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What is a solution of x^2+6x=-5

Sagot :

konrad509
[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

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