Question

For each equation, determine the number of solutions and place on the appropriate field in the table. 3x2 + 24x = -48 4x2 - 16x = 0 5x2 + 2 = 4x 3(x + 5)2 = -2

Answer 1

3x^2+24x=-48 -> one solution, x=-4

4x^2-16x=0 -> has two solutions, x=0 and x=4

5x^2+2=4x -> has no solutions,

3(x+5)^2=-2 -> has no solutions.

Explanation:

Answer 2

A). 3x² + 24x = -48 ---- one solution x = -4

B). 4x² - 16x = 0 --- two real solutions x = 0, 4

C). 5x² + 2 = 4x ---- has no real solution

D). 3(x + 5)² = -2 ---- has no real solution

Explanation:

We will solve each equation to get the solutions.

A).3x² + 24x = -48

we divide this equation by 3

x² + 8x = -16

x² + 8x + 16 = 0

(x + 4)² = 0

x = -4

There is only one solution, x = -4

B). 4x² - 16x = 0

4x(x - 4) = 0

x = 0 and x = 4 are the solutions.

C). 5x² + 2 = 4x

5x² - 4x + 2= 0

We will find the value of discriminant (b²- 4ac)

= (-4)²- 4(5)(2)

= 16 - 40

= -24 < 0

So this equation has no real solution.

D). 3(x + 5)² = -2

(x + 5)² =
`-(2)/(3)`

(x + 5) =
`\sqrt{-(2)/(3) }`

x = -5 +
`\sqrt{-(2)/(3) }`

So this equation has no real solutions.