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Solve for x
Question 1 of 3
Solution:
Step 1: Factorise the quadratic
\[x^2 + 7x - 8 = 0\]
We need two numbers that multiply to \(-8\) and add to \(7\).
Those numbers are \(8\) and \(-1\).
\[(x + 8)(x - 1) = 0\]
Step 2: Solve for x
Either \(x + 8 = 0\) or \(x - 1 = 0\)
\[x_1 = -8 \quad \text{or} \quad x_2 = 1\]
Solution:
Step 1: Factorise the quadratic
\[x^2 - 5x - 14 = 0\]
We need two numbers that multiply to \(-14\) and add to \(-5\).
Those numbers are \(-7\) and \(2\).
\[(x - 7)(x + 2) = 0\]
Step 2: Solve for x
Either \(x - 7 = 0\) or \(x + 2 = 0\)
\[x_1 = 7 \quad \text{or} \quad x_2 = -2\]
Solution:
Step 1: Rearrange to standard form
\[x^2 - 10 = -3x\]
\[x^2 + 3x - 10 = 0\]
Step 2: Factorise the quadratic
We need two numbers that multiply to \(-10\) and add to \(3\).
Those numbers are \(5\) and \(-2\).
\[(x + 5)(x - 2) = 0\]
Step 3: Solve for x
Either \(x + 5 = 0\) or \(x - 2 = 0\)
\[x_1 = -5 \quad \text{or} \quad x_2 = 2\]
