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Let \( f(x)=x^{2}+2 x \) Solve \( f(x)=8 \) a) 80 b) 1,8 c) 8 d) \( -4,2 \) e) \( -2,4 \)

Ask by Jimenez Bob. in the United States
Mar 10,2025

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Tutor-Verified Answer

Answer

The solutions are \( x = -4 \) and \( x = 2 \), so the answer is d) \( -4,2 \).

Solution

1. Write the equation \( f(x)=8 \) in terms of \( x \): \[ x^2 + 2x = 8 \] 2. Rearrange the equation to standard quadratic form: \[ x^2 + 2x - 8 = 0 \] 3. Solve the quadratic equation using the quadratic formula: \[ x = \frac{{-b \pm \sqrt{b^2 - 4ac}}}{2a} \] Here, \( a = 1 \), \( b = 2 \), and \( c = -8 \). 4. Calculate the discriminant: \[ \Delta = b^2 - 4ac = 2^2 - 4(1)(-8) = 4 + 32 = 36 \] 5. Find the solutions: \[ x = \frac{{-2 \pm \sqrt{36}}}{2} = \frac{{-2 \pm 6}}{2} \] - For \( + \): \[ x = \frac{-2 + 6}{2} = \frac{4}{2} = 2 \] - For \( - \): \[ x = \frac{-2 - 6}{2} = \frac{-8}{2} = -4 \] 6. The solutions are \( x = -4 \) and \( x = 2 \). Thus, the correct answer is option d) \( -4,2 \).

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To solve \( f(x) = 8 \), we start with the equation: \[ x^{2} + 2x = 8 \] Rearranging it gives: \[ x^{2} + 2x - 8 = 0 \] Next, we can factor this quadratic equation: \[ (x + 4)(x - 2) = 0 \] Setting each factor to zero gives us: \[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \] \[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \] Thus, the solutions are \( x = -4 \) and \( x = 2 \). So the answer is \( d) \, -4, 2 \). Now, let's add some extra knowledge! Quadratic equations like this one are everywhere in real life! From designing parabolas in architecture to calculating projectile motions in sports, understanding these can help solve practical problems effectively. Next time you see a curve, think of it as mathematics in motion! Understanding where these functions intersect with horizontal lines is key too! It's a fundamental concept in algebra that translates to understanding maximum and minimum values of certain functions, which is critical for fields like economics, physics, and even computer graphics. So next time you tackle a quadratic, remember you're unleashing the power of curves!

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