Question

Classify the quadratic equation based on the number of solutions

Answer 1

QUESTION 1

The given equation is


`3 {x}^(2) + 2x = - 4`

We rewrite in the form,


`a {x}^(2) + bx + c = 0`

This implies that,


`3 {x}^(2) + 2x + 4 = 0`


`a=3,b=2,c=4`

We can use the determinant to find the number of solutions without necessarily solving the equation.


`D=b^2-4ac`


`D=2^2-4(3)(4)`


`D=4-48`


`D=-44`

Since the determinant is negative the equation


`3 {x}^(2) + 2x = - 4`

has no solution.

QUESTION 2

The given equation is


`5 {x}^(2) + 14 = 19`


`5 {x}^(2) + 14 - 19 = 0`


`5 {x}^(2) - 5= 0`


`{x}^(2) - 1= 0`


`D=0^2-4(1)( - 1)`


`D=4`

Since the determinant is positive, the equation


`5 {x}^(2) + 14 = 19`

has two solutions.

QUESTION 3

The given equation is


`2 {x}^(2) + 5 = 2`

This implies that,


`2 {x}^(2) + 5 - 2 = 0`


`2 {x}^(2) + 3 = 0`


`a=2,b=0,c=3.`

The determinant is


`D= {0}^(2) - 4(2)(3)`


`D= - 24`

Since the determinant is negative, the equation


`2 {x}^(2) + 5 = 2`
has no solution.

QUESTION 4

The given equation is


`2 {x}^(2) + 3x = 5`

We rewrite to obtain,


`2 {x}^(2) + 3x - 5 = 0`


`a=2,b=3,c=-5`

The determinant is


`D= {3}^(2) - 4(2)( - 5)`


`D= 9 + 40`


`D= 49`

Since the determinant is positive the equation


`2 {x}^(2) + 3x = 5`

has two solutions.

QUESTION 5

The given equation is


`4 {x}^(2) + 12x = - 9`

We rewrite to obtain,


`4 {x}^(2) + 12x + 9 = 0`


`a=4,b=12,c=9`

We substitute in the determinant formula to obtain,


`D=12^2 - 4(4)(9)`

.

`D=144- 144`


`D=0`

Since the determinant is zero the equation


`4 {x}^(2) + 12x = - 9`
has only one root.