Becoming a calculator jockey is an under-rated strategy. A TI-84 Plus CE has functions in the APPS menu that bring high scores well within reach. You don’t need to be a math guru, just understand how to use the allowed tools.
Another very easy solution for question 2 I think would be to know that the Product = c / a so because c = ab and a = 57 , c/a = ab/57 = kab . The ab’s cancel out and we’re left with 1/57 = k
A shortcut for the second problem is to have the what's called "Vieta's Formula's" memorized, which are really simple: the product of the solutions (roots) of a quadratic ax^2 + bx + c is c/a, and the sum of the roots is -b :D
I solved these problems like this: 92) Use Discriminant D=0; b^2-4ac = 0. Solving for a gives a = 14.5 95) Use x1*x2 = c/a = ab/57; k= 1/57. 98) V(9/-14); f(x) = a(x-9)^2 - 14 = ax^2 - 18ax + 81a - 14; a+b+c = 64a - 14. Since the parabola touches the x-axis twice and yV = -14, the limit as n->∞ is ∞, and, thus, a>0, from which follows that 64a - 14 > -14. The solution, hence, must be -12.
@@azretomarrisbay8925 Admittedly, my notation was plain wrong there, sorry about that. I should have written a new equation after f(x) = ax^2 -18ax +81a -14. The polynomial equality theorem for polynomials says that if two polynomials are equal to each other for all x∈𝔻, that the coefficients cn of each degree n of x, where n∈ℕ0, have to be equal as well (you don't really need to know this theorem, it's intuitive that the coefficients must be equal). We know that f(x)=ax^2 + bx + c and we derived that f(x)=ax^2 -18ax + 81a - 14 using the vertex form. The vertex and standard form are equivalent, so that ax^2 + bx + c = ax^2 -18ax + 81a -14. Now apply the polynomial equality theorem, from which follows that a=a, b=18a, c=81a-14. It immediately follows that a+b+c= 64a -14. The parabola is open upwards (you realize this when visualizing the vertex below the x-axis and the fact the graph touches the x-axis twice, the limit was a mere notational formality), so a>0. Thus, 64a-14 can never be smaller than -14. We, hence, get (D) -12 as our only option.
The first one isn’t that bad really. If it says that it only has one solution then it means that the lines only hit once. For that to happen y=-1.5 (the straight line) has to be just touching the bottom part of the parabola (height at vertex). We know that the vertex is x= -b/2a, and height y = f(-b/2a). From there its making the line’s height = to the vertex height.
If you know a little bit of calculus, you don't even need desmos. y = -1.5 y = x^2 + 8x + a you can find the derivative of the second equation then set that equal to 0. then solve for x and that will give you the critical point. then insert that solution into x^2 + 8x + a = -1.5 then solve for a . and you've got it.
not trying to hate but you should stick to your eng the last explanation was just so confusing Also it is much easier to do the question in this example by your own ure gonna save a lot of time
Are y'all dump in America? That algebraic solutions are literally what every modern freshman in Kazakhstan would be shameful of not knowing. And it's Kazakhstan, man. Can't imagine what the Russian or Chinese students thinking about this.
Here's an alternative solution to the last problem: a+b+c is f(1). Since the parabola is open upwards, (9, -14) is the minimum value of f(x). Thus, f(1) (whatever it is) must be greater than 14 i.e. a+b+c > -14 Only choice d fulfills this criteria.
That is mad I am blown away by this method I happened to try this on my own and landed at College Board's explanation but I also thought of this Cause Ik that 2a = 2nd diff 3a +b = 1st diff and a + b + c = 1st Term Thanks Alot
I have my exams on Oct 7 🥹 And I am currently hitting only 640 in Maths and 630 in English .I need to improve at least 80 points each in the next 27 days 🙂 Totally scared
Thank you for what you do to help us, Laura. I've been watching your videos for a while now and I really appreciate what you do. But, I have a question on the second question. Why exactly did we set a and b as 1?
usa maths is crazy, they're not testing your actual mathematical abilities at all, all of these questions are stupidly easy, they're just testing speed/ tricks atp
Her answer is right, but the method is not. Instead what you can do is write the equation in the form of a(x+b/2a)^2 + c - a*(b/2a)^2 (Squared form to find the vertex) and you have the values where b/2a=-9 And c - a*(b/2a)^2 = -14 Now find b and c in terms of a Substitute in a + b + c to get an equation in a You will get 64a-14 = ... To get a +ve value for "a". The ... (which is the answer) has to be lesser than 14, since they are all -ve values. Hence D is the answer
Hey, i have never used the desmos and my exam is on saturday. Do you advise that i should just stick with using my calculator or get familiar with the desmos?
Hi random person, how did it end up going? In the same situation I'm wondering if the best is to use desmos as a last resort if I have no idea what to do.
@@allisonturner5571hi i saw some tutorials on how to use desmos for dsat! if you’re taking the august exam i suggest you to watch them, they’re very helpfullll! (Edit: the videos are so short and easy to understand so watching them may help since there is a short period of time left)
for the last question would it also work if I add the minimum plus the x intercept plus the y intercept? It also gives 12 but I don't know if that is a correct way to solve.
Please I want to buy your course but is only English I want to buy because I can't afford both with the price given but I really want to buy the English
