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


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Respuesta :

no. of real solns depends on the term b^2 - 4ac for y = ax^2 + bx + c
if its <0, no real soln
its =0, 1 real soln
its >0, 2 real solns

for
 y = -3x2 + x + 12, 1^2-4(-3)(12) > 0, two solns
 y = 2x2 - 6x + 5, -6^2-4(2)(5)  < 0, no soln
 y = x2 + 7x - 11, 7^2-4(1)(-11) > 0, two solns
 y = -x2 - 8x - 16, -8^2-4(-1)(-16) = 0, one soln

According to the discriminant:

  • [tex]y = -3x^2 + x + 12[/tex] has two real solutions.
  • [tex]y = 2x^2 - 6x + 5[/tex] has zero real solutions.
  • [tex]y = x^2 + 7x - 11[/tex] has two real solutions.
  • [tex]y = -x^2 - 8x - 16[/tex] has one real solution.

The number of real solutions of quadratic equation [tex]y = ax^2 + bx + c[/tex] depends on the discriminant [tex]\Delta = b^2 - 4ac[/tex].

  • If [tex]\Delta > 0[/tex], the equation has two real solutions.
  • If [tex]\Delta = 0[/tex], the equation has one real solution.
  • If [tex]\Delta < 0[/tex], the equation has zero real solutions.

Equation [tex]y = -3x^2 + x + 12[/tex]

  • The coefficients are: [tex]a = -3, b = 1, c = 12[/tex]
  • The discriminant is: [tex]\Delta = 1^2 - 4(-3)(12) = 145[/tex].
  • Positive, thus two real solutions.

Equation [tex]y = 2x^2 - 6x + 5[/tex]

  • The coefficients are: [tex]a = -2, b = -6, c = 5[/tex]
  • The discriminant is: [tex]\Delta = (-6)^2 - 4(2)(5) = -4[/tex].
  • Positive, thus zero real solutions.

Equation [tex]y = x^2 + 7x - 11[/tex]

  • The coefficients are: [tex]a = 1, b = 7, c = -11[/tex]
  • The discriminant is: [tex]\Delta = (7)^2 - 4(1)(-11) = 93[/tex].
  • Positive, thus two real solutions.

Equation [tex]y = -x^2 - 8x - 16[/tex]

  • The coefficients are: [tex]a = -1, b = -8, c = -16[/tex]
  • The discriminant is: [tex]\Delta = (-8)^2 - 4(-1)(-16) = 0[/tex].
  • Positive, thus one real solutions.

A similar problem is given at https://brainly.com/question/19776811