Question

Find the number of real solutions of the equation $x^2+rx-s=0\left(s>0\right)$by using discriminant.

Short answer

The equation has two distinct real solutions.

Step-by-step solution

Step 1. From discriminant method of quadratic equation.

An equation of the form$a{x}^2+bx+c=0$ where$a\ne 0$ has discriminant$\left(D\right)=b^2-4ac$ in which three particular cases arises.

1. If$D>0$, the equation has two distinct real solutions.
2. If$D<0$, the equation has no real solution.
3. If $D=0$, the equation has exactly one real solution.

Step 2. Evaluate discriminant of given quadratic.

The discriminant of given quadratic equation $x^2+rx-s=0\left(s>0\right)$is evaluated as,

role="math" localid="1643827623607" $$\begin{gathered} D={\left(r\right)}^2-4\left(1\right)\left(-s\right)x^2+rx-s=0,\left(s>0\right) \\ =r^2+4s \\ >0....\left(\text{As}{r}^2\ge 0\text{and}s>0\right) \\ >0....\left(1\right) \end{gathered}$$

Step 3. DESCRIPTION OF STEPS.

Since$D>0$ from (1), therefore the equation has two distinct real solutions.