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Using a suitable straight line, solve 2x2 − 2x − 1 = 0 given the graph of y = 2x2 − x − 1.

Mathematics

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Graph of y = 2x² − x − 1

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Key points of y = 2x² − x − 1
Pointxy
x-intercept−0.50
y-intercept0−1
Minimum0.25−1.12
x-intercept10

Step-by-step solution

4 steps

The image is still quite blurry, but I will continue solving the graphing problem (Question 20) from where we left off. I previously solved the first part of the "using a suitable straight line" section. I will now solve the second part, which I will label as (b).

Question 20 (continued):
Given the graph of y=2x2−x−1y = 2x^2 - x - 1.

b) Using a suitable straight line, solve 2x2−2x−1=02x^2 - 2x - 1 = 0.


b) Solve 2x2−2x−1=02x^2 - 2x - 1 = 0 using the graph of y=2x2−x−1y = 2x^2 - x - 1.

  1. Identify the given graph equation and the equation to be solved.

    Given graph: y=2x2−x−1y = 2x^2 - x - 1
    Equation to solve: 2x2−2x−1=02x^2 - 2x - 1 = 0

  2. Manipulate the equation to be solved to relate it to the given graph equation.

    We want to express 2x2−2x−1=02x^2 - 2x - 1 = 0 in terms of y=2x2−x−1y = 2x^2 - x - 1.
    Rewrite 2x2−2x−1=02x^2 - 2x - 1 = 0 as:

    (2x2−x−1)−x=0(2x^2 - x - 1) - x = 0

    Substitute yy for (2x2−x−1)(2x^2 - x - 1):

    y−x=0y - x = 0 y=xy = x

    This is the equation of the suitable straight line.

  3. Explain the graphical solution.

    To solve 2x2−2x−1=02x^2 - 2x - 1 = 0 graphically, you would draw the straight line y=xy = x on the same axes as the graph of y=2x2−x−1y = 2x^2 - x - 1. The solutions for xx are the x-coordinates of the points where the straight line intersects the parabola.

  4. Solve the equation algebraically to find the exact values (as a graph would provide approximations).

    Using the quadratic formula x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} for 2x2−2x−1=02x^2 - 2x - 1 = 0 (where a=2a=2, b=−2b=-2, c=−1c=-1):

    x=−(−2)±(−2)2−4(2)(−1)2(2)x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(2)(-1)}}{2(2)} x=2±4+84x = \frac{2 \pm \sqrt{4 + 8}}{4} x=2±124x = \frac{2 \pm \sqrt{12}}{4} x=2±234x = \frac{2 \pm 2\sqrt{3}}{4} x=1±32x = \frac{1 \pm \sqrt{3}}{2}

    The solutions are x=1+32x = \frac{1 + \sqrt{3}}{2} and x=1−32x = \frac{1 - \sqrt{3}}{2}.
    Approximately, x≈1+1.7322≈1.366x \approx \frac{1 + 1.732}{2} \approx 1.366 and x≈1−1.7322≈−0.366x \approx \frac{1 - 1.732}{2} \approx -0.366.

    The suitable straight line is y=x\boxed{y = x}.
    The solutions are x=1±32\boxed{x = \frac{1 \pm \sqrt{3}}{2}}.

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