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Determine the number of real solutions for each of the given equations.
Equation Number of Solutions
y = -3x2 + x + 12
y = 2x2 - 6x + 5
y = x2 + 7x - 11
y = -x2 - 8x - 16

Respuesta :

Answer:

1 2 real

2. No real

3. 2 Real

4.  One real Multiplicity 2.

Step-by-step explanation:

Find the value of the discriminant D  (b^2 - 4ac) :-

1 . D = (1)^1 -4 * -3 * 12 = 1 + 144 = 145

D is positive so there re 2 real solutions.

2.  D = (-6)^2 - 4*2*5 = 36 - 40 = -4

D is negative so there are NO real solutions.

3.  D = (7)^2 - 4 * 1 * -11 = 49 + 44 = 93

2 real

solutions

4.  D = (-8)^2 - 4* -1 * -16 = 64 - 64

1 solution. Multiplicity 2.

khuranashilpa76

The number of real solutions is option A,B and D.

How to find the equation  having real solution or not in a quadratic equation?

The number of real solution depends on the value of   b²-4ac

  • if b²-4ac is > o (equation will have  two real solution)
  • if b²-4ac is = o (equation will have  one real solution)
  • if b²-4ac is < o (equation will have  no real solution)

Solution for each equation:

Let y equals to 0 for all equations.

A) a= -3 , b=1 and c=12

    ⇒b²-4ac =1²-4×(-3)×12

    ⇒b²-4ac =-1²+144

   ⇒b²-4ac =143

∴ It has two real solutions.

B) a=2  , b=-6 and c=5

    ⇒b²-4ac =-6²-4×(-2)×5

    ⇒b²-4ac =-36+40

   ⇒b²-4ac =4

∴ It has two real solutions.

C) a=1  , b=7and c=-11

    ⇒b²-4ac =-7²-4×(1)×11

    ⇒b²-4ac =-49-44

   ⇒b²-4ac =-93

∴ It has no real solutions.

D) a=-1  , b=-8 and c=-16

    ⇒b²-4ac =-64²-4×(-1)×-16

    ⇒b²-4ac =-64+64

   ⇒b²-4ac =0

∴ It has one real solutions.

Learn more about quadratic equation here: https://brainly.com/question/1214333

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