Question

ap algebra 2
discriminant and quadratic formula \t\t\t\t\tname
determine the type and number of solutions.

1. $x^2 + 6x - 7 = 0$
2. $3x^2 - x = - 3$

find the number of x-intercepts.

1. $y = x^2 + 6x - 7$
2. $y = - x^2 + 6x - 9$

solve using quadratic formula. use the discriminant to determine the number and types of solutions.

1. $3n^2 - 5n - 8 = 0$
2. $x^2 + 10x = - 21$
3. $10x^2 + 6 = 9x$
4. $8x^2 - 3 = 0$
5. $12n^2 - 10n - 9 = 2n - 8$
6. $6p^2 = 8p - 6$

Explanation:

Let's solve problem 1: \(x^2 + 6x - 7 = 0\) using the discriminant method.

Step 1: Identify \(a\), \(b\), \(c\)

For a quadratic equation \(ax^2 + bx + c = 0\), here \(a = 1\), \(b = 6\), \(c = -7\).

Step 2: Calculate the discriminant \(D = b^2 - 4ac\)

Substitute the values: \(D = 6^2 - 4(1)(-7) = 36 + 28 = 64\).

Step 3: Analyze the discriminant

Since \(D = 64>0\) and a perfect square, the equation has two distinct real rational solutions.
To find the solutions, we can also factor: \(x^2 + 6x - 7=(x + 7)(x - 1)=0\), so \(x=-7\) or \(x = 1\).

Answer:

The equation \(x^2 + 6x - 7 = 0\) has two distinct real rational solutions, \(x=-7\) and \(x = 1\).