Chapter 9.6, Problem 44PPS

To determine

The value of the discriminant of the equation $x^2+\frac{2}{25}=\frac{3}{5}x$ and find the number of real solutions.

Answer to Problem 44PPS

  $D=\frac{1}{25},\text{ there are two distinct real roots}$

Explanation of Solution

Given information:

The given equation, $x^2+\frac{2}{25}=\frac{3}{5}x$

Concept used:

• The discriminant is given by, $D=b^2-4ac$
• For a quadratic equation $a{x}^2+bx+c=0$ the number of roots depend on:
• If discriminant is negative, $b^2-4ac<0$, then there is no real roots
• If discriminant is zero, $b^2-4ac=0$, then there are two same real roots
• If discriminant is positive, $b^2-4ac>0$, then there are two distinct real roots

Calculation:

To find the discriminant of the given quadratic equation first rewrite the given equation in standard form by subtract both sides by $\frac{3}{5}x$ So it gives

  $x^2-\frac{3}{5}x+\frac{2}{25}=0$

Now compare this equation with the general equation, $a{x}^2+bx+c=0$ to get the value of the variables, $a=1,b=\frac{-3}{5},c=\frac{2}{25}$ . So, the discriminant of the given equation is

  $\begin{array}{c}D=b^2-4ac\\ ={\left(\frac{-3}{5}\right)}^2-4\left(1\right)\left(\frac{2}{25}\right)\\ =\frac{9}{25}-\frac{8}{25}\\ =\frac{1}{25}\end{array}$

So, the discriminant of the given equation is $\frac{1}{25}$ .

Since the discriminant is positive and greater than zero, so there are there are two distinct roots of the given equation.