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Solve for \( x \) : 1. \( (x+3)(x-1)=0 \) 2. \( -2(x+2)(x+3)=0 \) 3. \( 5 x(2 x+1)=0 \) 4. \( x^{2}+14 x+48=0 \) 5. \( 2 x^{2}-18=0 \)

Ask by Macdonald Marsh. in South Africa
Mar 09,2025

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Answer

1. \( x = -3 \) and \( x = 1 \) 2. \( x = -2 \) and \( x = -3 \) 3. \( x = 0 \) and \( x = -\frac{1}{2} \) 4. \( x = -6 \) and \( x = -8 \) 5. \( x = 3 \) and \( x = -3 \)

Solution

**1. Solve:** \( (x+3)(x-1)=0 \) Since a product is zero, we set each factor equal to zero: - \( x+3=0 \) ⟹ \( x=-3 \) - \( x-1=0 \) ⟹ \( x=1 \) Thus, the solutions are: \( x=-3 \) and \( x=1 \) --- **2. Solve:** \( -2(x+2)(x+3)=0 \) The constant \(-2\) does not affect the roots. So we set: - \( x+2=0 \) ⟹ \( x=-2 \) - \( x+3=0 \) ⟹ \( x=-3 \) Thus, the solutions are: \( x=-2 \) and \( x=-3 \) --- **3. Solve:** \( 5x(2x+1)=0 \) Since \(5\) is nonzero, we have: - \( x=0 \) - \( 2x+1=0 \) ⟹ \( 2x=-1 \) ⟹ \( x=-\frac{1}{2} \) Thus, the solutions are: \( x=0 \) and \( x=-\frac{1}{2} \) --- **4. Solve:** \( x^{2}+14x+48=0 \) We factor the quadratic by searching for two numbers that multiply to \(48\) and add to \(14\): \( 6 \) and \( 8 \) satisfy these conditions. Factorization gives: \[ x^2 + 14x + 48 = (x+6)(x+8)=0 \] Set each factor equal to zero: - \( x+6=0 \) ⟹ \( x=-6 \) - \( x+8=0 \) ⟹ \( x=-8 \) Thus, the solutions are: \( x=-6 \) and \( x=-8 \) --- **5. Solve:** \( 2x^{2}-18=0 \) First, divide the entire equation by 2: \[ x^2 - 9 = 0 \] Recognize the difference of squares: \[ x^2 - 9 = (x-3)(x+3)=0 \] Set each factor equal to zero: - \( x-3=0 \) ⟹ \( x=3 \) - \( x+3=0 \) ⟹ \( x=-3 \) Thus, the solutions are: \( x=3 \) and \( x=-3 \)

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Bonus Knowledge

To solve for \( x \) in each equation: 1. For \( (x+3)(x-1)=0 \), set each factor to zero: - \( x+3=0 \) gives \( x=-3 \) - \( x-1=0 \) gives \( x=1 \) So, the solutions are \( x=-3 \) and \( x=1 \). 2. For \( -2(x+2)(x+3)=0 \), ignore the -2 and set each factor to zero: - \( x+2=0 \) gives \( x=-2 \) - \( x+3=0 \) gives \( x=-3 \) Thus, the solutions are \( x=-2 \) and \( x=-3 \). 3. For \( 5 x(2 x+1)=0 \), set each factor to zero: - \( 5x=0 \) gives \( x=0 \) - \( 2x+1=0 \) results in \( 2x=-1 \) or \( x=-\frac{1}{2} \) So, the solutions are \( x=0 \) and \( x=-\frac{1}{2} \). 4. For \( x^{2}+14 x+48=0 \), factor the quadratic: - The factors are \( (x+6)(x+8)=0 \) This gives us \( x=-6 \) and \( x=-8 \) as solutions. 5. For \( 2 x^{2}-18=0 \), first isolate \( x^2 \): - \( 2 x^2 = 18 \) then divide by 2, yielding \( x^2=9 \) - Taking the square root gives \( x=3 \) and \( x=-3 \) So, the solutions are \( x=3 \) and \( x=-3 \). Summarizing all solutions: 1. \( x=-3, 1 \) 2. \( x=-2, -3 \) 3. \( x=0, -\frac{1}{2} \) 4. \( x=-6, -8 \) 5. \( x=3, -3 \)

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