Question

find the zeros of f(x) = x² + 7x + 9. options: x = 3 ± √7; x = (3 ± √7)/2; x = (-7 ± √13)/2; x = -7 ± √13

Explanation:

Step1: Recall Quadratic Formula

For a quadratic equation \(ax^2 + bx + c = 0\), the solutions (zeros) are given by \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\).

Step2: Identify \(a\), \(b\), \(c\)

In the function \(f(x)=x^2 + 7x + 9\), we have \(a = 1\), \(b = 7\), \(c = 9\).

Step3: Calculate Discriminant

Discriminant \(D=b^2 - 4ac=(7)^2-4(1)(9)=49 - 36 = 13\).

Step4: Apply Quadratic Formula

Substitute \(a = 1\), \(b = 7\), \(D = 13\) into the formula: \(x=\frac{-7\pm\sqrt{13}}{2}\). Wait, no, wait: Wait, the function is \(x^2+7x + 9\)? Wait, no, wait the user's function: Wait, the image says \(f(x)=x^2 + 7x + 9\)? Wait, no, wait the options: Wait, maybe I misread. Wait, no, let's check again. Wait, the quadratic formula is \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). If the function is \(x^2 - 3x + 1\)? Wait, no, the options have \(3\pm\sqrt{7}\) over 2. Wait, maybe the function is \(x^2 - 3x + 1\)? Wait, no, the image's problem: Wait, the user's problem: "Find the zeros of \(f(x)=x^2 + 7x + 9\)"? Wait, no, maybe a typo. Wait, let's recalculate. Wait, if the function is \(x^2 - 3x + 1\), then \(a = 1\), \(b=-3\), \(c = 1\). Then discriminant is \(9 - 4 = 5\)? No. Wait, the options have \(3\pm\sqrt{7}\) over 2. Let's see: If \(a = 1\), \(b=-3\), \(c=-1\)? No. Wait, maybe the function is \(x^2 - 3x + 1\)? Wait, no, let's check the options. The second option is \(x=\frac{3\pm\sqrt{7}}{2}\). Let's see: If \(a = 1\), \(b=-3\), \(c=-1\)? No. Wait, let's compute for \(x^2 - 3x + 1\): discriminant \(9 - 4 = 5\). No. Wait, maybe the function is \(x^2 - 3x + 1\) is wrong. Wait, let's take the second option: \(x=\frac{3\pm\sqrt{7}}{2}\). Let's reverse: If \(x=\frac{3\pm\sqrt{7}}{2}\), then the quadratic is \((x - \frac{3+\sqrt{7}}{2})(x - \frac{3-\sqrt{7}}{2})=x^2 - 3x + \frac{9 - 7}{4}=x^2 - 3x + 0.5\)? No. Wait, maybe the function is \(x^2 - 3x + 1\): no. Wait, maybe the original function is \(x^2 - 3x + 1\) is incorrect. Wait, let's do the quadratic formula correctly. Wait, the user's problem: "Find the zeros of \(f(x)=x^2 + 7x + 9\)". Wait, \(a = 1\), \(b = 7\), \(c = 9\). Then discriminant is \(49 - 36 = 13\). So \(x=\frac{-7\pm\sqrt{13}}{2}\), which is the third option. Wait, the third option is \(x=\frac{-7\pm\sqrt{13}}{2}\). Let's check: \(a = 1\), \(b = 7\), \(c = 9\). Then \(x=\frac{-7\pm\sqrt{49 - 36}}{2}=\frac{-7\pm\sqrt{13}}{2}\), which matches the third option. Wait, but the options: the third option is \(x=\frac{-7\pm\sqrt{13}}{2}\). So that's the correct one. Wait, but let's confirm.

Wait, the quadratic formula: \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). For \(f(x)=x^2 + 7x + 9\), \(a = 1\), \(b = 7\), \(c = 9\). So \(x=\frac{-7\pm\sqrt{49 - 36}}{2}=\frac{-7\pm\sqrt{13}}{2}\), which is the third option (the one with \(-7\pm\sqrt{13}\) over 2).

Wait, but the options:

1. \(x = 3\pm\sqrt{7}\)

1. \(x=\frac{3\pm\sqrt{7}}{2}\)

1. \(x=\frac{-7\pm\sqrt{13}}{2}\)

1. \(x=-7\pm\sqrt{13}\)

So using the quadratic formula on \(x^2 + 7x + 9\), we get \(x=\frac{-7\pm\sqrt{13}}{2}\), which is the third option.

Answer:

The correct option is the third one: \(x=\frac{-7\pm\sqrt{13}}{2}\) (the option with \(x=\frac{-7\pm\sqrt{13}}{2}\))