Question

1. $4x^{2}-8x = - 3$

Explanation:

Step1: Rewrite the equation

First, we rewrite the given equation \(4x^{2}-8x = - 3\) in standard quadratic form \(ax^{2}+bx + c=0\) by adding 3 to both sides.
\(4x^{2}-8x + 3=0\)

Step2: Factor the quadratic (or use quadratic formula)

We can factor the quadratic equation. We need two numbers that multiply to \(4\times3 = 12\) and add up to \(- 8\). The numbers are \(-2\) and \(-6\).
Rewrite the middle term:
\(4x^{2}-2x-6x + 3=0\)
Group the terms:
\((4x^{2}-2x)-(6x - 3)=0\)
Factor out the common factors from each group:
\(2x(2x - 1)-3(2x - 1)=0\)
Now, factor out \((2x - 1)\):
\((2x - 1)(2x - 3)=0\)

Step3: Solve for x

Set each factor equal to zero and solve for \(x\):

• For \(2x-1 = 0\):

Add 1 to both sides: \(2x=1\)
Divide both sides by 2: \(x=\frac{1}{2}\)

• For \(2x - 3=0\):

Add 3 to both sides: \(2x=3\)
Divide both sides by 2: \(x=\frac{3}{2}\)

Answer:

\(x=\frac{1}{2}\) or \(x = \frac{3}{2}\)