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Question Solve the quadratic equation 2x2−8x−6=−92x^2 - 8x - 6 = -9 and round the solution to the nearest tenth.

Studdy Solution

STEP 1

Assumptions
1. The equation is 2x2−8x−6=−92x^{2} - 8x - 6 = -9
2. The solution should be to the nearest tenth

STEP 2

First, we need to rearrange the equation to a standard quadratic form, ax2+bx+c=0ax^{2} + bx + c = 0. We can do this by adding 9 to both sides of the equation.
2x2−8x−6+9=−9+92x^{2} - 8x - 6 + 9 = -9 + 9

STEP 3

Simplify the equation.
2x2−8x+3=02x^{2} - 8x + 3 = 0

STEP 4

Now, we can use the quadratic formula to solve for x. The quadratic formula is given by:
x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}

STEP 5

Identify the values of a, b, and c from the equation 2x2−8x+3=02x^{2} - 8x + 3 = 0. Here, a = 2, b = -8, and c = 3.

STEP 6

Substitute the values of a, b, and c into the quadratic formula.
x=−(−8)±(−8)2−4∗2∗32∗2x = \frac{-(-8) \pm \sqrt{(-8)^{2} - 4*2*3}}{2*2}

STEP 7

Simplify the equation.
x=8±64−244x = \frac{8 \pm \sqrt{64 - 24}}{4}

STEP 8

Continue to simplify the equation.
x=8±404x = \frac{8 \pm \sqrt{40}}{4}

STEP 9

The square root of 40 is approximately 6.3 (to the nearest tenth). Substitute this value back into the equation.
x=8±6.34x = \frac{8 \pm 6.3}{4}

STEP 10

Now, solve for the two possible values of x.
x=8+6.34orx=8−6.34x = \frac{8 + 6.3}{4} \quad or \quad x = \frac{8 - 6.3}{4}

STEP 11

Calculate the two possible values of x.
x=14.34≈3.6orx=1.74≈0.4x = \frac{14.3}{4} \approx 3.6 \quad or \quad x = \frac{1.7}{4} \approx 0.4
So, the solutions to the equation 2x2−8x−6=−92x^{2} - 8x - 6 = -9 to the nearest tenth are x≈3.6x \approx 3.6 and x≈0.4x \approx 0.4.

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