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Question Find the solutions to the equation 4w2−3w−3=04 w^{2}-3 w-3=0. Round each solution to two decimal places and enter them as a comma-separated list.

Studdy Solution

STEP 1

Assumptions
1. The equation is a quadratic equation of the form ax2+bx+c=0ax^2 + bx + c = 0
2. The coefficients are a=4a = 4, b=−3b = -3, and c=−3c = -3
3. The solutions are real numbers

STEP 2

The quadratic formula is used to find the solutions of a quadratic equation. The formula is:
w=−b±b2−4ac2aw = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

STEP 3

Plug in the values for aa, bb, and cc into the quadratic formula.
w=−(−3)±(−3)2−4∗4∗(−3)2∗4w = \frac{-(-3) \pm \sqrt{(-3)^2 - 4*4*(-3)}}{2*4}

STEP 4

Simplify the equation.
w=3±9+488w = \frac{3 \pm \sqrt{9 + 48}}{8}

STEP 5

Calculate the value under the square root.
w=3±578w = \frac{3 \pm \sqrt{57}}{8}

STEP 6

Now, separate the equation into two parts to find the two solutions.
w1=3+578w_1 = \frac{3 + \sqrt{57}}{8} w2=3−578w_2 = \frac{3 - \sqrt{57}}{8}

STEP 7

Calculate the approximate value of w1w_1 and w2w_2, rounding to two decimal places.
w1≈3+7.558=1.32w_1 \approx \frac{3 + 7.55}{8} = 1.32 w2≈3−7.558=−0.57w_2 \approx \frac{3 - 7.55}{8} = -0.57
So, the solutions to the equation are approximately w=1.32,−0.57w = 1.32, -0.57.

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