Question

Please help, and thank you!

Answer 1

Answer:

3. C

4. D

Step-by-step explanation:

3.

$x^2+25=0\\x^2=-25\\x=\pm\sqrt{-25}$

Once $\sqrt{-1}=i$

$x=\pm5i$

4.

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

Let's try all the equations:

(A)

$2x^2-7x-9=0\\\\\Delta=\sqrt{\left(-7\right)^2-4\cdot \:2\left(-9\right)}\\\Delta=\sqrt{\left(-7\right)^2+4\cdot \:2\cdot \:9}\\\Delta=\sqrt{121} \\\Delta=11$

(B)

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

$x=\frac{7}{2}$

(C)

$2x^2+7x-9=0\\\Delta = \sqrt{7^2-4\cdot \:2\left(-9\right)}\\\Delta = \sqrt{7^2+4\cdot \:2\cdot \:9}\\\Delta = \sqrt{121} \\\Delta= 11$

(D)

$2x^2-7x+9=0\\\Delta = \sqrt{\left(-7\right)^2-4\cdot \:2\cdot \:9}\\\\\Delta=\sqrt{-23} \\\Delta=\sqrt{23i}$