Solve for x
Question 1 of 3
Solution:
Step 1: Factor out the common factor
\[2x^2 - 4x - 30 = 0\]
\[2(x^2 - 2x - 15) = 0\]
Step 2: Factorise the quadratic
We need two numbers that multiply to \(-15\) and add to \(-2\).
Those numbers are \(-5\) and \(3\).
\[2(x - 5)(x + 3) = 0\]
Step 3: Solve for x
Either \(x - 5 = 0\) or \(x + 3 = 0\)
\[x_1 = 5 \quad \text{or} \quad x_2 = -3\]
Solution:
Step 1: Rearrange to standard form
\[3x^2 + 24 = 18x\]
\[3x^2 - 18x + 24 = 0\]
Step 2: Factor out the common factor
\[3(x^2 - 6x + 8) = 0\]
Step 3: Factorise the quadratic
We need two numbers that multiply to \(8\) and add to \(-6\).
Those numbers are \(-4\) and \(-2\).
\[3(x - 4)(x - 2) = 0\]
Step 4: Solve for x
Either \(x - 4 = 0\) or \(x - 2 = 0\)
\[x_1 = 4 \quad \text{or} \quad x_2 = 2\]
Solution:
Step 1: Rearrange to standard form
\[5x^2 - 50x = -125\]
\[5x^2 - 50x + 125 = 0\]
Step 2: Factor out the common factor
\[5(x^2 - 10x + 25) = 0\]
Step 3: Recognize perfect square trinomial
\(x^2 - 10x + 25\) is a perfect square: \((x - 5)^2\)
\[5(x - 5)^2 = 0\]
Step 4: Solve for x
\[(x - 5)^2 = 0\]
\[x - 5 = 0\]
\[x_1 = 5 \quad \text{and} \quad x_2 = 5\]
(This equation has a repeated root)
