Question

# Find the discriminant of each equation and determine whether the equation has (1) two nonreal complex solutions, (2) one real solution with a multiplicity of 2, or (3) two real solutions. Do not solve the equations. 7x^{2} - 2x - 14 = 0

Equations
Find the discriminant of each equation and determine whether the equation has (1) two nonreal complex solutions, (2) one real solution with a multiplicity of 2, or (3) two real solutions. Do not solve the equations. $$7x^{2} - 2x - 14 = 0$$

2021-06-04

Step 1
We know that, for a standard form of quadratic $$ax^{2}+bx+c=0$$, the value of discriminant D is calculated as
$$D=b^{2}-4ac$$...(1)
Now,
If $$D>0$$, then both the roots of the equation must be real and distinct.
If $$D=0$$, then both the roots must be real and equal.
If $$D<0$$, then both the roots must be imaginary.
Step 2
We have the given quadratic equation as
$$7x^{2}-2x-14=0$$...(2)
On comparing the equation (2) with standard equation $$ax^{2}+bx+c=0$$, we get the result as
$$a=7, b=−2$$ and $$c=−14$$
On using equation (1), we get the discriminant of equation $$7x^{2}−2x−14=0$$ as
$$D=(-2)^{2}-(4)(7)(-14)$$
$$D=4+392$$
$$D=396$$
$$D>0$$
Hence, there will be two real and distinct solution.