Chapter 4.9, Problem 48PPSE

Solution

To find: solution of the equations $\left\{\begin{array}{c}y=5x^2+9x+4\\ y=-5x+3\end{array}\right.$

Solution of the equations are $x=\frac{-14+\sqrt{5}}{10},y=\frac{-26-\sqrt{5}}{2}\&x=\frac{-14-\sqrt{5}}{10},y=\frac{-26+\sqrt{5}}{2}$

Given: the equation is $\left\{\begin{array}{c}y=5x^2+9x+4\\ y=-5x+3\end{array}\right.$

Concept Used:

Factoring method to find the roots of the quadratic equation.

Calculation:

  $\begin{array}{l}y=5x^2+9x+\mathrm{4............}(i)\\ y=-5x+\mathrm{3...................}(ii)\end{array}$

Equating equation (i) and (ii) we get:

  $\begin{array}{c}5x^2+9x+4=-5x+3\\ 5x^2+9x+5x+4-3=0\\ 5x^2+14x+1=0\end{array}$

Using quadratic formula: $a=5,b=14,c=1$ ,

  $\begin{array}{l}x=\frac{-14\pm \sqrt{5^2-4\times 5\times 1}}{2\times 5}\\ x=\frac{-14\pm \sqrt{25-20}}{10}\\ x=\frac{-14\pm \sqrt{5}}{10}\end{array}$

Now, in equation (ii), substituting the value of $x=\frac{-14+\sqrt{5}}{10}$ and solving for $y$ :

  $\begin{array}{l}y=-5\times \frac{-14+\sqrt{5}}{10}-4\\ y=-1\times \frac{-14+\sqrt{5}}{2}-4\\ y=\frac{14-\sqrt{5}-40}{2}\\ y=\frac{-26-\sqrt{5}}{2}\end{array}$

Now, if $x=\frac{-14-\sqrt{5}}{10}$

  $\begin{array}{l}y=-5\times \frac{-14-\sqrt{5}}{10}-4\\ y=-1\times \frac{-14-\sqrt{5}}{2}-4\\ y=\frac{14+\sqrt{5}-40}{2}\\ y=\frac{-26+\sqrt{5}}{2}\end{array}$

Solution of equations are $x=\frac{-14+\sqrt{5}}{10},y=\frac{-26-\sqrt{5}}{2}\&x=\frac{-14-\sqrt{5}}{10},y=\frac{-26+\sqrt{5}}{2}$

Conclusion:

Solution of equations are $x=\frac{-14+\sqrt{5}}{10},y=\frac{-26-\sqrt{5}}{2}\&x=\frac{-14-\sqrt{5}}{10},y=\frac{-26+\sqrt{5}}{2}$