Anyone can help me solve this equation using cross multiplying

Mathematics

1 answer:

Natali5045456 [20]2 years ago

8 0

9514 1404 393

Answer:

x = 1 or 5

Step-by-step explanation:

The notion of "cross-multiplying" is the idea that the numerator on the left is multiplied by the denominator on the right, and the numerator on the right is multiplied by the denominator on the left. This looks like ...

$\displaystyle \frac{x-1}{7}=\frac{2x-2}{3x-1}\ \longrightarrow\ (x-1)(3x-1)=(7)(2x-2)$

Then the solution proceeds by eliminating parentheses, and solving the resulting quadratic equation.

$3x^2-4x+1=14x-14\\\\3x^2-18x+15=0\qquad\text{subtract $14x-14$}\\\\x^2-6x+5=0 \qquad\text{divide by 3}\\\\(x-1)(x-5)=0\qquad\text{factor}\\\\x\in\{1,5\}$

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Comment on "cross multiply"

Like a lot of instructions in Algebra courses, the idea of "cross multiply" describes what the result looks like. It doesn't adequately describe how you get there. The one and only rule in solving Algebra problems is "whatever is done to one side of the equation must also be done to the other side of the equation." If you multiply one side by one thing and the other side by a different thing, you are violating this rule.

What looks like "cross multiply" is really "multiply by the product of the denominators and cancel like terms from numerator and denominator." Here's what that looks like with the intermediate steps added.

$\displaystyle \frac{x-1}{7}=\frac{2x-2}{3x-1}\\\\\frac{x-1}{7}\times7(3x-1)=\frac{2x-2}{3x-1}\times7(3x-1)\\\\(x-1)(3x-1)=(2x-2)(7)\qquad\textit{looks like}\text{ cross multiply}$

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These roots of the polynomial equation x^4-4x^3-2x^2+12x+9=0 are 3,-1,-1. Explain why the fourth root must be a real number. Fin

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Roots with imaginary parts always occur in conjugate pairs. Three of the four roots are known and they are all real, which means the fourth root must also be real.

Because we know 3 and -1 (multiplicity 2) are both roots, the last root $r$ is such that we can write

$x^4-4x^3-2x^2+12x+9=(x-3)(x+1)^2(x-r)$

There are a few ways we can go about finding $r$ , but the easiest way would be to consider only the constant term in the expansion of the right hand side. We don't have to actually compute the expansion, because we know by properties of multiplication that the constant term will be $(-3)(1)(1)(-r)=3r$ .

Meanwhile, on the left hand side, we see the constant term is supposed to be 9, which means we have

$3r=9\implies r=3$

so the missing root is 3.

Other things we could have tried that spring to mind:

- three rounds of division, dividing the quartic polynomial by $(x-3)$ , then by $(x+1)$ twice, and noting that the remainder upon each division should be 0