Hello candidate,

The best quadratic expression is to expanding it by multiplying both the terms which gives-

x^2+ x -2/ x+3, so the range of the quadratic equations is on the General range of all Real Numbers.

Hope that this information helps you!!

Hello!

Can you please check the question once again incase if the equation is x^2+6x+7

Then it would be like

x^2+6x-7=0

x^2+7x-x-7=0

x(x+7)-1(x+7)=0

(x+7)(x-1)=0

x = -7, 1

Incase the equation is right then check whether this method is a solution x^2+6x+7

x(x+6)+7 = 0              (taking x to the other side) (anything divided by 0 is 0)

(x+6)+7 = 0

x+6+7 = 0

x+13 = 0

x= -13

ALL THE BEST. Hope the answer was of some help.

Thank You

PROOF:-

As we know that;

x^1= x :- (LET it be :-1st series)

x^2= x^2 :-(LET it be :-2nd series) and,

x^3= x^3:- (LET it be :-3rd series)

Examining the above three series you can see that the product of all the three series is differ by x or you can say they have the common difference of the value x for your better understanding I am elaborating the above series:-

x ,x^2 , x^3(This are all the values of the above series and you can see that it differ by 'x' this means that the given series is in A.P)

Therefore, the value just before x will be 1 because ,

x^0 = x^(1 - 1)

x^0 = [x^1] x [x^(-1)]

x^0 = x  (1/x) [since x^1 = x and x^(-1) = 1/x]

x^0 = x / x

x^0 = 1

Therefore, X= 1 Or X^0= 1

Hence proved

Hope it helps!!