Question

Which equation could be solved using the graph above?

X^2+4x+3=0
X^2-4x+3=0
X^2-6x+9=0
X^2-1=0

Answer 1

Option A.

Explanation:

The vertex form of a parabola is

`y=a(x-h)^2+k`

where, (h,k) is vertex and a is a constant.

The vertex of the parabola is (-2,-1).

Substitute h=-2 and k=-1 in the above equation.

`y=a(x-(-2))^2+(-1)`

`y=a(x+2)^2-1` .... (1)

The parabola passes through the point (0,3). So, it must be satisfy by the point (0,3).

`3=a(0+2)^2-1`

`3+1=4a`

`a=1`

Substiturte a=1 in equation (1).

`y=1(x+2)^2-1`

On simplification we get

`y=x^2+4x+4-1`

`y=x^2+4x+3`

It means the equation
`x^2+4x+3=0` could be solved using the given graph.

Therefore, the correct option is A.

Answer 2

X²+4x+3=0

Explanation:

y = X² + 4x + 3

y = 0

Solution:

x = -3, x² = -1

Steps:

X² + 4x + 3 = 0

write 4x as a sum

X²+ 3x + x + 3 = 0

Factor out x from expression

X (x + 3) + x + 3 = 0

Factor out x + 3 from expression

(x + 3) x (x + 1) = 0

When the product of factors equals 0, at least one factor is 0

x + 3 = 0

x + 1 = 0

solve for x

x = -3

x + 1 = 0

solve for x

x = -3

x = -1

The final solutions are

X = -3, X² = -1