Solve for x
Question 1 of 3
Solution:
Step 1: Apply the zero product property
\[(x - 6)(x - 5) = 0\]
If the product of two factors equals zero, then at least one factor must be zero.
Step 2: Solve for x
Either \(x - 6 = 0\) or \(x - 5 = 0\)
\[x_1 = 6 \quad \text{or} \quad x_2 = 5\]
Solution:
Step 1: Divide both sides by 2
\[2(x - 3)(x - 1) = 16\]
\[(x - 3)(x - 1) = 8\]
Step 2: Expand the left side
\[x^2 - x - 3x + 3 = 8\]
\[x^2 - 4x + 3 = 8\]
Step 3: Rearrange to standard form
\[x^2 - 4x - 5 = 0\]
Step 4: Factorise the quadratic
We need two numbers that multiply to \(-5\) and add to \(-4\).
Those numbers are \(-5\) and \(1\).
\[(x - 5)(x + 1) = 0\]
Step 5: Solve for x
Either \(x - 5 = 0\) or \(x + 1 = 0\)
\[x_1 = 5 \quad \text{or} \quad x_2 = -1\]
Solution:
Step 1: Expand the left side
\[(x - 8)(11 - x) = 10\]
\[11x - x^2 - 88 + 8x = 10\]
\[-x^2 + 19x - 88 = 10\]
Step 2: Rearrange to standard form
\[-x^2 + 19x - 98 = 0\]
Multiply by \(-1\):
\[x^2 - 19x + 98 = 0\]
Step 3: Factorise the quadratic
We need two numbers that multiply to \(98\) and add to \(-19\).
Those numbers are \(-7\) and \(-14\).
\[(x - 7)(x - 14) = 0\]
Step 4: Solve for x
Either \(x - 7 = 0\) or \(x - 14 = 0\)
\[x_1 = 7 \quad \text{or} \quad x_2 = 14\]
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