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# How to solve this problem?

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4a^2-10a+6=0

Aug 9, 2017

#1
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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.

$$a = {-{\color{blue}(-5)} \pm \sqrt{{\color{blue}(-5)}^2-4{\color{red}(2)}{\color{green}(3)}} \over 2{\color{red}(2)}} \\~\\ a = {5 \pm \sqrt{25-24} \over 4} \\~\\ a={5 \pm \sqrt{1} \over 4} \\~\\ a={5 \pm 1 \over 4} \\~\\ \begin{array}\ a=\frac{5+1}{4}\qquad \text{ or }\qquad &a&=\frac{5-1}{4} \\~\\ a=\frac{6}{4}&a&=\frac44 \\~\\ a=\frac32&a&=1 \\~\\ \end{array}$$

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Aug 9, 2017
#2
+4

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.
 $$2a-3=0$$ $$a-1=0$$

Add the constants in both equations.
 $$2a=3$$ $$a=1$$

 $$a=\frac{3}{2}$$ $$a=1$$

Therefore, your solution set is the following:

$$a_1=\frac{3}{2}$$

$$a_2=1$$

Now, you are done!

Aug 9, 2017
#3
+1

We can just factorize it :)

4a^2 - 10a + 6 = 0

2a^2 - 5a + 3 = 0

(2a - something)(a - something)=0

We don't know what it is yet, but think of 2 numbers that satisfies:

$$-2x-y=-5\\ xy=3$$

We can easily see that x = 1 and y = 3.

(2a - 3)(a - 1) = 0

a = 3/2 or a = 1 :)

Aug 11, 2017