Question

If the discriminant of a quadratic equation is $ - 5$. What type of solutions does the equation have?

Answer 1

Hint: For a quadratic equation, there is a formula to find the discriminant of a quadratic equation. There are some conditions involved with the sign of discriminant regarding the solution of the equation.

Complete step by step answer:
We can find the solutions of a quadratic equation using the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. Here ${b^2} - 4ac$ gives us the value of discriminant.
Therefore, its value decides what type of solution we will get.
In a quadratic equation, there are three types of solutions based on the values of discriminant.
$\left( 1 \right)$ If the value of the discriminant is positive, then we have real and distinct solutions.
$\left( 2 \right)$ if the value of discriminant is zero, then we have real and equal solutions.
$\left( 3 \right)$ if the value of the discriminant is negative, then we have no real solution or imaginary solution.
Now,
In the above question it is given that the value of discriminant is $ - 5$. Here the value of discriminant is negative.
Therefore, the equation has no real solution.

Note:
The discriminant of a quadratic polynomial is the portion of the quadratic formula under the square root symbol ${b^2} - 4ac$:, that tells whether there are two solutions, one solution, or no solutions to the given equation. The discriminant value helps to determine the nature of the roots of the quadratic equation. The relationship between the discriminant value and the nature of roots are as follows:
$\left( 1 \right)$If \[discriminant{\text{ }} > {\text{ }}0\], then the roots are real and unequal
$\left( 2 \right)$If \[discriminant = 0\], then the roots are real and equal
$\left( 3 \right)$If \[discriminant{\text{ }} < {\text{ }}0\], then the roots are not real (we get a complex solution)
 Sometimes students get confused whether the determinant is ${b^2} - 4ac$ or $\sqrt{{b^2} - 4ac}$.