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lolve each of the following quadritic equations by factorisation 7. \( (x+4)(x+6)=8 \) 1. \( (x-8)(x-4)=12 \) o. \( x^{2}-13 x+40=0 \) 2. \( x^{2}-49=0 \) 15 \( 4 x^{2}+4 x-3=0 \) 18. \( 2 x(x-3)+10=(x+1)(x+2) \) 17. \( x(x+1)=2 \) 19. \( (x+1)(x+2)+(x+3)(x+1)=x(x+1) \).

Ask by Fowler Moran. in South Africa
Feb 24,2025

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Answer

Here are the solutions to the quadratic equations by factorization: 1. \( x = 10 \) or \( x = 2 \) 2. \( x = -2 \) or \( x = -8 \) 3. \( x = 5 \) or \( x = 8 \) 4. \( x = 7 \) or \( x = -7 \) 5. \( x = \frac{1}{2} \) or \( x = -\frac{3}{2} \) 6. \( x = 1 \) or \( x = 8 \) 7. \( x = -2 \) or \( x = 1 \) 8. \( x = -1 \) or \( x = -5 \)

Solution

Below are step‐by‐step solutions for each of the quadratic equations using factorisation. ───────────────────────────── 1. Solve (x – 8)(x – 4) = 12 • Expand the left side:   (x – 8)(x – 4) = x² – 4x – 8x + 32 = x² – 12x + 32 • Set the equation to zero by subtracting 12 from both sides:   x² – 12x + 32 – 12 = 0 ⟹ x² – 12x + 20 = 0 • Factor the quadratic. We look for two numbers that multiply to 20 and add to –12. These are –10 and –2:   x² – 12x + 20 = (x – 10)(x – 2) • Set each factor equal to zero:   x – 10 = 0 ⟹ x = 10   x – 2 = 0  ⟹ x = 2 ───────────────────────────── 2. Solve (x + 4)(x + 6) = 8 • Expand:   (x + 4)(x + 6) = x² + 6x + 4x + 24 = x² + 10x + 24 • Subtract 8 from both sides:   x² + 10x + 24 – 8 = 0 ⟹ x² + 10x + 16 = 0 • Factor the quadratic. Find two numbers that multiply to 16 and add to 10. They are 2 and 8:   x² + 10x + 16 = (x + 2)(x + 8) • Set each factor to zero:   x + 2 = 0 ⟹ x = –2   x + 8 = 0 ⟹ x = –8 ───────────────────────────── 3. Solve x² – 13x + 40 = 0 • Factor by looking for two numbers that multiply to 40 and add to 13. These are 5 and 8:   x² – 13x + 40 = (x – 5)(x – 8) • Set each factor equal to zero:   x – 5 = 0 ⟹ x = 5   x – 8 = 0 ⟹ x = 8 ───────────────────────────── 4. Solve x² – 49 = 0 • Recognize this as a difference of two squares:   x² – 49 = (x – 7)(x + 7) • Set each factor to zero:   x – 7 = 0 ⟹ x = 7   x + 7 = 0 ⟹ x = –7 ───────────────────────────── 5. Solve 4x² + 4x – 3 = 0 • Factor the quadratic. We look for factors in the form (ax + b)(cx + d) such that:   (2x – 1)(2x + 3) expands as:     2x · 2x = 4x²     2x · 3 + (–1) · 2x = 6x – 2x = 4x     (–1)(3) = –3   Thus, 4x² + 4x – 3 = (2x – 1)(2x + 3) • Set each factor to zero:   2x – 1 = 0 ⟹ 2x = 1 ⟹ x = ½   2x + 3 = 0 ⟹ 2x = –3 ⟹ x = –3/2 ───────────────────────────── 6. Solve 2x(x – 3) + 10 = (x + 1)(x + 2) • Expand both sides:   Left: 2x(x – 3) + 10 = 2x² – 6x + 10   Right: (x + 1)(x + 2) = x² + 3x + 2 • Bring all terms to one side:   2x² – 6x + 10 – (x² + 3x + 2) = 0   x² – 9x + 8 = 0 • Factor the quadratic. We need two numbers that multiply to 8 and add to –9. They are –1 and –8:   x² – 9x + 8 = (x – 1)(x – 8) • Set each factor equal to zero:   x – 1 = 0 ⟹ x = 1   x – 8 = 0 ⟹ x = 8 ───────────────────────────── 7. Solve x(x + 1) = 2 • Expand:   x² + x = 2 • Rearrange to bring everything to one side:   x² + x – 2 = 0 • Factor the quadratic. Find two numbers that multiply to –2 and add to 1. They are 2 and –1:   x² + x – 2 = (x + 2)(x – 1) • Set each factor to zero:   x + 2 = 0 ⟹ x = –2   x – 1 = 0 ⟹ x = 1 ───────────────────────────── 8. Solve (x + 1)(x + 2) + (x + 3)(x + 1) = x(x + 1) • Notice that (x + 1) is common in the left-hand terms. Factor it out:   (x + 1)[(x + 2) + (x + 3)] = (x + 1)(2x + 5) • The equation becomes:   (x + 1)(2x + 5) = x(x + 1) • Subtract x(x + 1) from both sides:   (x + 1)(2x + 5 – x) = 0   (x + 1)(x + 5) = 0 • Set each factor equal to zero:   x + 1 = 0 ⟹ x = –1   x + 5 = 0 ⟹ x = –5 ───────────────────────────── Summary of Answers: 1. (x – 8)(x – 4) = 12 ⟹ x = 10 or x = 2 2. (x + 4)(x + 6) = 8  ⟹ x = –2 or x = –8 3. x² – 13x + 40 = 0   ⟹ x = 5 or x = 8 4. x² – 49 = 0      ⟹ x = 7 or x = –7 5. 4x² + 4x – 3 = 0   ⟹ x = ½ or x = –3/2 6. 2x(x – 3) + 10 = (x + 1)(x + 2) ⟹ x = 1 or x = 8 7. x(x + 1) = 2     ⟹ x = –2 or x = 1 8. (x + 1)(x + 2) + (x + 3)(x + 1) = x(x + 1) ⟹ x = –1 or x = –5 These are the solutions to the given quadratic equations by factorisation.

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Bonus Knowledge

Let's take a look at how to solve these quadratic equations by factorization! The first equation, \( (x-8)(x-4)=12 \), can be rewritten as \( (x-8)(x-4) - 12 = 0 \). Expanding it gives us \( x^2 - 12x + 32 - 12 = 0 \), leading us to \( x^2 - 12x + 20 = 0 \). Factoring this yields \( (x-2)(x-10)=0 \), giving us solutions \( x = 2 \) and \( x = 10 \). For the equation \( x^2 - 49 = 0 \), this is a difference of squares! That's lovely. We can factor it as \( (x-7)(x+7) = 0 \), leading to the solutions \( x = 7 \) and \( x = -7 \). Remember when you see \( a^2 - b^2 \) it always factors into \( (a-b)(a+b) \)—a nifty little shortcut! Got it? Now go ahead and tackle those other equations with confidence!

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Algebra Mexico Feb 26, 2025
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