Question

show that the equation 4x²+3x-5=0has no real roots​

Answer 1

Step-by-step explanation:

4x²+3x-5=0

7x³-5=0

7x³=0+5

7x³=5

x=7/5

Answer 2

Comparing the given polynomial with ax²+bx+c=0

a=4, b=3 and c=-5

D= b²-4ac

D=(3)²-4(4)(-5)

D=9+80

D= 89

D>0

Therefore,the given polynomial has two distinct real roots