CTET 2 Math and Science 07 Jul 2024 Paper

Step 1: Simplify the Left-Hand Side (LHS):
Expand the LHS:
${\left(2\mathit{x}+5\mathit{y}\right)}^2-5\left(2\mathit{x}+5\mathit{y}\right)-14$
Let $\mathit{z}=2\mathit{x}+5\mathit{y}$
So, the equation becomes: ${\mathit{z}}^2-5\mathit{z}-14$
Step 2: Factorize the Quadratic Expression:
We need to factorize ${\mathit{z}}^2-5\mathit{z}-14$ into the form $\left(\mathit{z}+\mathit{p}\right)\left(\mathit{z}+\mathit{q}\right)$
To do this, we need to find two numbers whose product is -14 and whose sum is -5 .
These numbers are -7 and 2 .
Therefore, ${\mathit{z}}^2-5\mathit{z}-14$ can be written as: $\left(\mathit{z}-7\right)\left(\mathit{z}+2\right)$
Step 3: Substitute Back:
Recall that $\mathit{z}=2\mathit{x}+5\mathit{y}$
So, $\left(\mathit{z}-7\right)\left(\mathit{z}+2\right)$ becomes:
$\left(2\mathit{x}+5\mathit{y}-7\right)\left(2\mathit{x}+5\mathit{y}+2\right)$
Step 4: Identify $\mathit{p}$ and $\mathit{q}$ :
From the factorization, we have:

Step 5: Calculate $\mathit{p}+\mathit{q}$ :
$\mathit{p}+\mathit{q}=-7+2=-5$