Question

find nature of roots of quadratic equation 4 root 5 x square + 7 x minus 3 root 5 equal to zero.

Answer 1

Here is your answer

root 5 x² + 7x- 3 root 5

discriminant = b²-4ac
= (7)^2 - 4×4✓5×3✓5
= 49 - 240
= -191

-191<0
so the given has no real roots.

Hope it helps you ^_^

Answer 2

Answer:

Two real roots.

Step-by-step explanation:

A quadratic equation ax²+ bx + c has real roots,

If the value of b² - 4ac ≥ 0,

And, it has imaginary roots,

If the value of b² - 4ac < 0

Here, the given equation,

4√5 x² + 7x - 3√5 = 0,

By comparing it with the above equation,

a = 4√5, b = 7 and c = -3√5,

Since, (7)² - 4 × 4√5 × -3√5

= 49 + 240

= 289 > 0

Thus, the given equation has two real roots.