Chapter 1.8, Problem 19E

Solution

With the help of quadratic formula solve the equation $4x^2+3=x^2-7x$ for $\left(x\right)$ .

The solution for the quadratic equation is $x=\left\{\frac{-7+\sqrt{13}}{6},\frac{-7-\sqrt{13}}{6}\right\}$ .

Given:

The quadratic equation is $4x^2+3=x^2-7x$ .

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 $4x^2+3=x^2-7x$ is re-written as:

  $\begin{array}{c}4x^2+3=x^2-7x\mathrm{......}\left(\text{Given}\right)\\ \text{or, }3x^2+7x+3=0\\ 3x^2+7x+3=0\mathrm{......}\left(2\right)\end{array}$

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

  $\begin{array}{c}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\mathrm{......}\left(\text{From }\left(1\right)\right)\\ =\frac{-\left(7\right)\pm \sqrt{{\left(7\right)}^2-4\left(3\right)\left(3\right)}}{2\left(3\right)}\mathrm{......}\left(\because 3x^2+7x+3=0\right)\\ =\frac{-7\pm \sqrt{13}}{6}\\ =\left\{\frac{-7+\sqrt{13}}{6},\frac{-7-\sqrt{13}}{6}\right\}\end{array}$

Hence the solution for the quadratic equation is $x=\left\{\frac{-7+\sqrt{13}}{6},\frac{-7-\sqrt{13}}{6}\right\}$ .