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SolvePlot showing the uploaded quadratic question and explanation on the left, with Desmos data lists, a fitted parabola, and answer 143 on the right
A problem image is dragged into SolvePlot. After a brief loading indicator, SolvePlot displays the question and recommended regression method beside its automatically populated Desmos calculator. Data lists supply the points (1, 8), (2, 15), and (4, 35). The regression y₁ ~ ax₁² + bx₁ + c gives a = 1, b = 4, c = 3. Define f(x) = ax² + bx + c, then enter f(10) to get 143.

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Your problem

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3x2−16x+2=0.3 x^{2} - 16 x + 2 = 0\text{.} One solution can be written as a+b6,\frac{a + \sqrt{b}}{6}\text{,} where a and bb are integers. What is a+b?a + b\text{?}

Your solution

Answer
248
Choice C

Bracket regression

Strategy

Use the graph’s two xx-intercepts to fit a and bb in the given root form.

In Desmos

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Answer choices
  • A
    6666
  • B
    132132
  • C
    248248
  • D
    296296
Question text

3x2−16x+2=0.3 x^{2} - 16 x + 2 = 0\text{.} One solution can be written as a+b6,\frac{a + \sqrt{b}}{6}\text{,} where a and bb are integers. What is a+b?a + b\text{?}

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In Desmos

3 lines
  1. 2TypeEnter line 2
    [(a+b)/6,(a−b)/6]∼[5.20526,0.128073][(a+\sqrt{b})/6,(a-\sqrt{b})/6]\sim[5.20526,0.128073]

    Match the plus form to the larger root and the minus form to the smaller root. The ~ sign asks Desmos to fit a and bb to those two values.

Fitting both unknowns

About calculator line 2

The ~ sign tells Desmos to find values of a and bb that make the two lists match. There are two unknowns, so both roots are needed.

Why it’s used here

Line 22 matches the plus form to 5.205265.20526 and the minus form to 0.128073.0.128073\text{.} Desmos fits a≈16a \approx 16 and b≈232b \approx 232 together.

Example

The fitted values shown by Desmos are:

a≈16,b≈232a\approx16,\quad b\approx232

History and Saved tricks

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Saved tricks

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Saved tricks

2 tricks saved

  • Bracket regression

    Graph the quadratic, then fit both root forms to the two xx-intercepts.

    Review trick · 3 lines

    One root of 3x2−16x+2=03 x^{2} - 16 x + 2 = 0 is (a + √bb) / 6,6\text{,} where a and bb are integers. Find a+b.a + b\text{.}

    1. y=3x2−16x+2y=3x^2-16x+2
    2. [(a+b)/6,(a−b)/6]∼[5.20526,0.128073][(a+\sqrt{b})/6,(a-\sqrt{b})/6]\sim[5.20526,0.128073]
    3. round⁡(a)+round⁡(b)\operatorname{round}(a)+\operatorname{round}(b)
    Answer 248248View solution
  • Three-point regression

    Fit a quadratic to the three given points, then evaluate the fitted function.

    Review trick · 5 lines

    A quadratic has f(1)=8,f \left(1\right) = 8\text{,} f(2)=15,f \left(2\right) = 15\text{,} and f(4)=35.f \left(4\right) = 35\text{.} Find f(10).f \left(10\right)\text{.}

    1. x1=[1,2,4]x_1=[1,2,4]
    2. y1=[8,15,35]y_1=[8,15,35]
    3. y1∼ax12+bx1+cy_1\sim ax_1^2+bx_1+c
    4. f(x)=ax2+bx+cf(x)=ax^2+bx+c
    5. f(10)f(10)
    Answer 143143

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Your solutionAnswer 248
3x2−16x+2=03x^2-16x+2=0
Bracket regression
Similar problem1 / 2
2x2−9x+1=02x^2-9x+1=0

One root is (a+b)/4(a+\sqrt b)/4.
Find a + b.

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