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find the number of x-intercepts. 3. $y = x^2 + 6x - 7$ 4. $y = -x^2 + 6…

Question

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:

Problem 3: \( y = x^2 + 6x - 7 \)

Step 1: Recall x-intercept definition

To find the number of \( x \)-intercepts, we set \( y = 0 \) and solve the quadratic equation \( x^2 + 6x - 7 = 0 \). We can use the discriminant \( D = b^2 - 4ac \) for a quadratic \( ax^2 + bx + c = 0 \). Here, \( a = 1 \), \( b = 6 \), \( c = -7 \).

Step 2: Calculate the discriminant

$$ D = 6^2 - 4(1)(-7) = 36 + 28 = 64 $$

Since \( D = 64 > 0 \), the quadratic equation has two distinct real roots. So the number of \( x \)-intercepts is 2.

Problem 4: \( y = -x^2 + 6x - 9 \)

Step 1: Set \( y = 0 \)

We solve \( -x^2 + 6x - 9 = 0 \), or multiply by -1: \( x^2 - 6x + 9 = 0 \). For \( ax^2 + bx + c = 0 \), here \( a = 1 \), \( b = -6 \), \( c = 9 \).

Step 2: Calculate the discriminant

$$ D = (-6)^2 - 4(1)(9) = 36 - 36 = 0 $$

Since \( D = 0 \), the quadratic equation has one real root (a repeated root). So the number of \( x \)-intercepts is 1.

Problem 5: \( 3n^2 - 5n - 8 = 0 \)

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

For \( 3n^2 - 5n - 8 = 0 \), \( a = 3 \), \( b = -5 \), \( c = -8 \).

Step 2: Quadratic Formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

First, calculate discriminant \( D = (-5)^2 - 4(3)(-8) = 25 + 96 = 121 \)
Then,

$$ n = \frac{-(-5) \pm \sqrt{121}}{2(3)} = \frac{5 \pm 11}{6} $$

Step 3: Find the two solutions

For \( + \): \( \frac{5 + 11}{6} = \frac{16}{6} = \frac{8}{3} \)
For \( - \): \( \frac{5 - 11}{6} = \frac{-6}{6} = -1 \)
So solutions are \( n = \frac{8}{3} \) and \( n = -1 \). Discriminant \( D = 121 > 0 \), so two distinct real solutions.

Problem 6: \( x^2 + 10x = -21 \) (rewrite as \( x^2 + 10x + 21 = 0 \))

Answer:

s:

  1. Number of \( x \)-intercepts: \( \boldsymbol{2} \)
  2. Number of \( x \)-intercepts: \( \boldsymbol{1} \)
  3. Solutions: \( \boldsymbol{n = \frac{8}{3}, n = -1} \) (two real solutions)
  4. Solutions: \( \boldsymbol{x = -3, x = -7} \) (two real solutions)
  5. Solutions: \( \boldsymbol{\text{No real solutions (complex: } \frac{9 \pm i\sqrt{159}}{20}\text{)}} \)
  6. Solutions: \( \boldsymbol{x = \frac{\sqrt{6}}{4}, x = -\frac{\sqrt{6}}{4}} \) (two real solutions)
  7. Solutions: \( \boldsymbol{n = \frac{3 + 2\sqrt{3}}{6}, n = \frac{3 - 2\sqrt{3}}{6}} \) (two real solutions)
  8. Solutions: \( \boldsymbol{\text{No real solutions (complex: } \frac{2 \pm i\sqrt{5}}{3}\text{)}} \)