Solve the following quadratic equations by factoring using Greatest Common Factor. 1. 2x²+4x= 이 2. 12x²-6x= 0 3. 15x² + 25x = 0 4. 12x²-18x = 0 5. 28x² = 14x ==​

Answer:Step-by-step explanation:You are solving for the values of \(x\) that satisfy the given quadratic equations by factoring out the Greatest Common Factor (GCF).               What's given in the problem      Five quadratic equations are given in the form \(ax^{2}+bx=0\).           How to solve               Factor out the GCF from each equation and then set each factor equal to zero to find the solutions for \(x\).      Step 1 . Factor the first equation: \(2x^{2}+4x=0\) Find the GCF of \(2x^{2}\) and \(4x\): \(2x\). Factor out the GCF: \(2x(x+2)=0\). Set each factor to zero: \(2x=0\) or \(x+2=0\). Solve for \(x\): \(x=0\) or \(x=-2\). Step 2 . Factor the second equation: \(12x^{2}-6x=0\) Find the GCF of \(12x^{2}\) and \(-6x\): \(6x\). Factor out the GCF: \(6x(2x-1)=0\). Set each factor to zero: \(6x=0\) or \(2x-1=0\). Solve for \(x\): \(x=0\) or \(x=\frac{1}{2}\). Step 3 . Factor the third equation: \(15x^{2}+25x=0\) Find the GCF of \(15x^{2}\) and \(25x\): \(5x\). Factor out the GCF: \(5x(3x+5)=0\). Set each factor to zero: \(5x=0\) or \(3x+5=0\). Solve for \(x\): \(x=0\) or \(x=-\frac{5}{3}\). Step 4 . Factor the fourth equation: \(12x^{2}-18x=0\) Find the GCF of \(12x^{2}\) and \(-18x\): \(6x\). Factor out the GCF: \(6x(2x-3)=0\). Set each factor to zero: \(6x=0\) or \(2x-3=0\). Solve for \(x\): \(x=0\) or \(x=\frac{3}{2}\). Step 5 . Factor the fifth equation: \(28x^{2}=14x\) Rewrite the equation: \(28x^{2}-14x=0\). Find the GCF of \(28x^{2}\) and \(-14x\): \(14x\). Factor out the GCF: \(14x(2x-1)=0\). Set each factor to zero: \(14x=0\) or \(2x-1=0\). Solve for \(x\): \(x=0\) or \(x=\frac{1}{2}\).          Solution               The solutions for the equations are: \(x=0,-2\); \(x=0,\frac{1}{2}\); \(x=0,-\frac{5}{3}\); \(x=0,\frac{3}{2}\); and \(x=0,\frac{1}{2}\).

Answered by lakshmi12102008 | 2025-07-18