Question

Determine the real life solutions of 9x^2+6x+1=0

Answer 1

Answer:

1  (double root)

Step-by-step explanation:

Given quadratic equation:

$9x^2+6x+1=0$

To find the solutions to the given quadratic equation, factor the equation.

To factor a quadratic in the form $ax^2+bx+c$, find two numbers that multiply to ac and sum to b.

$\implies ac=9 \cdot 1=9$

$\implies b=6$

Therefore, the two numbers are: 3 and 3.

Rewrite the middle term as the sum of these two numbers:

$\implies 9x^2+3x+3x+1=0$

Factor the first two terms and the last two terms separately:

$\implies 3x(3x+1)+1(3x+1)=0$

Factor out the common term (3x + 1):

$\implies (3x+1)(3x+1)=0$

Therefore:

$\implies(3x+1)^2=0$

This means that the curve has one root with multiplicity 2. So the curve touches the x-axis and bounces off the axis at one point. (See attached graph).

Apply the zero-product property:

$\implies 3x+1=0 \implies x=-\dfrac{1}{3}$

Therefore there is one real solution of the given quadratic equation.

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