Chapter 9.6, Problem 51HP

To determine

The value of the discriminant of the equation $2x^2-3x+5k=0$ and find the number of real solutions.

Answer to Problem 51HP

  $k\le \frac{9}{40}$

Explanation of Solution

Given information:

The given equation, $2x^2-3x+5k=0$

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 So it gives

  $2x^2-3x+5k=0$

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

  $\begin{array}{c}D=b^2-4ac\\ ={\left(-3\right)}^2-4\left(2\right)\left(5k\right)\\ =9-40k\end{array}$

Assume $k\in Q$

For the equation have two roots (real and complex)the discrimnant must not equal zero

  $\begin{array}{l}\Rightarrow 9-40k\ne 0\\ k\ne \frac{9}{40}\end{array}$

If the roots are to be real only the discriminant must be greater than zero

  $\begin{array}{c}\Rightarrow 9-40k\ge 0\\ \Rightarrow 40k\le 9\\ \Rightarrow k\le \frac{9}{40}\end{array}$

So, the discriminant of the given equation is $k\le \frac{9}{40}$ .

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