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

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asked 2021-06-03
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. \(7x^{2} - 2x - 14 = 0\)

Answers (1)

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.

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