+9466 4a2 - 10a + 6 = 0 First, let's divide through by 2 .
2a2 - 5a + 3 = 0
We can use the Quadratic formula to solve this problem for a .
The Quadratic formula says, if you have an equation written in the form
Ax2 + Bx + C = 0 , then... \(x = {-{\color{blue}B} \pm \sqrt{{\color{blue}B}^2-4{\color{red}A}{\color{green}C}} \over 2{\color{red}A}}\)
So, for our problem...
2a2 + -5a + 3 = 0
x = a , A = 2 , B = -5 , C = 3 . Substitute these values into the Quadratic formula.
\(a = {-{\color{blue}(-5)} \pm \sqrt{{\color{blue}(-5)}^2-4{\color{red}(2)}{\color{green}(3)}} \over 2{\color{red}(2)}} \\~\\ a = {5 \pm \sqrt{25-24} \over 4} \\~\\ a={5 \pm \sqrt{1} \over 4} \\~\\ a={5 \pm 1 \over 4} \\~\\ \begin{array}\ a=\frac{5+1}{4}\qquad \text{ or }\qquad &a&=\frac{5-1}{4} \\~\\ a=\frac{6}{4}&a&=\frac44 \\~\\ a=\frac32&a&=1 \\~\\ \end{array}\)
+2441 Hectictar's method is perfectly executed, but this equation is also factorable. How do I know? There is a universal test that determines if a quadratic is factorable. It is the following:
If \(\sqrt{b^2-4ac}\in\mathbb{Q}\hspace{1mm}\text{and}\sqrt{b^2-4ac}\geq0\), then the quadratic is factorable. In other words, the standard form of a quadratic equation is \(ax^2+bx+c\). If you apply the rule above and get a rational number (integers and decimals that terminate or repeat) and is greater than or equal to 0, then the quadratic is factorable. Let's try it:
\(4a^2-10a+6=0\)
Before, we start, let's determine our a's, b's, and c's. Our a is the coefficient of the quadratic term. In this case, that is 4. OUr b is the coefficient in front of the linear term, which is -10. Make sure to include the sign when doing this calculation. Our c is our constant term, which is 6. Let's apply the rule and see if this equation is factorable:
| \(\sqrt{b^2-4ac}\in\mathbb{Q}\hspace{1mm}\text{and}\sqrt{b^2-4ac}\geq0\) | Check to see if the condition is true by plugging in the appropriate values for a, b, and c. |
| \(\sqrt{(-10)^2-4(4)(6)}\) | Simplify inside the radical first. |
| \(\sqrt{100-96}\) | |
| \(\sqrt{4}\) | |
| \(2\) | This meets the condition of being apart of the set of rational numbers and being greater than or equal to 0. |
Great! Now that we have determined that this quadratic can be factored, let's factor it! I'll use a picture to demonstrate what I am doing:
Our job is to find 2 factors that both multiply to get -24 and add to get -10. If you toy with the numbers for some time, you will eventually figure this out:
Now that we have figured out which two numbers satisfy the above problem, let's split the b-term:
| \(4a^2-10a+6=0\) | Split the -10a into -6a and -4a. | ||
| \(4a^2-4a-6a+6=0\) | Solve this by grouping. | ||
| \((4a^2-4a)+(-6a+6)=0\) | Factor out the GCF of both in the parentheses. | ||
| \(4a(a-1)-6(a-1)=0\) | Now, use the rule that \(a*c\pm b*c=(a\pm b)(c)\). | ||
| \((4a-6)(a-1)=0\) | In the first set of parentheses, you can factor out a GCF of 2 from the equation, so let's do that. | ||
| \(2(2a-3)(a-1)=0\) | Set both factors equal to zero and solve each set by using the zero-product theorem. | ||
| Add the constants in both equations. | ||
| |||
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Therefore, your solution set is the following:
\(a_1=\frac{3}{2}\)
\(a_2=1\)
Now, you are done!
