Solve the following equation using the method | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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: $(x+2)^{2}-14(x+2)+45=0$

Similar Questions

Verify whether the given value of $x$ is a solution of the quadratic equation or not: $6x^{2}-x-2=0, x=3$.

Find the discriminant of the following quadratic equation: $x^{2}+5x+1=0$.

Find the roots of the following quadratic equation by the factorisation method:
$3x^{2} + 5\sqrt{5}x - 10 = 0$

Obtain the roots of the following equation using the method of 'completing the square': $5x^{2} - 4x - 10 = 0$.

Find the roots of the following quadratic equation by using the quadratic formula,if they exist: $x^{2}-5x-1=0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module