Respuesta :
Answer and Explanation :
Given : Quadratic equations.
To find : Determine the number of real solutions each quadratic equation has.
Solution :
To determine the number of solution of quadratic equation [tex]ax^2+bx+c=0[/tex] we find the discriminant.
If [tex]D=b^2-4ac=0[/tex] the quadratic has one real solution.
If [tex]D=b^2-4ac<0[/tex] the quadratic has no real solution.
If [tex]D=b^2-4ac>0[/tex] the quadratic has two real solution.
Now, We find one by one
1) [tex]y=12x^2-9x+4[/tex]
a=12 , b=-9 , c=4
[tex]D=(-9)^2-4(12)(4)=81-192=-111<0[/tex]
[tex]y=12x^2-9x+4[/tex] has zero real solution(s).
2) [tex]10x+y=-x^2+2[/tex]
[tex]y=-x^2-10x+2[/tex]
a=-1 , b=-10 , c=2
[tex]D=(-10)^2-4(-1)(2)=100+8=108>0[/tex]
[tex]10x+y=-x^2+2[/tex] has two real solution(s).
3) [tex]4y-7=5x^2-x+2+3y[/tex]
[tex]y=5x^2-x+9[/tex]
a=5 , b=-1 , c=9
[tex]D=(-1)^2-4(5)(9)=1-180=-179<0[/tex]
[tex]4y-7=5x^2-x+2+3y[/tex] has zero real solution(s).
4) [tex]y=(-x+4)^2[/tex]
[tex]y=x^2-8x+16[/tex]
a=1 , b=-8 , c=16
[tex]D=(-8)^2-4(1)(16)=64-64=0[/tex]
[tex]y=(-x+4)^2[/tex] has one real solution(s).
