Answer:
1 (double root)
Step-by-step explanation:
Given quadratic equation:
[tex]9x^2+6x+1=0[/tex]
To find the solutions to the given quadratic equation, factor the equation.
To factor a quadratic in the form [tex]ax^2+bx+c[/tex], find two numbers that multiply to ac and sum to b.
[tex]\implies ac=9 \cdot 1=9[/tex]
[tex]\implies b=6[/tex]
Therefore, the two numbers are: 3 and 3.
Rewrite the middle term as the sum of these two numbers:
[tex]\implies 9x^2+3x+3x+1=0[/tex]
Factor the first two terms and the last two terms separately:
[tex]\implies 3x(3x+1)+1(3x+1)=0[/tex]
Factor out the common term (3x + 1):
[tex]\implies (3x+1)(3x+1)=0[/tex]
Therefore:
[tex]\implies(3x+1)^2=0[/tex]
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:
[tex]\implies 3x+1=0 \implies x=-\dfrac{1}{3}[/tex]
Therefore there is one real solution of the given quadratic equation.
Learn more about factoring quadratic equations here:
https://brainly.com/question/27956741
https://brainly.com/question/27947331
