Question

1. solve the equation using the quadratic formula.

3x² + 2x - 1 = 0

1. determine if each quadratic equation has real or non - real solutions.

a.) x² + 2x + 3 = 0 b.) x² - 2x = - 6 c.) x² + 4x = - 3
real.
real.
real.
non - real
non - real
non - real
for 7 - 9, consider the equation:
(x - 4)² + (y + 2)² = 9.

1. what is the center of the circle?
2. what is the radius of the circle?

r=

1. graph the circle.

Explanation:

Step1: Identify the quadratic formula

The quadratic formula for \(ax^{2}+bx + c=0\) is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).

Step2: Determine \(a\), \(b\), and \(c\)

For the equation \(3x^{2}+2x - 1=0\), \(a = 3\), \(b=2\), \(c=-1\).

Step3: Calculate the discriminant

\(b^{2}-4ac=(2)^{2}-4\times3\times(-1)=4 + 12=16\).

Step4: Substitute into the quadratic formula

\(x=\frac{-2\pm\sqrt{16}}{2\times3}=\frac{-2\pm4}{6}\).

Step5: Find the two solutions

For \(x=\frac{-2 + 4}{6}\), \(x=\frac{2}{6}=\frac{1}{3}\).
For \(x=\frac{-2-4}{6}\), \(x=\frac{-6}{6}=-1\).

Step1: Recall the standard form of a circle's equation

The standard form of a circle's equation is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius.

Step2: Identify the center for the equation \((x - 4)^{2}+(y + 2)^{2}=9\)

Comparing with the standard form, \(h = 4\), \(k=-2\).

Step1: Recall the standard form of a circle's equation

The standard form of a circle's equation is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \(r\) is the radius.

Step2: Identify the radius for the equation \((x - 4)^{2}+(y + 2)^{2}=9\)

Since \(r^{2}=9\), then \(r = 3\) (we take the positive value as radius is non - negative).

Step1: Recall the discriminant formula

For a quadratic equation \(ax^{2}+bx + c=0\), the discriminant is \(\Delta=b^{2}-4ac\). If \(\Delta>0\), real solutions; if \(\Delta<0\), non - real solutions.

Step2: For equation \(A\): \(x^{2}+2x + 3=0\)

Here \(a = 1\), \(b = 2\), \(c = 3\). \(\Delta=(2)^{2}-4\times1\times3=4 - 12=-8<0\).

Step3: For equation \(B\): \(x^{2}-2x+6 = 0\) (rewritten from \(x^{2}-2x=-6\))

Here \(a = 1\), \(b=-2\), \(c = 6\). \(\Delta=(-2)^{2}-4\times1\times6=4 - 24=-20<0\).

Step4: For equation \(C\): \(x^{2}+4x + 3=0\) (rewritten from \(x^{2}+4x=-3\))

Here \(a = 1\), \(b = 4\), \(c = 3\). \(\Delta=(4)^{2}-4\times1\times3=16 - 12 = 4>0\).

Answer:

\(x=-1\) or \(x=\frac{1}{3}\)