QUESTION IMAGE
Question
question 1 (5 points)
solve the quadratic equation (4x^2 - x = 8) using the quadratic formula.
a) (x = \frac{1 pm sqrt{129}}{4})
b) (x = \frac{-1 pm sqrt{129}}{8})
c) (x = \frac{-40 pm sqrt{129}}{8})
d) (x = \frac{1 pm sqrt{129}}{8})
question 2 (5 points)
solve the quadratic equation (x^2 - 10x + 14 = -4) by completing the square.
a) (x = 5 pm sqrt{7})
b) (x = -5 pm sqrt{7})
c) (x = 7 pm sqrt{5})
d) (x = -7 pm sqrt{5})
Write in standard form
To solve the quadratic equation \(4x^2 - x = 8\) using the quadratic formula, we first rewrite it in standard form \(ax^2 + bx + c = 0\) by subtracting \(8\) from both sides:
Here, the coefficients are:
Apply the quadratic formula
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we substitute our coefficients:
Simplify the terms inside and outside the radical:
This matches option D for Question 1.
Prepare for completing the square
For Question 2, we need to solve \(x^2 - 10x + 14 = -4\) by completing the square. First, we isolate the variable terms by subtracting \(14\) from both sides:
Complete the square
To complete the square, we take half of the coefficient of \(x\), square it, and add it to both sides.
Half of \(-10\) is \(-5\), and \((-5)^2 = 25\).
Write the left side as a perfect square and simplify the right side:
Solve for x
Take the square root of both sides:
Add \(5\) to both sides to isolate \(x\):
This matches option A for Question 2.
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Question 1
- A) \(x = \frac{1 \pm \sqrt{129}}{4}\)
- B) \(x = \frac{-1 \pm \sqrt{129}}{8}\)
- C) \(x = \frac{-40 \pm \sqrt{129}}{8}\)
- D) \(x = \frac{1 \pm \sqrt{129}}{8}\) (Correct answer)
Question 2
- A) \(x = 5 \pm \sqrt{7}\) (Correct answer)
- B) \(x = -5 \pm \sqrt{7}\)
- C) \(x = 7 \pm \sqrt{5}\)
- D) \(x = -7 \pm \sqrt{5}\)