Chapter P.6, Problem 43E

Solution

The solutions of the equation $4x^2-6x+5=x+1$ .

  $x=\frac{7}{8}\pm \frac{\sqrt{15}}{8}i$

Given:

The equation, $4x^2-6x+5=x+1$ .

Concept used:

• Product of powers property, $a^n\cdot a^m=a^{n+m}$
• FOIL to multiply binomials, $\left(a+b\right)\left(c+d\right)=a\cdot c+a\cdot d+b\cdot c+b\cdot d$
• Properties of complex number: $\sqrt{-1}=i,i^2=-1,i^3=-i$
• The complex conjugate of $a+bi$ is $a-bi$ .
• Difference of squares formula, $\left(a+b\right)\left(a-b\right)=a^2-b^2$
• Quadratic formula: The roots of a quadratic equation $a{x}^2+bx+c=0;\text{ where }a\ne 0$ are:
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

To solve the given equation, first rewrite it in standard form as shown below

  $\begin{array}{c}4x^2-6x+5=x+1\\ 4x^2-6x+5-x-1=x+1-x-1\\ 4x^2-7x+4=0\end{array}$

On comparing the given equation with the general equation $a{x}^2+bx+c=0$ , it follows

  $a=4,b=-7,c=4$

In order to solve the given equation, use the quadratic formula and simplify further as shown below,

  $\begin{array}{c}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-\left(-7\right)\pm \sqrt{{\left(-7\right)}^2-4\cdot 4\cdot 4}}{2\cdot 4}\\ =\frac{7\pm \sqrt{49-64}}{8}\\ =\frac{7\pm \sqrt{-15}}{8}\\ =\frac{7\pm \sqrt{15}\cdot \sqrt{-1}}{8}\\ =\frac{7\pm \sqrt{15}\cdot i}{8}\\ =\frac{7}{8}\pm \frac{\sqrt{15}}{8}i\end{array}$

Thus, the solutions of the equation are $x=\frac{7}{8}\pm \frac{\sqrt{15}}{8}i$ .