We will show how to solve the shown below by using three different methods: graphing, factoring, and using the . Then, we will conclude which method works better for this case and why.
3x^2+11x+6 = 0
We can solve a quadratic equation of the form ax^2+bx+c = 0 by graphing its associated function f(x) = ax^2+bx+c. The of the graph are the solutions to the quadratic equation. Let's find the associated function for our equation.
&Original Equation &&Associated Function
& 3x^2+11x+6 = 0 && f(x) = 3x^2+11x+6
Now, we can graph the associated function to identify the solutions.
As we can see, the graph intersects the x-axis at (- 3, 0) and (- 23,0). Therefore, the equation has two real solutions, x=-3 and x=- 23.
b) Solving the Equation by Factoring
Notice that the expression of the left-hand side 3x^2+11x+6 does not have a , and it is of the form ax^2+bx+c.
ax^2+ bx+ c
3x^2+ 11x+ 6
Because a= 1, b= 11, and c= 6 are all positive, we need to look for positive factors of a and c. We can use a table to organize the information about the factors of a and c.
| Factors of 3
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Factors of 6
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Possible Factorization
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Middle Term
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| 1,3
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1,6
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(x+1)(3x+6)
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6x + 3x = 9x *
|
| 1,3
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6,1
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(x+6)(3x+1)
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x + 18x = 19x *
|
| 1,3
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2,3
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(x+2)(3x+3)
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3x + 6x = 9x *
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| 1,3
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3,2
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(x+3)(3x+2)
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2x + 9x = 11x âś“
|
With this information, we can write the quadratic equation in factored form.
(x+3)(3x+2) = 0
By the we get two conditions from which we can solve for the solutions.
x+3 = 0 ⇔ x= -3
3x+2 = 0 ⇔ x= - 2/3
As we can see above, there are two real solutions, x = - 3 and x = - 23.
c) Solving the Equation by Using the Quadratic Formula
Let's review the form of Quadratic Formula.
x = - b ± sqrt(b^2-4ac)/2a
To solve the quadratic equation by using the Quadratic Formula, we first need to identify the value of the parameters. We can do this by comparing our equation to the standard form of a quadratic equation.
ax^2+ bx+ c = 0
3x^2+ 11x+ 6 = 0
We can see that a= 3, b = 11, and c= 3. Now, we can substitute these values in the Quadratic Formula to solve the equation.
x = - b ± sqrt(b^2-4ac)/2a
x = - 11 ± sqrt(11^2-4( 3)( 6))/2 (3)
x = - 11 ± sqrt(121-4(3)(6))/2(3)
x = - 11 ± sqrt(121-72)/6
x = - 11 ± sqrt(49)/6
x = - 11 ± 7/6
We have two solutions, x = - 11 - 76, and x = - 11 +76.
| Solution 1
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Solution 2
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| x = - 11 - 7/6
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x = - 11 + 7/6
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| Add / Subtract Terms
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| x = - 18/6
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x = - 4/6
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| Calculate quotient and simplify
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| x = - 3
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x = - 2/3
|
Once more, we have found that the equation has two real solutions, x = - 3 and x = - 23.
Conclusion
Preferences may vary, but for this case, solving by using the Quadratic Formula is more convenient. The expression is not easily factorable and this method does not require graphing, which can be complicated if we do not have a graphing calculator at hand.
