Answer: Option 1,3,4,6 are correct
Explanation
we have formula for discriminant that is [tex]D=b^{2} -4ac[/tex]
when D>0 the roots are real and unequal
when D= 0 roots are real and equal
and when D< 0 roots are imaginary or not real and unequal
In 1st equation: [tex]x^2+6x+8[/tex]
a=1,b=6,c=8 on substituting the values in the formula we will get
[tex]D=6^2-4(1)(8)=4>0[/tex] hence real and unequal roots.
In 2nd equation:[tex]x^2+4x+8[/tex]
a=1,b=4,c=8 on substituting values in formula
[tex]D=4^2-4(1)(8)=-16<0[/tex] so roots are not real
In 3rd equation [tex]x^2-12x+3[/tex]
a=1,b=-12,c=3 on substituting values
D=[tex]D=(-12)^2-4(1)(3)=132>0[/tex] roots are real and unequal
In 4th equation: [tex]x^2+4x-1[/tex]
a=1,b=4,c=-1 on substituting the values
[tex]D= 4^2-4(1)(-1)=20>0[/tex] real roots
In 5th equation:[tex]5x^2+5x+4[/tex]
a=5,b=5,c=4 on substituting the values
[tex]D=5^2-4(5)(4)=-55<0[/tex] roots are not real
In 6th Equation: [tex]x^2-2x-15[/tex]
a=1,b=-2,c=-15 on substituting the values
[tex]D=(-2)^2-4(1)(-15)=64>0[/tex] roots are real.
An equation has real roots when the discriminant returns a positive value.
We can use the discriminant to determine the type of roots that quadratic equation has. We shall now proceed to use the formula; D = b^2 – 4ac on each equation;
f(x) = x^2 + 6x + 8
D = (6)^2 - 4 (1) (8)
D = 4
The equation has real roots
x^2 + 4x + 8
D = (4)^2 - 4(1) (8)
D = -16
The equation does not have real roots
x^2 – 12x + 32
D = (-12)^2 - 4 (1) (32)
D = 16
The equation has real roots
x^2 + 4x – 1
D = (4)^2 - 4 (1) (-1)
D = 20
The equation has real roots
5x^2 + 5x + 4
D = (5)^2 - 4 (5) (4)
D = -55
The equation does not have real roots
x^2 – 2x – 15
D = (-2)^2 - 4 (1) (-15)
D = 64
The equation has real roots
An equation has real roots when the discriminant returns a positive value.
Learn more about quadratic discriminant: https://brainly.com/question/7669100
