SAT math: a quadratic problem
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- Опубликовано: 12 май 2024
- The function f(x)=ax^2+bx+c, where a is a positive integer greater than one, contains the points (7,0) and (-3,0). We would like to know a+b could be which of the following?
(A) -6 (B) -3 (C) 4 (D) 5
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The roots are 7 and -3, so the axis of symmetry is x = 2 and therefore -b/2a = 2. Isolating b, we get b = -4a. So a + b = a + -4a = -3a. Plugging in 2 for “a” (since it’s the smallest positive integer greater than 1) gives a result of -6.
Ah this way is much cleaner!
@@bprpmathbasics thanks for the reply! By the way…love your channel!
actually, even the given "a is an integer" is not needed, only that a>1 is enough to exclude all three wrong answers, since a+b cannot be a positive number as a+b = -3a while "a" is a positive number itself, being larger than 1, thus C. an D. are automatically incorrect. The larger than one is needed to exclude the answer B. as this answer needs "a" to be exactly one.
a + b should be -9a, not -3a. If you plug (7,0) and (-3,0) into the f(x) = ax^2 + bx + c, you get a(7)^2 + b(7) + c = 0 and a(-3)^2 + b(-3) + c = 0. This gives you 49a + 7b + c = 0 and 9a - 3b + c = 0. If you multiply the 2nd equation by negative 1 and do elimination, you get 40a + 4b = 0. Or 10a + b = 0. This can be written as a + b = -9a.
@@maxhagenauer24 you've a mistake, when you multiply 9a -3b +c = 0 by negative 1 you get -9a PLUS 3b - c = 0, and when you add to first equation you have +7b PLUS 3b which is +10b, not +4b. Thus 4a + b =0
@z000ey Ohhh I see, thank you. I'm stupid for messing up on that.
@@maxhagenauer24 np, I did those mistakes myself millions of times :)
@@maxhagenauer24 ..No..ur not stupid at all for doing that ... its very easy to miss .. and it tought u to be more carefull, if u learned something or gained new knowlege its a win!..
Yeah essentially vieta's formulae used here, two unknowns (a and b) with two equations (a+b and -b/a) which are both linearly independent.
In short, -b/a = 7-3 = 4
Or, b = -4a
So, a+b = a-4a = -3a
Since a>1, so only option is *A*
Technically it doesn't end there since the answer could have been "none of them could be a+b", since you'd have to check whether you'd get a value of c that gives the output of the function as 0 for both 7 & -3.
@@trumpgaming5998that’s isn’t a choice?
@@CyberFlare-fn9kn it's not a choice, you have to do it
You could also sub in the 2 points and then subtract to eliminate c. You get 4a+b = 0. Next check each choice:
A) b = -6-a, B) b=-3-a, C) b=4-a D)b=5-a and A) is the only choice which gives a an integer>1. Not saying it's better just slightly different.
Well done you are a good teacher 👍👍👍👍👍
Great videos!
-b / 2a = 2 => b/a = -4 => b=-4a => a+b = -3a, as a>=2, is -6 the only option.
I first somehow in mind rotated the roots to -7 and 3 and then got that a=1 and was stuck.
Yeah, even faster, although dependent on how the question was put, giving the 2 intersections of the x-axis immediately, it's easy to see the midpoint being at x=2. Had not both points where y=0 been given, yopu'd have to plug in the x and y variables into the full quadratic.
While it is quicker to use properties of polynomials (even quadratic polynomials), this can also be done in a more brute-force way. Put the coordinates of the two given points into the equation to get 49a+7b+c=0 and 9a−3b+c=0. We don't care about c, so subtract these equations to eliminate it: 40a+10b=0 (or you could solve one equation for c and substitute into the other, or solve both equations for c and set them equal, whatever comes most naturally to you). Now we can solve for b to get b=−4a and continue as in the video.
could you do a higher maths paper from the SQA in scotland? Its like the scottish version of the a-level paper u done
I can look into it!
I did it a little bit differently though......
Since y=f(x)=ax^2+bx+c, (7,0) and (-3,0) are the solutions, I just substituted 0 in place of f(x) and 7 and -3 instead of x and got two linear equations in 3 variables.
The equations are 49a+7b+c=0 and 9a-3b+c=0.
Solving both I got 4a+b=0 and by given condition a>1 and is an integer I substituted a=2 in 4a+b=0.
Doing that I got b= -8. So a+b= -8+2= -6. Is this a right method or not BPRP?
Oh hey! That works, too!
why is f(x)=a(x-r1)(x-r2)?
That’s the factored form of a quadratic function
maybe a more difficult way but we can get -3a from a linear system
49a + 7b +c = 0
9a - 3b + c = 0
solving it we get
4a + b = 0
b = -4a
a - 4a = 3a
That one is not more difficult, it's easier, but the easiest one is just to see that both point given are the two possible intersections and from the x's get that the inflexion point is exactly where x is the midpoint of the two intersections, thus it is at x=2, and then as okaroo6595 pointed use -b/2a=2 => a+b=-3a
a-4a is not 3a, you made a mistake.
yeah -3a
How to approximate radicals easily??
One trick is to find the nearest number that has a rational root, and use a tangent line to find it.
For instance, consider cbrt(9). We can calculate cbrt(8) mentally, as 2. So we'll use that as a reference point.
Find the slope of cbrt(x) at x=8.
d/dx cbrt(x) = d/dx x^(1/3) = 1/3*x^(-2/3)
Plug in x = 8:
1/3*27^(-2/3) = 1/(3*2^2) = 1/12
Thus, the tangent line is:
L(x) = 1/12*(x - 8) + 2
Plug in x = 9:
L(9) = 2 + 1/12 is approx 2.08333
Actual cbrt(9) = 2.08008
@@carultch oh thank youuu!!
Is it good to plus the number with the nearest number under radical and devide to the radical multiplied by two?
Radical 5= 5+4/2(2)=9/4=2.5
@@borushiki1464 9/4 is 2.25, but yes, the idea is the important part.
There are more efficient methods, but this is the easiest to do by hand, to get a first order approximation.
Don’t ignore it
My other comment
That you reply me
Hello
Please stop using "negative" instead of "minus" when writing down an equation.
Better yet, stop using it at all.
no 😊