Yes, because a y-intercept can always be a fraction.
Answer:
the answer is 4
Step-by-step explanation:
Answer:
7x+ 14
Step-by-step explanation:
Answer:
Please check the explanation.
Step-by-step explanation:
The general quadratic equation is
ax²+bx+c=0
The discriminant = D = b² - 4ac
When b² - 4ac = 0 there is one real root.
When b² - 4ac > 0 there are two real roots.
When b² - 4ac < 0 there are two complex roots.
1) x² -6x + 9=0
On comparing with given quadratic equation x² -6x + 9=0
a = 1, b=-6, c=9
D = b² - 4ac
= (-6)² - 4(1)(9)
= 36 - 36
= 0
D = 0
Thus, there is one real root of quadratic equation x² -6x + 9=0.
2) x² -4x + 3=0
On comparing with given quadratic equation x² -4x + 3=0
a = 1, b=-4, c=3
D = b² - 4ac
= (-4)² - 4(1)(3)
= 16 - 12
= 4
D > 0
Thus, there are two real roots of quadratic equation x² -4x + 3=0.
3) x² -7x - 4=0
On comparing with given quadratic equation x² -7x - 4=0
a = 1, b=-7, c=-4
D = b² - 4ac
= (-7)² - 4(1)(-4)
= 49 + 16
= 65
D > 0
Thus, there are two real roots of quadratic equation x² -7x - 4=0
4) 2x² +3x +5=0
On comparing with given quadratic equation 2x² +3x +5=0
a = 2, b=3, c=5
D = b² - 4ac
= (3)² - 4(2)(5)
= 9 - 40
= -31
D < 0
Thus, there are two complex roots of quadratic equation 2x² +3x +5=0
