How do you solve $5 x^{2} - 2 x + 16 = 4 x^{2} + 6 x$ $5x^2 - 2x + 16 = 4x^2 + 6x$ ?

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Answer 1 · 2016-09-16T20:14:32

$x = 4$ $x = 4$

We have: $5 x^{2} - 2 x + 16 = 4 x^{2} + 6 x$ $5 x^(2) - 2 x + 16 = 4 x^(2) + 6 x$

First, let's subtract $4 x^{2} + 6 x$ $4 x^(2) + 6 x$ from both sides of the equation:

$\Rightarrow x^{2} - 8 x + 16 = 0$ $=> x^(2) - 8 x + 16 = 0$

Then, let's factorise the equation using the middle-term break:

$\Rightarrow x^{2} - 4 x - 4 x + 16 = 0$ $=> x^(2) - 4 x - 4 x + 16 = 0$

$\Rightarrow x (x - 4) - 4 (x - 4) = 0$ $=> x (x - 4) - 4 (x - 4) = 0$

$\Rightarrow (x - 4) (x - 4) = 0$ $=> (x - 4) (x - 4) = 0$

$\Rightarrow {(x - 4)}^{2} = 0$ $=> (x - 4)^(2) = 0$

$\Rightarrow x - 4 = 0$ $=> x - 4 = 0$

$\Rightarrow x = 4$ $=> x = 4$

Therefore, the solution to the equation is $x = 4$ $x = 4$ .

How do you solve 5x^2 - 2x + 16 = 4x^2 + 6x5x2−2x+16=4x2+6x5x^2 - 2x + 16 = 4x^2 + 6x?