Question

Which statement about the following equation is true? 2x2 – 9x + 2 = –1 The discriminant is less than 0, so there are two real roots. The discriminant is less than 0, so there are two complex roots. The discriminant is greater than 0, so there are two real roots. The discriminant is greater than 0, so there are two complex roots.

Answer 1

Answer:

The correct statement is The discriminant is greater than 0, so there are two real roots.

Step-by-step explanation:

Given equation is $2x^{2}-9x+2=-1$

⇒ $2x^{2}-9x+3=0$

we know that if the equation has discriminant greater than than 0 then the real roots exists. So, we find the discriminant first which can be calculated as

$D= b^{2}-4ac$  for the equation $ax^{2}+bx+c=0$

∴  $D=(-9)^{2}-4(2)(3)$  

             =$81-24=57>0$

Hence, for the given equation the correct statement is

The discriminant is greater than 0, so there are two real roots.

Answer 2

Answer:

There are two real roots

Step-by-step explanation:

2x^2 - 9x + 2 = -1

2x^2 - 9x + 2 + 1 =0

2x^2 - 9x + 3 = 0

Discriminant = b^2 - 4ac

a= 2

b=(-9)

c=3

Discriminant = b^2 - 4ac

                     = (-9)^2 - 4* 2* 3

                     = 81 - 24

                     =57

If the value of determinant is greater than 0 and is a perfect square then roots are real , unequal and rational.

If the value of determinant is greater than 0 and is not a perfect square then roots are real , unequal and irrational.

So, there are two real roots

#SPJ2