Chapter 4.5, Problem 61HP

To determine

To find: the perfect square trinomial equation in which linear coefficient is negative and constant term is fraction and solve the equation.

Answer to Problem 61HP

The value of $\text{x}$ is $\frac{3}{4}$ .

Explanation of Solution

Given:

Let’s consider the equation:

  ${\text{4x}}^{\text{2}}\text{-6x+}\frac{\text{9}}{\text{4}}\text{=0}$ .

Concept used:

An expression obtained from the square of binomial equation is a perfect square trinomial is in the form of ${\text{ax}}^{\text{2}}\text{+bx+c}$ is said to be a perfect square, if only if it satisfies the condition ${\text{b}}^{\text{2}}\text{=4ac}$ .

The perfect squares trinomial formula is given as:

  ${\left(\text{ax}\right)}^{\text{2}}{\text{+2abx+b}}^{\text{2}}\text{=}{\left(\text{ax+b}\right)}^{\text{2}}$ .

  ${\left(\text{ax}\right)}^{\text{2}}{\text{-2abx+b}}^{\text{2}}\text{=}{\left(\text{ax-b}\right)}^{\text{2}}$ .

Calculation:

Let’s consider the equation:

  ${\text{4x}}^{\text{2}}\text{-6x+}\frac{\text{9}}{\text{4}}\text{=0}$ .

  ${\left(\text{2x}\right)}^{\text{2}}\text{-2}\left(\text{2x}\right)\left(\frac{\text{3}}{\text{2}}\right)\text{+}{\left(\frac{\text{3}}{\text{2}}\right)}^{\text{2}}\text{-}{\left(\frac{\text{3}}{\text{2}}\right)}^{\text{2}}\text{+}\frac{\text{9}}{\text{4}}\text{=0}$ .

  ${\left(\text{2x-}\frac{3}{2}\right)}^{\text{2}}\text{-}{\left(\frac{\text{3}}{\text{2}}\right)}^{\text{2}}\text{+}\frac{\text{9}}{\text{4}}\text{=0}$ .

  ${\left(\text{2x-}\frac{3}{2}\right)}^{\text{2}}\text{=0}$ .

  $\text{2x=}\frac{\text{3}}{\text{2}}$ .

  $\text{x=}\frac{\text{3}}{\text{4}}$ .

Hence value of $\text{x}$ is $\frac{3}{4}$ .