Find the Solutions of a Quadratic Equation Using the Quadratic Formula

[mathlearn]

By completing the square or by any other method , find the solutions of the quadratic equation , to two decimal places $x^2+4x-8=0$.

Using the quadratic formula,(Smile)

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.

$x=\frac{-4\pm\sqrt{16+32}}{2}$.

$x=\frac{-4\pm\sqrt{48}}{2}$.

$x=\frac{-4\pm\sqrt{3}\sqrt{16}}{2}$.

$x=\frac{-4\pm 4\sqrt{3}}{2}$.

$x={-2\pm 2\sqrt{3}}$.

$x= 1.46$.

As the problem states there is another value for x which is -5.46, How to derive it using the quadratic formula ? (Happy)

 

[topsquark]

By completing the square or by any other method , find the solutions of the quadratic equation , to two decimal places $x^2+4x-8=0$.

Using the quadratic formula,(Smile)

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.

$x=\frac{-4\pm\sqrt{16+32}}{2}$.

$x=\frac{-4\pm\sqrt{48}}{2}$.

$x=\frac{-4\pm\sqrt{3}\sqrt{16}}{2}$.

$x=\frac{-4\pm 4\sqrt{3}}{2}$.

$x={-2\pm 2\sqrt{3}}$.

$x= 1.46$.

As the problem states there is another value for x which is -5.46, How to derive it using the quadratic formula ? (Happy)

$x={-2\pm 2\sqrt{3}}$.

$x= 1.46$.

You have another solution: \(\displaystyle -2 - 2 \sqrt{3}\)

-Dan

 

[mathlearn]

You have another solution: \(\displaystyle -2 - 2 \sqrt{2}\)

-Dan

(Happy) Thank you top squark

(Nod) A typing mistake I guess, \(\displaystyle -2 - 2 \sqrt{3}\) , I guess it should be . Correct ? (Thinking)

 

[topsquark]

(Happy) Thank you top squark

(Nod) A typing mistake I guess, \(\displaystyle -2 - 2 \sqrt{3}\) , I guess it should be . Correct ? (Thinking)

Yes. Thanks for the catch!

-Dan