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Anon25 [30]
3 years ago
9

Please help, and thank you!

Mathematics
1 answer:
lawyer [7]3 years ago
8 0

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}

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