Answer:
3. C
4. D
Step-by-step explanation:
3.
[tex]x^2+25=0\\x^2=-25\\x=\pm\sqrt{-25}[/tex]
Once [tex]\sqrt{-1}=i[/tex]
[tex]x=\pm5i[/tex]
4.
A quadratic equation has no real roots when the discriminant ([tex]\Delta[/tex]) is negative. [tex]\Delta<0[/tex].
Let's try all the equations:
(A)
[tex]2x^2-7x-9=0\\\\\Delta=\sqrt{\left(-7\right)^2-4\cdot \:2\left(-9\right)}\\\Delta=\sqrt{\left(-7\right)^2+4\cdot \:2\cdot \:9}\\\Delta=\sqrt{121} \\\Delta=11[/tex]
(B)
[tex]2x^2=7x\\2x^2-7x=0\\x(2x-7)=0\\x=0\\\text{or}\\2x-7=0\\2x=7\\\\[/tex]
[tex]$x=\frac{7}{2} $[/tex]
(C)
[tex]2x^2+7x-9=0\\\Delta = \sqrt{7^2-4\cdot \:2\left(-9\right)}\\\Delta = \sqrt{7^2+4\cdot \:2\cdot \:9}\\\Delta = \sqrt{121} \\\Delta= 11[/tex]
(D)
[tex]2x^2-7x+9=0\\\Delta = \sqrt{\left(-7\right)^2-4\cdot \:2\cdot \:9}\\\\\Delta=\sqrt{-23} \\\Delta=\sqrt{23i}[/tex]
