How do you solve $5x^2+98=0$ using the quadratic formula?

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Answer 1 · 2016-09-11T21:26:05

You don't need the quadratic formula. But here goes:

Answer is:

Equation not valid in $RR$

Quadratic solving

$5x^2+98=0$

This is the same as:

$5x^2+0x+98=0$

For $a=5$ $b=0$ $c=98$ :

$Δ=b^2-4*a*c=0^2-4*5*98=-1960$

Since $Δ<0$ the equation in impossible inside $RR$ .

Simpler solution

$5x^2+98=0$

$5x^2=-98$

$x^2=-98/5$

No real number can have a negative square, therefore the equation in impossible inside $RR$ .

How do you solve 5x^2+98=05x^2+98=0 using the quadratic formula?