show that the equation X2+6X+6 = 0 has real roots and solve it
Answers
The given equation is X²+6X+6 = 0
Comparing it with aX²+bX + C = 0
a= 1 , b = 6 , C = 6.
THEREFORE,
D =(b²-4ac) = 36-4×1×6 = 12 greater than 0.
So,
the given equation has real roots.
Now,
✓D = ✓12 = 2✓3
Therefore,
Alpha = -b+✓D/2a = (-6+2✓3)/2×1 = (-6+2✓3)/2 = (-3+✓3)
Beta = -b-✓D/2a = (-6-2✓3)/2×1 = (-6-2✓3)/2 = (-3-✓3) .
Hence,
(-3-✓3) and (-3-✓3) are the roots of the given equation.
HOPE IT WILL HELP YOU... :-)
Given:
Equation: x²+6x+6=0
To find:
If the equation has real roots and the value of the roots
Solution:
The value of the real roots of the equation is (√3-3) and -(√3+3).
We can find the roots by following the given steps-
We know that an equation has real roots when the value of the discriminant of the equation is greater than or equal to 0.
The equation is x²+6x+6=0.
Here, the coefficient of x² is a
The coefficient of x is b
The constant term is c.
So, a=1, b=6, c=6.
The value of discriminant of the equation=b²-4ac
On putting the values of a, b, c, we get
Discriminant, D= 6²- 4×1×6
D=36-24
D=12
So, D>0.
The value of D is greater than 0 and thus, has real roots.
Now, we know that the value of the roots can be obtained by using the value of D.
Let us assume that the roots of the equation are A and B.
A= (-b+√D)/2a and B= (-b-√D)/2a
On putting the values, we get
A= (-6 +√12)/2×1
=(-6+2√3)/2= 2(-3+√3)/2
A=(√3-3)
Similarly, B=(-6-√12)/2×1
=(-6-2√3)/2= -2(3+√3)/2
B= -(3+√3)
Therefore, the value of the real roots of the equation is (√3-3) and -(√3+3).