Mysterious Quadratic Equation

By Vieta's, the sum of the roots is $2$ . Since $6$ is one of the roots, the other one is $2-6 = \boxed{-4}$

$6^2 - 2 (6) + a = 0\\ a = - 24\\ x^2 - 2 x - 24 = 0\\ (x + 4)(x - 6) = 0$ So $x = -4$ or $x = 6$

The other root is -4

plug in 6 into x, and solve for a we get that a=-24. Plug in this value of a into the original equation and solve for x we get x=-4 or 6. Therefore -4 is the answer

From the Gerenal equation of quadratic equation ax^2+bx +c =0 From the rules in finding the roots in quadratic equation... The sum of two roots r1 + r2 =-b/a and the products of two roots r1.r2=c/a So you have a given equation of quadratic equation which is x^2-2x +a =0 Your a=1;b=-2; c=a one root is given let say that is r1=6 to find the 2nd root. So r1 + r2 =-b/a 6 + r2 =-(-2)/1 r2= 2-6 r2=-4 ans. To find a=>c Take the product of the roots r1.r2=c/a 6.-4=c=>a a=-24 So the equation would be x^2 -2x -24=0 To check by factoring the equation you have( x-6)(×+4)=0 X1=6 which is the first root X2=-4 which is the second root.

we know that x^2+2x+1 has a line of symmetry at x=1 and that changing the last coefficient won't change the line of symmetry so if one root of x^2+2x+a is 6 then the other is an equal distance from 1 and since 6-1=5 then the other root is 1-5 which is -4

We know that:

${ x }_{ 1 }+{ x }_{ 2 }=\frac { -b }{ a }$ And we have:

${ x }_{ 1 }=6$ $b=-2$ $a=1$ So:

${ x }_{ 2 }=\frac { -b }{ a } -{ x }_{ 1 }$ ${ x }_{ 2 }=\frac { 2 }{ 1 } -6$ $\boxed{{ x }_{ 2 }=-4}$

(x-6)(x+4) = 0; So, x - 6 = 0 & x + 4=0; x=6 & x=-4

First, use 6 to substitute x and find the value of a

Then 6^2-2x6+a=0

36-12+a=0

24+a=0

a=-24

Use the quadratic formula to find the other root

......(Can't type radicals)

The other root is -4

(6)^2 - 2 (6) + a = 0

a = - 24

x^2 - 2 x - 24 = 0

(x + 4)(x - 6) = 0

x = -4 or x = 6

The other root is -4

Knowing that 6 is one of the roots, you can plug it into the equation and solve for a first. After that, solve this quadratic equation.