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Find the zeros of the function f(x)=1.1x^(2)+4x+0.7. Round values to the nearest hundredth (if necessary). Answer: x=

Find the zeros of the function f(x)=1.1x2+4x+0.7 f(x)=1.1 x^{2}+4 x+0.7 . Round values to the nearest hundredth (if necessary).\newlineAnswer: x= x=

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Q. Find the zeros of the function f(x)=1.1x2+4x+0.7 f(x)=1.1 x^{2}+4 x+0.7 . Round values to the nearest hundredth (if necessary).\newlineAnswer: x= x=
  1. Identify type of equation: Identify the type of equation.\newlineWe have a quadratic equation in the form of ax2+bx+c=0ax^2 + bx + c = 0, where a=1.1a = 1.1, b=4b = 4, and c=0.7c = 0.7.
  2. Apply quadratic formula: Apply the quadratic formula to find the zeros of the function.\newlineThe quadratic formula is x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. We will use this formula to find the values of xx that make f(x)=0f(x) = 0.
  3. Substitute values into formula: Substitute the values of aa, bb, and cc into the quadratic formula.x=−4±42−4⋅1.1⋅0.72⋅1.1x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1.1 \cdot 0.7}}{2 \cdot 1.1}
  4. Simplify discriminant: Simplify the expression under the square root (the discriminant).\newlineDiscriminant = 42−4×1.1×0.7=16−3.08=12.924^2 - 4\times 1.1\times 0.7 = 16 - 3.08 = 12.92
  5. Calculate square root: Calculate the square root of the discriminant. 12.92≈3.595\sqrt{12.92} \approx 3.595
  6. Continue with quadratic formula: Continue with the quadratic formula using the value of the square root. x=−4±3.5952.2x = \frac{{-4 \pm 3.595}}{{2.2}}
  7. Solve for possible values: Solve for the two possible values of xx.x1=−4+3.5952.2≈−0.184x_1 = \frac{{-4 + 3.595}}{{2.2}} \approx -0.184x2=−4−3.5952.2≈−3.452x_2 = \frac{{-4 - 3.595}}{{2.2}} \approx -3.452
  8. Round values to nearest hundredth: Round the values of xx to the nearest hundredth.x1≈−0.18x_1 \approx -0.18x2≈−3.45x_2 \approx -3.45

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