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Solve the equation x^(2)+17 x+45=0 to the nearest tenth. Answer: x=

Solve the equation x2+17x+45=0 x^{2}+17 x+45=0 to the nearest tenth.\newlineAnswer: x= x=

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Full solution

Q. Solve the equation x2+17x+45=0 x^{2}+17 x+45=0 to the nearest tenth.\newlineAnswer: x= x=
  1. Identify the quadratic equation: Identify the quadratic equation.\newlineThe given equation is x2+17x+45=0x^2 + 17x + 45 = 0, which is a quadratic equation in the standard form ax2+bx+c=0ax^2 + bx + c = 0, where a=1a = 1, b=17b = 17, and c=45c = 45.
  2. Apply the quadratic formula: Apply the quadratic formula to find the solutions for xx. The quadratic formula is x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. For our equation, a=1a = 1, b=17b = 17, and c=45c = 45.
  3. Calculate the discriminant: Calculate the discriminant b2−4acb^2 - 4ac.\newlineDiscriminant = b2−4ac=(17)2−4(1)(45)=289−180=109b^2 - 4ac = (17)^2 - 4(1)(45) = 289 - 180 = 109.
  4. Calculate the two solutions: Calculate the two solutions using the quadratic formula.\newlinex=−b±discriminant2ax = \frac{-b \pm \sqrt{\text{discriminant}}}{2a}\newlinex=−17±1092x = \frac{-17 \pm \sqrt{109}}{2}
  5. Calculate the first solution: Calculate the first solution (using the '+' in '±\pm').\newlinex1=−17+1092x_1 = \frac{-17 + \sqrt{109}}{2}\newlinex1≈−17+10.44032x_1 \approx \frac{-17 + 10.4403}{2}\newlinex1≈−6.55972x_1 \approx \frac{-6.5597}{2}\newlinex1≈−3.27985x_1 \approx -3.27985\newlineRound to the nearest tenth: x1≈−3.3x_1 \approx -3.3
  6. Calculate the second solution: Calculate the second solution (using the '-' in '±\pm').\newlinex2=−17−1092x_2 = \frac{{-17 - \sqrt{109}}}{{2}}\newlinex2≈−17−10.44032x_2 \approx \frac{{-17 - 10.4403}}{{2}}\newlinex2≈−27.44032x_2 \approx \frac{{-27.4403}}{{2}}\newlinex2≈−13.72015x_2 \approx -13.72015\newlineRound to the nearest tenth: x2≈−13.7x_2 \approx -13.7

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