Chapter 1.8, Problem 28E

Solution

Solve the equation $45x^2+57x+1=5$ by quadratic formula and then by factoring to check solution.

The solution for the quadratic equation is $x=\left\{-\frac{4}{3},\frac{1}{15}\right\}$ .

Given:

The quadratic equation is $45x^2+57x+1=5$ .

Concept used:

Any quadratic equation of the form $a{x}^2+bx+c=0\text{ where }a\ne 0$ has the roots $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\mathrm{......}\left(1\right)$

Calculation:

The quadratic equation $45x^2+57x+1=5$ is re-written as:

  $\begin{array}{c}45x^2+57x+1=5\mathrm{......}\left(\text{Given}\right)\\ 45x^2+57x-4=0\mathrm{......}\left(2\right)\end{array}$

The solution for the quadratic equation $\left(2\right)$ can be evaluated with the help of formula $\left(1\right)$ . Therefore,

  $\begin{array}{c}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\mathrm{......}\left(\text{From }\left(1\right)\right)\\ =\frac{-\left(57\right)\pm \sqrt{{\left(57\right)}^2-4\left(45\right)\left(-4\right)}}{2\left(45\right)}\mathrm{......}\left(\because 45x^2+57x-4=0\right)\\ =\frac{-57\pm 63}{90}\\ =\frac{-57+63}{90},\frac{-57-63}{90}\\ =\left\{\frac{1}{15},-\frac{4}{3}\right\}\end{array}$

Hence the solution for the quadratic equation is $x=\left\{\frac{1}{15},-\frac{4}{3}\right\}$ .

Now, factoring the equation $\left(2\right)$ gives,

  $\begin{array}{c}45x^2+57x-4=0\mathrm{......}\left(\text{From }\left(2\right)\right)\\ 45x^2+60x-3x-4=0\\ 15x\left(3x+4\right)-1\left(3x+4\right)=0\\ \left(3x+4\right)\left(15x-1\right)=0\\ 3x+4=0\text{ or }15x-1=0\\ x=-\frac{4}{3}\text{ or }x=\frac{1}{15}\\ x=\left\{-\frac{4}{3},\frac{1}{15}\right\}\end{array}$

From both the techniques the solutions are same for of the quadratic equation. Hence the required solution is $x=\left\{-\frac{4}{3},\frac{1}{15}\right\}$ of quadratic equation.