Question

What are the solutions of this quadratic equation? x2 + 10 = 0

Answer 1

Answer:

x = ± i $\sqrt{10}$

Step-by-step explanation:

$x^{2} + 10 = 0$

Since this isn't a perfect square, we solve for x using the square root property.

Subtract 10 from each side.

$x^{2} + 10 - 10 = 0 - 10$

$x^{2} = - 10$

To solve fo x we need to take the square root of each side.

$\sqrt{ x^{2} } = \sqrt{-10}$

$x = \sqrt{-10}$

Because there is a negative under the square root sign we know there will be an imaginary number, i.

± i = $\sqrt{-1}$

So we break it down more.

$x = \sqrt{-1} *\sqrt{10}$

Because we know $\sqrt{-1}$ =  i then we know x.

x = ± i $\sqrt{10}$

Here we solve using the quadratic formula:

$x = \frac{-b ± \sqrt{b^{2} -4ac} }{2a}$  (ignore the A with the line over it)

a = 1  b = 0   c = 10

Plug in the values.

$x = \frac{-0 ± \sqrt{0^{2} - 4*1*10} }{2*1}$ (ignore the A again)

$x = \frac{ ± \sqrt{- 40} }{2}$ (keep ignoring the A with the hat)

If possible, you always want to look at your square root and see if any multiples of the number have a square root. In this case 4 has a square root of 2.

x = ± $\frac{\sqrt{4}* \sqrt{-10} }{2}$  

x = ± $\frac{2\sqrt{-10} }{2}$   2/2 = 1 so it cancels

x = ± $\sqrt{-10}$

Again we will have an imaginary number because $\sqrt{-1}$ =  i

x = ±$\sqrt{-1} \sqrt{10}$

x = ± i $\sqrt{10}$

Answer 2

Answer:

x=-5

Step-by-step explanation:

don't think thats a quadratic equation, in that case the answer is above, just subtract from both sides