How do you solve the rational equation $\frac{x^{2} - 7}{x - 4} = 0$ $(x^2-7)/(x-4)=0$ ?

Question

1480 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-11T21:39:55

$x = \pm \sqrt{7}$ $x = ±sqrt7$

for this rational function to equate to zero means that the numerator must be zero as the denominator ≠ 0

$\Rightarrow x^{2} - 7 = 0 \Rightarrow x^{2} = 7 \Rightarrow x = \pm \sqrt{7}$ $rArr x^2 - 7 = 0 rArr x^2 = 7 rArr x = ±sqrt7$

Answer 2 · 2016-01-11T21:55:47

$x = \pm \sqrt{7}$ $x=+-sqrt(7)$

Strong guidance given on manipulation!

Given: $... . \frac{x^{2} - 7}{x - 4} = 0$ $color(brown)(color(white)(....) (x^2-7)/(x-4)=0)$

Multiply both sides by $color(blue)((x-4))$

$color(brown)((x^2-7)/(x-4)color(blue)(xx(x-4))=0xxcolor(blue)((x-4))$

$(x^2-7) xx(x-4)/(x-4)=0$

But $(x-4)/(x-4)" is another way of writing " 1$ giving:

$(x^2-7) xx1=0$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Add $color(blue)(7)$ to both sides

$color(brown)( x^2-7 color(blue)(+7) =0color(blue)(+7)$

$x^2=7$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

so $color(white)(...) x= sqrt(x^2) = +-sqrt(7)$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The square root has to be $+-$ because:

$x^2=( + sqrt(7))xx (+sqrt(7) )-> "positive "x^2$

And

$x^2= (-sqrt(7))xx(-sqrt(7))-> "positive "x^2$

How do you solve the rational equation (x^2-7)/(x-4)=0x2−7x−4=0(x^2-7)/(x-4)=0?