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.

Answer 2

Answer:

Step-by-step explanation:

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