Question

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

Answer 1

Answer:

x = 3 + √6 ; x = 3 - √6 ; $x = \frac{2+3\sqrt{2}}{2}$ ;  $x = \frac{2-(3)\sqrt{2}}{2}$

Step-by-step 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:

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

for the equation ax² + bx + c = 0

thus,

the roots are

$x = \frac{-(-6)\pm\sqrt{(-6)^2-4\times1\times(3)}}{2\times(1)}$

or

$x = \frac{6\pm\sqrt{36-12}}{2}$

or

$x = \frac{6+\sqrt{24}}{2}$ and, x = $x = \frac{6-\sqrt{24}}{2}$

or

$x = \frac{6+2\sqrt{6}}{2}$ and, x = $x = \frac{6-2\sqrt{6}}{2}$

or

x = 3 + √6 and x = 3 - √6

similarly for (2x² − 4x − 7) = 0.

we have

the roots are

$x = \frac{-(-4)\pm\sqrt{(-4)^2-4\times2\times(-7)}}{2\times(2)}$

or

$x = \frac{4\pm\sqrt{16+56}}{4}$

or

$x = \frac{4+\sqrt{72}}{4}$ and, x = $x = \frac{4-\sqrt{72}}{4}$

or

$x = \frac{4+\sqrt{2^2\times3^2\times2}}{2}$ and, x = $x = \frac{4-\sqrt{2^2\times3^2\times2}}{4}$

or

$x = \frac{4+(2\times3)\sqrt{2}}{2}$ and, x = $x = \frac{4-(2\times3)\sqrt{2}}{4}$

or

$x = \frac{2+3\sqrt{2}}{2}$ and, $x = \frac{2-(3)\sqrt{2}}{2}$

Hence, the possible roots are

x = 3 + √6 ; x = 3 - √6 ; $x = \frac{2+3\sqrt{2}}{2}$ ; $x = \frac{2-(3)\sqrt{2}}{2}$