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(A) Let $(x+2) = y$. Substituting this into the equation,we get:$y^{2} - 14y + 45 = 0$To factorize,we look for two numbers whose product is $45$ and s

Vedclass · Accepted Answer

$3, 7$

Vedclass · Answer

(A) Let $(x+2) = y$. Substituting this into the equation,we get:$y^{2} - 14y + 45 = 0$To factorize,we look for two numbers whose product is $45$ and sum is $-14$. These numbers are $-9$ and $-5$.$y^{2} - 9y - 5y + 45 = 0$$y(y - 9) - 5(y - 9) = 0$$(y - 9)(y - 5) = 0$So,$y = 9$ or $y = 5$.Now,substitute $y = x + 2$ back:Case $1$: $x + 2 = 9 \implies x = 7$Case $2$: $x + 2 = 5 \implies x = 3$Thus,the solutions are $x = 3, 7$.

Solve the following equation using the method of factorization: $(x+2)^{2}-14(x+2)+45=0$

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