To calculate: The solution set of the polynomial equation 51−14x+4=x−2.

Question

Keith Dooley · Accepted Answer

Given information:
The polynomial equation is $\sqrt{51 - 14 x} + 4 = x - 2$ .
Formula used:
The square root property,
$a^{2} = b$
where ais the square root of b
Calculation:
Simply the equation $\sqrt{51 - 14 x} + 4 = x - 2$ ,
$\sqrt{51 - 14 x} = x - 2 - 4$
$\sqrt{51 - 14 x} = x - 6$
Square both the side and simplify,
${(\sqrt{51 - 14 x})}^{2} = {(x - 6)}^{2}$
$51 - 14 x = x^{2} + 36 - 12 x$
$x^{2} - 12 x + 14 x + 36 - 51 = 0$
$x^{2} + 2 x - 15 = 0$
Simplity it,
$x^{2} + 5 x - 3 x - 15 = 0$
$x (x + 5) - 3 (x + 5) = 0$
$(x - 3) (x + 5) = 0$
Set each factor equal to zero,
$x - 3 = 0$
$x = 3$
Or
$x + 5 = 01$
$x = - 5$
Check at $x = 3$
$\sqrt{51 - 14 (3)} + 43 - 2$
$\sqrt{51 - 14} + 41$
$\sqrt{9} + 41$ Simplify it,
$3 + 41$
$7 \neq 1$
The left side value is not equal to the right-side value. Thus, the value $x = 3$ is not verified.
Check at $x = - 5$
$\sqrt{51 - 14 (- 5)} + 4 - 5 - 2$
$\sqrt{51 + 70} + 4 - 7$
$\sqrt{121} + 4 - 7$
simplify it,
$11 + 4 - 7$
$15 \neq - 7$
The left side value is not equal to the right-side value. Thus, the value $x = - 5$ Sis not verified.
Therefore, all together the solution set of polynomial equation
$\sqrt{51 - 14 x} + 4 = x - 2 i s {ϕ}$ .

To calculate: The solution set of the polynomial equation \sqrt{51-

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