1. cranberry_juice_01 says:

For 23, there's a couple ways you could approach it. You were so close to one of them. When you try to rationalize a denominator like that, when you multiply numerator and denominator, you have to flip the sign in the middle. So when you multiplied by x - sqrt(2), you should have multiplied by x + sqrt(2).

2. cranberry_juice_01 says:

There's one other thing I'd like to point out, and as a tutor, this is an extremely common mistake I see, so don't feel bad about it (it's poorly taught in general). You can't cancel in a fraction when things are added together. They have to be multiplied. Just remember that a fraction is just a big division, and the opposite of division is multiplication. All we do when we cancel is realize that when you multiply by a thing and then divide by that same thing, you're just back where you started.

3. Rude-Repeat-4348 says:

for number 22,

4. cranberry_juice_01 says:

For 22, all the choices are in vertex form. The general form of a quadratic in vertex form is a(x - h)2 + k, where (h, k) is the vertex. Notice that there is a minus sign in the parentheses. That means that whatever sign h has will be flipped in the parentheses.

5. P_boluri says:

In this fucking shit that I still havnt lend the English name of, you have a general for of:

6. NoAhH_1228 says:

It’s to the right so b must be negative

7. [deleted] says:

Simple. Learn the formula of the parabola in vertex form. Once you know, the question is a breeze.

8. CuriousElectrons says:

Vertex form of a quadratic equation is defined by f(x) = a(x-b)2 + c, where (b, c) is the vertex. For a parabola to have a maximum point and open downwards, the a value would have to be negative. Note that this particular parabola’s vertex point is in the first quadrant, so the x and y values of the vertex are positive.

9. Bulky_Internet1952 says:

22) for concavity of the graph we check the 2nd Derivative of any eqn in case of quadratic (y=ax²+bx+c) the 2nd derivative comes out to be the coefficient of x² hence from the given figure the parabola is open downwards which means coeff of x² must be negative I.e -ax² .Now check the x coordinate of vertex which is +ve so it should be of form -a(x-b)²[do equal to 0 you must get x=b] now the curve is c units upward (as in all option) hence our eqn is y=-a(x-b)²+c

10. Anshul7509 says:

Correct answer is D DM me for the solution

11. GnomeUnknown07 says:

For the second question, use the property x2 -a2 =(x-a) (x+a), you'll get the answer x- root(2).

12. cranberry_juice_01 says:

You cancelled the wrong factor. The answer is x + sqrt(2).

13. GnomeUnknown07 says:

Well, first off if a, b,c are positive, that means for the graph to face downwards i.e.. for the graph to open up in the downward direction like shown in the figure, the coefficient of x square will need to be negative. Thus you eliminate option 3 and 4 since they show x square having a positive coefficient.

14. meloneberry says:

you don’t need the discriminant buddy the vertex is positive meaning the b has to be negative 