GKT101: General Knowledge for Teachers – Math

1.

$\text { lesser } x=-7$

$\text { lesser } x=-3$

$4 x^{2}+40 x+84=0$

$4\left(x^{2}+10 x+21\right)=0$

Now let's factor the expression in the parentheses.

$x^{2}+10 x+21$ can be factored as $(x+7)(x+3)$.

$4(x+7)(x+3)=0$

$x+7=0$ or $x+3=0$

$x = -7$ or $x=-3$

In conclusion,

$\text { lesser } x=-7$

$\text { lesser } x=-3$

2. $x = 6$

Dividing both sides by $4$ gives

$x^{2}-12 x+36=0$

The coefficient on the $x$ term is $-12$ and the constant term is $36$, so we need to find two numbers that add up to $-12$ and multiply to $36$.

The number $-6$ used twice satisfies both conditions:

$−6+−6=−12$

$−6×−6=36$

So $(x-6)^{2}=0$.

$x−6=0$

Thus, $x = 6$ is the solution.

3.

$\text { lesser } x=1$

$\text { lesser } x=3$

$5 x^{2}-20 x+15=0$

$5\left(x^{2}-4 x+3\right)=0$

Now let's factor the expression in the parentheses.

$x^{2}-4 x+3$ can be factored as $(x-1)(x-3)$.

$5(x−1)(x−3) = 0$

$x-1 = 0$ or $x -3 = 0$

$x = 1$ or $x= 3$

In conclusion,

$\text { lesser } x=1$

$\text { lesser } x=3$

4. $x =9$

Dividing both sides by $3$ gives $x^{2}-18 x+81=0$

The coefficient on the $x$ term is $-18$ and the constant term is $81$, so we need to find two numbers that add up to $-18$ and multiply to $81$.

The number $-9$ used twice satisfies both conditions:

$−9+−9=−18$

$−9×−9=81$

So $(x-9)^{2}=0$.

$x−9=0$

Thus, $x =9$ is the solution.