Question

Please help, and thank you!

Answer 1

3. C

4. D

Explanation:

3.

`x^2+25=0\\x^2=-25\\x=\pm√(-25)`

Once
`√(-1)=i`

`x=\pm5i`

4.

A quadratic equation has no real roots when the discriminant (
`\Delta`) is negative.
`\Delta<0`.

Let's try all the equations:

(A)

`2x^2-7x-9=0\\\\\Delta=√(\left(-7\right)^2-4\cdot \:2\left(-9\right))\\\Delta=√(\left(-7\right)^2+4\cdot \:2\cdot \:9)\\\Delta=√(121) \\\Delta=11`

(B)

`2x^2=7x\\2x^2-7x=0\\x(2x-7)=0\\x=0\\\text{or}\\2x-7=0\\2x=7\\\\`

`$x=(7)/(2) $`

(C)

`2x^2+7x-9=0\\\Delta = √(7^2-4\cdot \:2\left(-9\right))\\\Delta = √(7^2+4\cdot \:2\cdot \:9)\\\Delta = √(121) \\\Delta= 11`

(D)

`2x^2-7x+9=0\\\Delta = √(\left(-7\right)^2-4\cdot \:2\cdot \:9)\\\\\Delta=√(-23) \\\Delta=√(23i)`