Answer :
[tex]x^2+6x=-5\\
x^2+6x+5=0\\
x^2+x+5x+5=0\\
x(x+1)+5(x+1)=0\\
(x+5)(x+1)=0\\
x=-5 \vee x=-1[/tex]
Answer:
Solutions / zeroes of x² + 6x + 5 = 0 are -5 , -1
Step-by-step explanation:
Given: Quadratic equation, x² + 6x = -5
To find: Value of x or zeroes of equation
Rewrite the given equation in standard form,
x² + 6x + 5 = 0
now comparing it with standard form of quadratic equation
ax² + bx + c = 0
we get, a = 1 , b = 6 , c = 5
we first find discriminant, D to check if zeroes exit or not and nature of zeroes.
D = b² - 4ac
= 6² - 4 × 1 × 5
= 36 - 20
D = 16
Since, D > 0 i.e., positive
⇒ zeroes exist and they are real & distinct.
Now e can solve it by any method,
By middle term split method, we get
x² + 6x + 5 = 0
x² + 5x + x + 5 = 0
x ( x + 5) + ( x + 5 ) = 0
( x + 5 ) ( x + 1 ) = 0
now equating both factor with 0 we get,
x + 5 = 0 and x + 1 = 0
x = -5 , -1
Therefore, Solutions / zeroes of x² + 6x + 5 = 0 are -5 , -1
