Question

find the zeros of the function $f(x) = x^2 - 6x + 4$. round values to the nearest thousandth (if necessary). answer attempt 1 out of 2

Explanation:

Step1: Recall quadratic formula

For a quadratic function \( f(x) = ax^2 + bx + c \), the zeros are found using \( x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b=-6 \), \( c = 4 \).

Step2: Calculate discriminant

Discriminant \( D=b^2 - 4ac=(-6)^2-4(1)(4)=36 - 16 = 20 \).

Step3: Apply quadratic formula

\( x=\frac{-(-6)\pm\sqrt{20}}{2(1)}=\frac{6\pm2\sqrt{5}}{2}=3\pm\sqrt{5} \).

Step4: Evaluate numerical values

\( \sqrt{5}\approx2.23607 \), so \( 3+\sqrt{5}\approx3 + 2.23607 = 5.23607\approx5.236 \), \( 3-\sqrt{5}\approx3 - 2.23607 = 0.76393\approx0.764 \).

Answer:

The zeros are \( x\approx0.764 \) and \( x\approx5.236 \)