Question

Find all real solutions to the equation (x² − 6x +3)(2x² − 4x − 7) = 0.

Answer 1

x = 3 + √6 ; x = 3 - √6 ;
`x = (2+3√(2))/(2)` ;
`x = (2-(3)√(2))/(2)`

Explanation:

Relation given in the question:

(x² − 6x +3)(2x² − 4x − 7) = 0

Now,

for the above relation to be true the following condition must be followed:

Either (x² − 6x +3) = 0 ............(1)

or

(2x² − 4x − 7) = 0 ..........(2)

now considering the equation (1)

(x² − 6x +3) = 0

the roots can be found out as:

`x = (-b\pm√(b^2-4ac))/(2a)`

for the equation ax² + bx + c = 0

thus,

the roots are

`x = (-(-6)\pm√((-6)^2-4*1*(3)))/(2*(1))`

or

`x = (6\pm√(36-12))/(2)`

or

`x = (6+√(24))/(2)` and, x =
`x = (6-√(24))/(2)`

or

`x = (6+2√(6))/(2)` and, x =
`x = (6-2√(6))/(2)`

or

x = 3 + √6 and x = 3 - √6

similarly for (2x² − 4x − 7) = 0.

we have

the roots are

`x = (-(-4)\pm√((-4)^2-4*2*(-7)))/(2*(2))`

or

`x = (4\pm√(16+56))/(4)`

or

`x = (4+√(72))/(4)` and, x =
`x = (4-√(72))/(4)`

or

`x = (4+√(2^2*3^2*2))/(2)` and, x =
`x = (4-√(2^2*3^2*2))/(4)`

or

`x = (4+(2*3)√(2))/(2)` and, x =
`x = (4-(2*3)√(2))/(4)`

or

`x = (2+3√(2))/(2)` and,
`x = (2-(3)√(2))/(2)`

Hence, the possible roots are

x = 3 + √6 ; x = 3 - √6 ;
`x = (2+3√(2))/(2)` ;
`x = (2-(3)√(2))/(2)`