Answer with Step-by-step explanation:
We are given that
Discriminant, D=[tex]b^2-4ac[/tex]
When [tex]D\geq 0[/tex]
Then, the function have two real zeroes.
1.[tex]f(x)=x^2+6x+8[/tex]
By comparing withe general quadratic equation
[tex]ax^2+bx+c=0[/tex]
We get a=1,b=6,c=8
Using the discriminant formula
[tex]D=(6)^2-4(1)(8)=36-32=4>0[/tex]
Hence, function have two real zeroes.
2.[tex]g(x)=x^2+4x+8[/tex]
[tex]D=(4)^2-4(1)(8)=16-32[/tex]
[tex]D=-16<0[/tex]
Hence, the function have no two real number zeroes.
3.[tex]h(x)=x^2-12x+32[/tex]
[tex]D=(-12)^2-4(1)(32)=144-128=16[/tex]
[tex]D>0[/tex]
Hence, function have two real zeroes.
4.[tex]k(x)=x^2+4x-1[/tex]
[tex]D=(4)^2-4(1)(-1)=16+4=20[/tex]
[tex]D>0[/tex]
Hence, function have two real zeroes.
5.[tex]p(x)=5x^2+5x+4[/tex]
[tex]D=(5)^2-4(5)(4)=25-80[/tex]
[tex]D=-55<0[/tex]
Hence, the function have no two real number zeroes.
6.[tex]r(x)=x^2-2x-15[/tex]
[tex]D=(-2)^2-4(1)(-15)=4+60=64[/tex]
[tex]D>0[/tex]
Hence, function have two real zeroes.
