Chapter 7.3, Problem 12WE

To determine

Find the nature of roots of the equation $7y^2+2=2\sqrt{14}y$ without solving it.

Answer to Problem 12WE

The equation has real and equal roots.

Explanation of Solution

Given information: The given quadratic equation is,

  $7y^2+2=2\sqrt{14}y\mathrm{....}\left(\text{Given}\right)$

Calculation: The given equation can be written as,

  $7y^2-2\sqrt{14}y+2=\mathrm{0....}\left(\text{1}\right)$

Now if $\left(D\right)$ is the discriminant of quadratic equation (1) then using the formula $D=b^2-4ac$ ,

  $\begin{array}{c}D={\left(-2\sqrt{14}\right)}^2-4\left(7\right)\left(2\right)\mathrm{....}\left\{\begin{array}{l}a=7,b=-2\sqrt{14}\\ \text{and }c=2\end{array}\right\}\\ =56-56\\ =0\end{array}$

Since the value of discriminant is zero, therefore the roots are real and equal to each other. In other words the equation has double root.