Lewis Harvey

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. $7{x}^{2}-2x-14=0$

tabuordy

Step 1
We know that, for a standard form of quadratic $a{x}^{2}+bx+c=0$, 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
$7{x}^{2}-2x-14=0$...(2)
On comparing the equation (2) with standard equation $a{x}^{2}+bx+c=0$, we get the result as
$a=7,b=-2$ and $c=-14$
On using equation (1), we get the discriminant of equation $7{x}^{2}-2x-14=0$ as
$D=\left(-2{\right)}^{2}-\left(4\right)\left(7\right)\left(-14\right)$
$D=4+392$
$D=396$
$D>0$
Hence, there will be two real and distinct solution.

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