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

Easy question. \(x^2 - 6x + 5 = 0\) factors into \((x-1)(x-5)=0\). Thus, roots are x=1 and x=5.

Do the same for each option:

A. \(x^2 + 1 = 0\)
\(x^2 = -1\)
roots are imaginary

B. \(x^2 - x - 2 = 0\)
\((x-2)(x+1) = 0\)
roots are x=2 and x=-1
none match question

C. \(2x^2 - 2 = 0\)
\(2(x^2-1)=0\)
\((x^2-1) = 0\)
\((x-1)(x+1) = 0\)
roots are x=1 and x=-1
x=1 matches the root of original equation

No need to do the rest, we have our answer. However, if you have lots of time you may want to do them anyway to make sure you didn't make a mistake above. You'll know you made a mistake somewhere if any of the questions below also yield a root in common.

D. \(x^2 - 2x - 3 = 0\)
\((x-3)(x+1) = 0\)
roots are x=3 and x=-1, no roots in common

E. \(x^2 - 10x - 5 = 0\)
doesn't factor into integers, no roots in common

Kudos if you like the explanation and/or nice formatting.

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

Substitute these roots in all the equations;

1.x^2+1=0
x=1; 1^2+1=1+1=2!=0. Not a root.
x=5; 5^2+1=25+1=26!=0. Not a root.

2.x^2-x-2=0
x=1; 1^2-1-2=-2!=0. Not a root.
x=5; 5^2-5-2=25-7=18!=0. Not a root.

3.x^2-10x-5=0
x=1; 1^2-10*1-5=-14!=0. Not a root.
x=5; 5^2-10*5-5=25-55=-30!=0. Not a root.

4.2x^2-2=0
x=1; 2*1^2-2=2-2=0=0. 1 is a root. We can stop here.
x=5; 2*5^2-2=50-2=48!=0. Not a root.

5. x^2-2x-3=0
x=1; 1^2-2*1-3=-3!=0. Not a root.
x=5; 5^2-2*5-3=12!=0. Not a root.

Ans: "D"

Let's expand the correct answer algebraically;
2x^2-2=0
2*x^2=2
x^2=1
x=+1 and x=-1

even your answer seems correct;
(2x-2)(x+1) = 0
So; 2x-2=0; x=2/2=1. "1 is also one of the roots in the equation(x^2-6x+5=0)"
x+1=0; x=-1.

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.
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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

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

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