Question

ivantes.
(2x - 1)(2x - 1) - (x + 4)(x + 4)
b) $(2x - 1)^2 - (x + 4)^2 = 5$
e) $4x^3(x^2 + 4)(x^2 - 16) = 0$
h) $9^{2x} = 27^{2 - 3x}$

Explanation:

Part b)

Step 1: Apply difference of squares

We know that \(a^2 - b^2=(a - b)(a + b)\). Here \(a = 2x-1\) and \(b=x + 4\). So \((2x-1)^2-(x + 4)^2=[(2x-1)-(x + 4)][(2x-1)+(x + 4)]\)
Simplify the two factors:
First factor: \((2x-1)-(x + 4)=2x-1-x - 4=x-5\)
Second factor: \((2x-1)+(x + 4)=2x-1+x + 4=3x + 3 = 3(x + 1)\)
So the equation becomes \((x - 5)\times3(x + 1)=5\)

Step 2: Expand the left - hand side

\(3(x - 5)(x + 1)=5\)
\(3(x^{2}+x-5x - 5)=5\)
\(3(x^{2}-4x - 5)=5\)
\(3x^{2}-12x-15 = 5\)

Step 3: Rearrange to standard quadratic form

\(3x^{2}-12x-15 - 5=0\)
\(3x^{2}-12x-20 = 0\)

Step 4: Use quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\)

For \(ax^{2}+bx + c = 0\) (here \(a = 3\), \(b=-12\), \(c=-20\))
\(\Delta=b^{2}-4ac=(-12)^{2}-4\times3\times(-20)=144 + 240=384\)
\(\sqrt{\Delta}=\sqrt{384}=\sqrt{64\times6}=8\sqrt{6}\)
\(x=\frac{12\pm8\sqrt{6}}{2\times3}=\frac{12\pm8\sqrt{6}}{6}=\frac{6\pm4\sqrt{6}}{3}=2\pm\frac{4\sqrt{6}}{3}\)

Part e)

Step 1: Analyze the zero - product property

If \(ab = 0\), then either \(a = 0\) or \(b = 0\). For the equation \(4x^{3}(x^{2}+4)(x^{2}-16)=0\), we set each factor equal to zero:

• Case 1: \(4x^{3}=0\)

Divide both sides by 4: \(x^{3}=0\), so \(x = 0\) (with multiplicity 3)

• Case 2: \(x^{2}+4=0\)

\(x^{2}=-4\), since the square of a real number is non - negative, there are no real solutions for this factor.

• Case 3: \(x^{2}-16=0\)

Using the difference of squares \(a^{2}-b^{2}=(a - b)(a + b)\), where \(a=x\) and \(b = 4\), we have \((x - 4)(x + 4)=0\)
So \(x-4=0\) or \(x + 4=0\), which gives \(x = 4\) or \(x=-4\)

Part h)

Step 1: Express both sides with the same base

We know that \(9 = 3^{2}\) and \(27=3^{3}\)
So \(9^{2x}=(3^{2})^{2x}=3^{4x}\) and \(27^{2 - 3x}=(3^{3})^{2-3x}=3^{3(2 - 3x)}=3^{6-9x}\)
The equation \(9^{2x}=27^{2 - 3x}\) becomes \(3^{4x}=3^{6-9x}\)

Step 2: Set the exponents equal (since if \(a^{m}=a^{n}\) and \(a>0,a

eq1\), then \(m = n\))
\(4x=6-9x\)

Step 3: Solve for \(x\)

\(4x + 9x=6\)
\(13x=6\)
\(x=\frac{6}{13}\)

Part b) Answer: \(x = 2+\frac{4\sqrt{6}}{3}\) or \(x = 2-\frac{4\sqrt{6}}{3}\)

Part e) Answer: \(x = 0\), \(x = 4\), \(x=-4\)

Part h) Answer: \(x=\frac{6}{13}\)

Answer:

Step 1: Express both sides with the same base

We know that \(9 = 3^{2}\) and \(27=3^{3}\)
So \(9^{2x}=(3^{2})^{2x}=3^{4x}\) and \(27^{2 - 3x}=(3^{3})^{2-3x}=3^{3(2 - 3x)}=3^{6-9x}\)
The equation \(9^{2x}=27^{2 - 3x}\) becomes \(3^{4x}=3^{6-9x}\)

Step 2: Set the exponents equal (since if \(a^{m}=a^{n}\) and \(a>0,a

eq1\), then \(m = n\))
\(4x=6-9x\)

Step 3: Solve for \(x\)

\(4x + 9x=6\)
\(13x=6\)
\(x=\frac{6}{13}\)

Part b) Answer: \(x = 2+\frac{4\sqrt{6}}{3}\) or \(x = 2-\frac{4\sqrt{6}}{3}\)

Part e) Answer: \(x = 0\), \(x = 4\), \(x=-4\)

Part h) Answer: \(x=\frac{6}{13}\)