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3 years ago

Craig writes 3x + 5y - x - 4y = 3 +9y. Can you spot the 2 mistakes he has made?

Mathematics

2 answers:

fenix001 [56]3 years ago

6 0

Answer:

first 5y-4y does not equal 9y.

second 3x-x is 2x but he put 3.

Step-by-step explanation:

Math - Answered Sun.Sept.19.2021

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sattari [20]3 years ago

4 0

craig forgor how subtraction works

mistake 1: instead of doing 3x - x, he did like multiplication and also forgor about the negative sign and somehow got 3.

mistake 2: instead of doing 5y - 4y, he did 5y + 4y

if he did it right, the answer would be 2x + y

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$6x^2-x=2\implies 6x^2-1x-2=0 \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{quadratic formula} \\\\ \stackrel{\stackrel{a}{\downarrow }}{6}x^2\stackrel{\stackrel{b}{\downarrow }}{-1}x\stackrel{\stackrel{c}{\downarrow }}{-2}=0 \qquad \qquad x= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}$

$x = \cfrac{ -(-1) \pm \sqrt { (-1)^2 -4(6)(-2)}}{2(6)}\implies x = \cfrac{1\pm\sqrt{1+48}}{12} \\\\\\ x = \cfrac{1\pm\sqrt{49}}{12}\implies x = \cfrac{1\pm 7}{12}\implies x = \begin{cases} \frac{8}{12}\to &\frac{2}{3}\\[1em] -\frac{6}{12}\to &-\frac{1}{2} \end{cases}$

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3 years ago

Find all roots: x^3 + 7x^2 + 12x = 0 Show all work and check your answer.

Aliun [14]

The three roots of x^3 + 7x^2 + 12x = 0 is 0,-3 and -4

Solution:

We have been given a cubic polynomial.

$x^{3}+7 x^{2}+12 x=0$

We need to find the three roots of the given polynomial.

Since it is a cubic polynomial, we can start by taking ‘x’ common from the equation.

This gives us:

$x^{3}+7 x^{2}+12 x=0$

$x\left(x^{2}+7 x+12\right)=0$ ----- eqn 1

So, from the above eq1 we can find the first root of the polynomial, which will be:

x = 0

Now, we need to find the remaining two roots which are taken from the remaining part of the equation which is:

$x^{2}+7 x+12=0$

we have to use the quadratic equation to solve this polynomial. The quadratic formula is:

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Now, a = 1, b = 7 and c = 12

By substituting the values of a,b and c in the quadratic equation we get;

$\begin{array}{l}{x=\frac{-7 \pm \sqrt{7^{2}-4 \times 1 \times 12}}{2 \times 1}} \\\\{x=\frac{-7 \pm \sqrt{1}}{2}}\end{array}$

Therefore, the two roots are:

$\begin{array}{l}{x=\frac{-7+\sqrt{1}}{2}=\frac{-7+1}{2}=\frac{-6}{2}} \\\\ {x=-3}\end{array}$

And,

$\begin{array}{c}{x=\frac{-7-\sqrt{1}}{2}} \\\\ {x=-4}\end{array}$

Hence, the three roots of the given cubic polynomial is 0, -3 and -4

4 0

3 years ago

Craig writes 3x + 5y - x - 4y = 3 +9y. Can you spot the 2 mistakes he has made?​

Craig writes 3x + 5y - x - 4y = 3 +9y. Can you spot the 2 mistakes he has made?