Solve for all real values of x. x^2+6=0
2 answers:
Hey there !
There are not atleast 1 real root of the given polynomial see the attachment for full explanation .
Also You can here see that here discriminant ( D ) = b² - 4ac = ( 0 )² - 4 × 1 × 6 = - 24 < 0 . So , By using definition of quadratic equation we can say that their is no real root of the given equation .
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Multiply both sides of the equation by 12
Step-by-step explanation:
ED is greater than BC
the line from B to ED ^ (which i drew ) the point it touches ED name it X. so EX will be 2 ( ED-BC)(6-4).
then u have a triangle. EX, XB and EB.
you have length of EX(2) and u have hypotenuse. so u can calculate XB using Pythagoras theorem.
15.1²=2²+XB²
15.1²-2²=XB²
224.01=XB²
XB=14.97
since XB and DC are parallel ( a rectangle is forming XBCD) so DC is also 14.97
