How do you solve $x^{2} + 4 = 5 x$ $x^2+4=5x$ ?

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Answer 1 · 2016-05-08T22:12:55

Make it equal to 0.
Factorise, then let each factor in turn be equal to 0.
$x = 4, x = 1$ $x = 4, x = 1$

$x^{2} - 5 x + 4 = 0$ $x^2 - 5x + 4 = 0 " "$ make a quadratic equation = 0

$(x - 4) (x - 1) = 0$ $(x - 4)(x - 1) = 0" "$ factors of 4 which add to 5 are 4 x 1

If $x - 4 = 0 \Rightarrow x = 4$ $x - 4 = 0 rArr x = 4" "$ if $x - 1 = 0 \Rightarrow x = 1$ $x - 1 = 0 rArr x = 1$

How do you solve x2+4=5xx^2+4=5x?