Question

Find the x for the given secind degree equation x²-50x-600=0

Answer 1

$\textsf{\large{\underline{Solution}:}}$

Given Equation:

$\rm: \longmapsto {x}^{2} - 50x - 600 = 0$

By splitting the middle term, we get:

$\rm: \longmapsto {x}^{2} - (60 - 10)x - 600 = 0$

$\rm: \longmapsto {x}^{2} - 60x + 10x - 600 = 0$

$\rm: \longmapsto x(x - 60) + 10(x - 60)= 0$

$\rm: \longmapsto (x+ 10)(x - 60)= 0$

By Zero Product Rule, we can say that:

$\rm: \longmapsto \begin{cases} \rm (x+ 10) = 0 \\ \rm(x - 60)= 0 \end{cases}$

$\rm: \longmapsto x = 60, - 10$

★ So, the values of x satisfying the given equation are 60 and -10.

$\textsf{\large{\underline{Verification}:}}$

Put x = 60 in the equation, we get:

$\rm = {(60)}^{2} - 50 \times 60 - 600$

$\rm = 3600 - 3000 - 600$

$\rm = 600- 600$

$\rm =0$

Put x = -10 in the equation, we get:

$\rm = {( - 10)}^{2} - 50 \times ( - 10) - 600$

$\rm = 100 + 500- 600$

$\rm =0$

★ Hence, our answers are correct (Verified).

$\textsf{\large{\underline{Answer}:}}$

• The values of x are 60 and -10.

Answer 2

Answer:

x = -10 or 60

Step-by-step explanation:

we have,

$\rm x^2 -50x-600=0$

$\rm \Rightarrow x^2 -(60 - 10)x-600=0$

$\rm \Rightarrow x^2 -60x+10x-600=0$

$\rm \Rightarrow x(x-60)+10(x-60)=0$

$\rm \Rightarrow (x+10)(x-60)=0$

$\rm \Rightarrow (x+10)=0\: or \:(x-60)=0$

$\rm \Rightarrow x=-10\:or\: x=60$