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const2013 [10]
4 years ago
6

Factor the following quadratic equation;

d="TexFormula1" title="12 {x}^{2} + 22x - 14" alt="12 {x}^{2} + 22x - 14" align="absmiddle" class="latex-formula">

Mathematics
1 answer:
Bogdan [553]4 years ago
5 0

Answer:

2(3x + 7)(2x - 1)

Step-by-step explanation:

You can see it a little easier if you take out a common factor of 2

2(6x^2 + 11x - 7)

The 6 leaves you with a lot of factors, the 7 does not. It only has 2 factors.

Let 6 factor into 2 and 3 and the 7 into 7 and 1

2(3x - 1 )(2x + 7)

Now remove the brackets.

2(6x^2 + 21x - 2x - 7) This obviously does not work but we'll combine like terms anyway.

2(6x^2 + 19x - 7)

So we'll try it again

2(3x + 7)(2x - 1)

2(6x^2 + 14x - 3x - 7)    Looks like we have it.

2(6x^2 + 11x - 7)

So the right factors are

2(3x + 7)(2x - 1)

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A piece of wire 19 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria
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Answer: 8.26 m

Step-by-step explanation:

$$Let s be the length of the wire used for the square. \\Let $t$ be the length of the wire used for the triangle. \\Let $A_{S}$ be the area of the square. \\Let ${A}_{T}}$ be the area of the triangle. \\One side of the square is $\frac{s}{4}$ \\Therefore,we know that,$$A_{S}=\left(\frac{s}{4}\right)^{2}=\frac{s^{2}}{16}$$

$$The formula for the area of an equilateral triangle is, $A=\frac{\sqrt{3}}{4} a^{2}$ where $a$ is the length of one side,And one side of our triangle is $\frac{t}{3}$So,We know that,$$A_{T}=\frac{\sqrt{3}}{4}\left(\frac{t}{3}\right)^{2}$$We have to find the value of "s" such that,$\mathrm{s}+\mathrm{t}=19$ hence, $\mathrm{t}=19-\mathrm{s}$And$$A_{S}+A_{T}=A_{S+T}$$

$$Therefore,$$\begin{aligned}&A_{T}=\frac{\sqrt{3}}{4}\left(\frac{(19-s)}{3}\right)^{2}=\frac{\sqrt{3}(19-s)^{2}}{36} \\&A_{T+S}=\frac{s^{2}}{16}+\frac{\sqrt{3}(19-s)^{2}}{36}\end{aligned}

$$Differentiating the above equation with respect to s we get,$$A^{\prime}{ }_{T+S}=\frac{s}{8}-\frac{\sqrt{3}(19-s)}{18}$$Now we solve $A_{S+T}^{\prime}=0$$$\begin{aligned}&\Rightarrow \frac{s}{8}-\frac{\sqrt{3}(19-s)}{18}=0 \\&\Rightarrow \frac{s}{8}=\frac{\sqrt{3}(19-s)}{18}\end{aligned}$$Cross multiply,$$\begin{aligned}&18 s=8 \sqrt{3}(19-s) \\&18 s=152 \sqrt{3}-8 \sqrt{3} s \\&(18+8 \sqrt{3}) s=152 \sqrt{3} \\&s=\frac{152 \sqrt{3}}{(18+8 \sqrt{3})} \approx 8.26\end{aligned}$$

$$The domain of $s$ is $[0,19]$.So the endpoints are 0 and 19$$\begin{aligned}&A_{T+S}(0)=\frac{0^{2}}{16}+\frac{\sqrt{3}(19-0)^{2}}{36} \approx 17.36 \\&A_{T+S}(8.26)=\frac{8.26^{2}}{16}+\frac{\sqrt{3}(19-8.26)^{2}}{36} \approx 9.81 \\&A_{T+S}(19)=\frac{19^{2}}{16}+\frac{\sqrt{3}(19-19)^{2}}{36}=22.56\end{aligned}$$

$$Therefore, for the minimum area, $8.26 \mathrm{~m}$ should be used for the square

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