Question

1. the base of a triangle is $\frac{2x^{2}+5x - 3}{x^{2}-x - 12}$ and the height is $\frac{x^{2}-x - 2}{2x^{2}-5x + 2}$

Explanation:

1. First, factor the polynomials:

• Factor the base - related polynomials:
• Factor \(2x^{2}+5x - 3\):
• We use the formula \(ax^{2}+bx + c=a(x - x_1)(x - x_2)\), where \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(2x^{2}+5x - 3\), \(a = 2\), \(b = 5\), \(c=-3\). Then \(x=\frac{-5\pm\sqrt{25+24}}{4}=\frac{-5\pm7}{4}\). The roots are \(x_1=\frac{-5 + 7}{4}=\frac{1}{2}\) and \(x_2=-3\). So \(2x^{2}+5x - 3=(2x - 1)(x + 3)\).
• Factor \(x^{2}-x - 12\):
• For \(x^{2}-x - 12\), where \(a = 1\), \(b=-1\), \(c=-12\), \(x=\frac{1\pm\sqrt{1 + 48}}{2}=\frac{1\pm7}{2}\). The roots are \(x_1 = 4\) and \(x_2=-3\). So \(x^{2}-x - 12=(x - 4)(x+3)\).
• Factor the height - related polynomials:
• Factor \(x^{2}-x - 2\):
• For \(x^{2}-x - 2\), where \(a = 1\), \(b=-1\), \(c=-2\), \(x=\frac{1\pm\sqrt{1 + 8}}{2}=\frac{1\pm3}{2}\). The roots are \(x_1 = 2\) and \(x_2=-1\). So \(x^{2}-x - 2=(x - 2)(x + 1)\).
• Factor \(2x^{2}-5x + 2\):
• For \(2x^{2}-5x + 2\), where \(a = 2\), \(b=-5\), \(c = 2\), \(x=\frac{5\pm\sqrt{25 - 16}}{4}=\frac{5\pm3}{4}\). The roots are \(x_1 = 2\) and \(x_2=\frac{1}{2}\). So \(2x^{2}-5x + 2=(2x - 1)(x - 2)\).
• The base \(b=\frac{2x^{2}+5x - 3}{x^{2}-x - 12}=\frac{(2x - 1)(x + 3)}{(x - 4)(x + 3)}=\frac{2x - 1}{x - 4}\), \(x

eq - 3\).

• The height \(h=\frac{x^{2}-x - 2}{2x^{2}-5x + 2}=\frac{(x - 2)(x + 1)}{(2x - 1)(x - 2)}=\frac{x + 1}{2x - 1}\), \(x

eq2\).

1. Then, use the formula for the area of a triangle \(A=\frac{1}{2}bh\):

• Substitute \(b=\frac{2x - 1}{x - 4}\) and \(h=\frac{x + 1}{2x - 1}\) into the area formula.
• \(A=\frac{1}{2}\times\frac{2x - 1}{x - 4}\times\frac{x + 1}{2x - 1}\).
• Cancel out the common factor \((2x - 1)\):
• \(A=\frac{1}{2}\times\frac{x + 1}{x - 4}=\frac{x + 1}{2(x - 4)}\), \(x

eq\frac{1}{2},2, - 3,4\).

Step1: Factor the polynomials

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

Step2: Simplify the base and height expressions

\(b=\frac{(2x - 1)(x + 3)}{(x - 4)(x + 3)}=\frac{2x - 1}{x - 4}(x
eq - 3)\), \(h=\frac{(x - 2)(x + 1)}{(2x - 1)(x - 2)}=\frac{x + 1}{2x - 1}(x
eq2)\).

Step3: Calculate the area of the triangle

\(A=\frac{1}{2}bh=\frac{1}{2}\times\frac{2x - 1}{x - 4}\times\frac{x + 1}{2x - 1}=\frac{x + 1}{2(x - 4)}(x
eq\frac{1}{2},2, - 3,4)\).

Answer:

\(\frac{x + 1}{2(x - 4)}\), \(x
eq\frac{1}{2},2, - 3,4\)