Question

solve the equation $5x^2 - 11x + 4 = 0$ to the nearest tenth.

Explanation:

Step1: Identify coefficients for quadratic formula

For quadratic equation \(ax^2 + bx + c = 0\), here \(a = 5\), \(b=-11\), \(c = 4\).
Quadratic formula: \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\)

Step2: Calculate discriminant \(D = b^{2}-4ac\)

\(D=(-11)^{2}-4\times5\times4=121 - 80 = 41\)

Step3: Substitute into quadratic formula

\(x=\frac{-(-11)\pm\sqrt{41}}{2\times5}=\frac{11\pm\sqrt{41}}{10}\)

Step4: Calculate two solutions

First solution: \(x_1=\frac{11+\sqrt{41}}{10}\approx\frac{11 + 6.403}{10}=\frac{17.403}{10}=1.7403\approx1.7\) (to nearest tenth)
Second solution: \(x_2=\frac{11-\sqrt{41}}{10}\approx\frac{11 - 6.403}{10}=\frac{4.597}{10}=0.4597\approx0.5\) (to nearest tenth)

Answer:

\(x\approx1.7\) and \(x\approx0.5\)