Question

Determine the natural of root of the quadratic equation 2x² - 3x = 4​

Answer 1

Answer:

b²-4ac=41

b²-4ac>0

The roots are real and unequal

we get two distinct roots

$\frac{-b+\sqrt{ b^{2}-4ac }}{2a}$ and $\frac{-b-\sqrt{ b^{2}-4ac }}{2a}$

Step-by-step explanation:

2x²-3x-4=0

a=2

b=-3

c=-4

To find Nature of roots

b²-4ac

(-3)²-4(2)(-4)  

=9+32

b²-4ac=41

41>0

b²-4ac>0

The roots are real and unequal

we get two distinct roots

$\frac{-b+\sqrt{ b^{2}-4ac }}{2a}$ and $\frac{-b-\sqrt{ b^{2}-4ac }}{2a}$

(-b±$\sqrt{b^{2}-4ac }$)/2a

3±$\sqrt{(-3)^{2}-4(2)(-4) }$)/2(2)

3±$\sqrt{(-3)^{2}-4(2)(-4) }$)/4

3±$\sqrt{9+32 }$)/4

3±$\sqrt{41}$)/4

The roots are$\frac{3+\sqrt{41}}{4}$  and $\frac{3-\sqrt{41}}{4}$