How to use the factored form of a quadratic function (SAT math)

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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#math #algebra #mathbasics

[jstarks]

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.

[zey]

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.

[trumpgaming]

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.

[ianfowler]

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.

[soyanshumohapatra]

In short, -b/a = 7-3 = 4
Or, b = -4a
So, a+b = a-4a = -3a

Since a>1, so only option is *A*

[niceboiiz]

Well done you are a good teacher 👍👍👍👍👍

[tobybartels]

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.

[okaro]

-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.

[tethys]

could you do a higher maths paper from the SQA in scotland? Its like the scottish version of the a-level paper u done

[sinekavi]

I did it a little bit differently
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?

[edwin]

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

[leonardobarrera]

Don’t ignore it
My other comment
That you reply me

[Kyrelel]

Please stop using "negative" instead of "minus" when writing down an equation.
Better yet, stop using it at all.