How do you solve for x in this problem? $4{x}^{2}-3=-x$?

Milton Anderson
2022-09-12
Answered

How do you solve for x in this problem? $4{x}^{2}-3=-x$?

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Adrienne Harper

Answered 2022-09-13
Author has **14** answers

$\text{rearrange into standard form}\phantom{\rule{1ex}{0ex}}{x}a{x}^{2}+bx+c=0$

add x to both sides

$\Rightarrow 4{x}^{2}+x-3=0\leftarrow {\text{in standard form}}$

factor the quadratic using the a-c method

$\text{the factors of the product}\phantom{\rule{1ex}{0ex}}4\times -3=-12$

which sum to + 1 are + 4 and - 3

split the middle term using these factors

$4{x}^{2}+4x-3x-3=0\leftarrow {\text{factor by grouping}}$

4x(x+1)−3(x+1)=0

take out the common factor (x+1)

$\Rightarrow (x+1)\left({4x-3}\right)=0$

equate each factor to zero and solve for x

$x+1=0\Rightarrow x=-1$

$4x-3=0\Rightarrow x=\frac{3}{4}$

add x to both sides

$\Rightarrow 4{x}^{2}+x-3=0\leftarrow {\text{in standard form}}$

factor the quadratic using the a-c method

$\text{the factors of the product}\phantom{\rule{1ex}{0ex}}4\times -3=-12$

which sum to + 1 are + 4 and - 3

split the middle term using these factors

$4{x}^{2}+4x-3x-3=0\leftarrow {\text{factor by grouping}}$

4x(x+1)−3(x+1)=0

take out the common factor (x+1)

$\Rightarrow (x+1)\left({4x-3}\right)=0$

equate each factor to zero and solve for x

$x+1=0\Rightarrow x=-1$

$4x-3=0\Rightarrow x=\frac{3}{4}$

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Why are free variables used?

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So what is the point in writing the solutions as a set of vector additions using free variables?

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Where , denotes a new row

Solving for the pivot variables x and y and writing into decomposed vector form

[x,y] = [4,0] - [0,3] + z[1,2]

Z is said to be any real number zeR But looking at the graph, when z = 4, y=11/8 is not on the line of intersection So why do they say that z is a free variable when it isnt?

Using row reduction you get a system of linear equations which still satisfies the intersection

Why are free variables used?

The values in which the free variable can be are limited in a range as for if the solution is a line and not a plane

So what is the point in writing the solutions as a set of vector additions using free variables?

If z in this case is a free variable, writing the solution in a vector equation

Where , denotes a new row

Solving for the pivot variables x and y and writing into decomposed vector form

[x,y] = [4,0] - [0,3] + z[1,2]

Z is said to be any real number zeR But looking at the graph, when z = 4, y=11/8 is not on the line of intersection So why do they say that z is a free variable when it isnt?

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