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Question Find the solutions to the quadratic equation 1=12x2+11x1=12x^2+11x. Round answers to two decimal places and separate with a comma.

Studdy Solution

STEP 1

1. The equation 1=12x2+11x1=12x^2+11x is a quadratic equation and can be graphed as a parabola.
2. The solutions to the equation are the x-values where the graph of the equation crosses the x-axis.
3. The graph can be sketched or plotted using a graphing tool or by calculating key points, such as the vertex and x-intercepts.
4. The solutions should be rounded to two decimal places.

STEP 2

1. Rewrite the equation in standard form.
2. Determine the key points of the parabola, including the vertex and x-intercepts.
3. Sketch the graph of the parabola.
4. Identify the x-values where the parabola crosses the x-axis (the solutions).
5. Round the solutions to two decimal places.

STEP 3

Rewrite the equation in standard form by subtracting 1 from both sides to set the equation to zero.
0=12x2+11x−1 0 = 12x^2 + 11x - 1

STEP 4

Calculate the vertex of the parabola using the formula x=−b2ax = -\frac{b}{2a}, where a=12a = 12 and b=11b = 11.
xvertex=−112⋅12 x_{\text{vertex}} = -\frac{11}{2 \cdot 12}

STEP 5

Simplify the expression to find the x-coordinate of the vertex.
xvertex=−1124 x_{\text{vertex}} = -\frac{11}{24}

STEP 6

Substitute xvertexx_{\text{vertex}} into the original equation to find the y-coordinate of the vertex.
yvertex=12(−1124)2+11(−1124)−1 y_{\text{vertex}} = 12\left(-\frac{11}{24}\right)^2 + 11\left(-\frac{11}{24}\right) - 1

STEP 7

Simplify the expression to find the y-coordinate of the vertex.
yvertex=12(121576)−12124−1 y_{\text{vertex}} = 12\left(\frac{121}{576}\right) - \frac{121}{24} - 1

STEP 8

Further simplify the expression to get the exact value of yvertexy_{\text{vertex}}.
yvertex=2.42−5.04−1 y_{\text{vertex}} = 2.42 - 5.04 - 1

STEP 9

Complete the calculation to find the y-coordinate of the vertex.
yvertex=−3.62 y_{\text{vertex}} = -3.62

STEP 10

Sketch the graph of the parabola using the vertex and the direction of the parabola (which opens upwards since a=12>0a=12>0).
Note: This step assumes the use of graphing software or paper and pencil. The actual sketch is not provided here.

STEP 11

Identify the x-values where the parabola crosses the x-axis by finding the roots of the equation 0=12x2+11x−10 = 12x^2 + 11x - 1.
Note: This can be done by using the quadratic formula, factoring (if possible), or graphically finding the intercepts.

STEP 12

Use the quadratic formula to find the x-intercepts:
x=−b±b2−4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
where a=12a = 12, b=11b = 11, and c=−1c = -1.

STEP 13

Calculate the discriminant b2−4acb^2 - 4ac.
Δ=112−4(12)(−1) \Delta = 11^2 - 4(12)(-1)

STEP 14

Simplify the discriminant.
Δ=121+48 \Delta = 121 + 48

STEP 15

Find the value of the discriminant.
Δ=169 \Delta = 169

STEP 16

Substitute the values of aa, bb, and Δ\Delta into the quadratic formula to find the x-intercepts.
x=−11±16924 x = \frac{-11 \pm \sqrt{169}}{24}

STEP 17

Calculate the square root of the discriminant.
Δ=169=13 \sqrt{\Delta} = \sqrt{169} = 13

STEP 18

Find the two solutions for xx using the quadratic formula.
x1=−11+1324,x2=−11−1324 x_1 = \frac{-11 + 13}{24}, \quad x_2 = \frac{-11 - 13}{24}

STEP 19

Simplify the expressions to find the exact values of the x-intercepts.
x1=224,x2=−2424 x_1 = \frac{2}{24}, \quad x_2 = \frac{-24}{24}

STEP 20

Reduce the fractions to find the exact x-intercepts.
x1=112,x2=−1 x_1 = \frac{1}{12}, \quad x_2 = -1

STEP 21

Round the solutions to two decimal places.
x1≈0.08,x2≈−1.00 x_1 \approx 0.08, \quad x_2 \approx -1.00
The solutions are 0.08,−1.000.08, -1.00.

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