x^2 - 6x + 5 = 0
(x-5)(x-1) = 0
The roots are 1 and 5

We need to check for each option to calculate the roots. This can be done by plugging in the values in the equations

(A) x2 + 1 = 0. None of 1 or 5 satisfies this equation
(B) x2 - x - 2 =0. None of 1 or 5 satisfies this equation
(C) 2x2 - 2 =0. x = 1 satisfies this equation. Hence this has a common root
(D) x2 - 2x - 3 =0. None of 1 or 5 satisfies this equation
(E) x^2 - 10x - 5 = 0. None of 1 or 5 satisfies this equation

Correct Option: C

Step 1: Solve the given equation: x² – 6x + 5 = 0
This is a quadratic set equal to zero, so let's factor to get:
(x-1)(x-5)=0
So, we have two solutions (roots): x=1 or x=5

Step 2: Solve the other 5 equations to see which one has a root (solution) of x=1 or x=5

IMPORTANT: It appears that the only way to answer this question is to keep checking every single answer choice until we find that one that has a solution of either x=1 or x=5. Given this, where do you think the test-maker would hide the correct answer? In these situations, I always start at E and work my way up. Is the answer to these questions always E (or perhaps D)? No, but it's more likely that the correct answer is near the bottom.

Okay, E: x² – 2x – 3 =0
Factor to get: (x-3)(x+1)=0
So, x=3 or x=-1
No shared solutions (roots) so keep going.

D: x² – 2x -3 =0
Factor: (x - 3)(x + 1) = 0
So, x = 3 or x = -1
No shared solutions (roots) so keep going.

C: 2x² – 2 =0
Factor: 2(x² - 1) = 0
Keep factoring: 2(x+1)(x-1)=0
So, x=1 or x=-1

We have a common solution, so the correct answer must be C

Cheers,
Brent

is my approach correct ?

so from here \(x^2 - 6x + 5 = 0\) we have:

Root one = 1
Root two = 5

so i simply plugged one of these roots ( i took 1 ) in all answer choices and only C yielded ZERO

is it correct?

not sure about 5 though Am i just lucky to answer correctly?:)

Perfect approach!

Cheers,
Brent

x^2-6x+5=0

x=5 or 1

3. 2x^2-2=0
x = 1 or -1

Root x=1 is common

IMO C

Posted from my mobile device

First, let’s solve x^2 - 6x + 5 = 0 by factoring:

(x - 5)(x - 1) = 0

x = 5 or x = 1

We can solve the equations in the given choices by factoring also. However, we see that the equation in choice A is not factorable (the equation is a sum of squares), so we can start with choice B.

B. (x - 2)(x + 1) = 0

x = 2 or x = -1

We see that the equation in choice B does not have a root in common with x^2 - 6x + 5 = 0.

C. 2(x^2 - 1) = 0

2(x - 1)(x + 1) = 0

x = 1 or x = -1

We see that the equation in choice C does have a root in common with x^2 - 6x + 5 = 0 (namely, x = 1).

Answer: C

This approach takes less time . Thanks this is exactly how I too have solved the problem.

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