Answer:
x = 3 + √6 ; x = 3 - √6 ;
; ![x = \frac{2-(3)\sqrt{2}}{2}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B2-%283%29%5Csqrt%7B2%7D%7D%7B2%7D)
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}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B-b%5Cpm%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D)
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)}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B-%28-6%29%5Cpm%5Csqrt%7B%28-6%29%5E2-4%5Ctimes1%5Ctimes%283%29%7D%7D%7B2%5Ctimes%281%29%7D)
or
![x = \frac{6\pm\sqrt{36-12}}{2}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B6%5Cpm%5Csqrt%7B36-12%7D%7D%7B2%7D)
or
and, x = ![x = \frac{6-\sqrt{24}}{2}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B6-%5Csqrt%7B24%7D%7D%7B2%7D)
or
and, x = ![x = \frac{6-2\sqrt{6}}{2}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B6-2%5Csqrt%7B6%7D%7D%7B2%7D)
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)}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B-%28-4%29%5Cpm%5Csqrt%7B%28-4%29%5E2-4%5Ctimes2%5Ctimes%28-7%29%7D%7D%7B2%5Ctimes%282%29%7D)
or
![x = \frac{4\pm\sqrt{16+56}}{4}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B4%5Cpm%5Csqrt%7B16%2B56%7D%7D%7B4%7D)
or
and, x = ![x = \frac{4-\sqrt{72}}{4}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B4-%5Csqrt%7B72%7D%7D%7B4%7D)
or
and, x = ![x = \frac{4-\sqrt{2^2\times3^2\times2}}{4}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B4-%5Csqrt%7B2%5E2%5Ctimes3%5E2%5Ctimes2%7D%7D%7B4%7D)
or
and, x = ![x = \frac{4-(2\times3)\sqrt{2}}{4}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B4-%282%5Ctimes3%29%5Csqrt%7B2%7D%7D%7B4%7D)
or
and, ![x = \frac{2-(3)\sqrt{2}}{2}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B2-%283%29%5Csqrt%7B2%7D%7D%7B2%7D)
Hence, the possible roots are
x = 3 + √6 ; x = 3 - √6 ;
; ![x = \frac{2-(3)\sqrt{2}}{2}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B2-%283%29%5Csqrt%7B2%7D%7D%7B2%7D)
Answer:
Square roots are when you multiply the same number by itself like for example, the square root of 5 is 25, because 5*5=25
Step-by-step explanation:
True or False: The relation {(3, 2), (1, 2), (7, –5), (11, 6), (17, –4), (13, 8)} is a function.
8090 [49]
Answer:
False
Step-by-step explanation:
This is because if you have a function which is Y=2x-4 , then trying to give an example using the given points which (3,2) , here we will have Y=2 upon replacing 3 in the function cause 3 is the value of x and vice versa if we have 2 as the value of Y then x will be 3 . However,if we get the (1,2) the answer for Y will be -2 which is different from the one we have in the question hence this concludes false
Answer:
its either a or b but like im pretty sure its A
Step-by-step explanation:
Answer:
EH = 10
Step-by-step explanation:
Because F is a midpoint, that means that FG and EF are congruent to each other. So, since FG = 2, that means that EF also equals 2. Because FH also includes FG, you can take the 2 from the FG, and then take that 2 away from 8. Now you know that GH = 6. All that is left is to add all 3 of the units together (2+2+6=10).