GKT101: General Knowledge for Teachers – Math

1.

• $x=\sqrt{7}$
• $x= -\sqrt{7}$
• $x=\sqrt{5}$
• $x=-\sqrt{5}$

We could solve for $x$ by expanding $\left(x^{2}-7\right)^{2}$, combining terms that are alike, and using the quadratic formula or factoring to solve for $x$. However there is a shorter and more elegant way to approach this problem. Let's use structural features to rewrite the equation in a simpler form.

Note that $2 x^{2}-14=2\left(x^{2}-7\right)$. This means that we can rewrite the equation as:

$\left(x^{2}-7\right)^{2}+2\left(x^{2}-7\right)=0$

If we let $p=x^{2}-7$, we can see that this equation is in the form:

$p^{2}+2 p=0$

Let's solve this equation in terms of $p$:

$p^{2}+2 p=0$

$p(p+2)=0$

$p=0$ or $p=-2$

Since $p=x^{2}-7$, let's substitute this value back into our two solutions in order to solve for $x$.

$x^{2}-7=0$ or $x^{2}-7=-2$

When we solve $x^{2}-7=0$, we find that $x=\pm \sqrt{7}$.

When we solve $x^{2}-7=-2$, we find that $x=\pm \sqrt{5}$.

In conclusion, the four solutions of the equation $\left(x^{2}-7\right)^{2}+2 x^{2}-14=0$ are:

• $x=\sqrt{7}$
• $x= -\sqrt{7}$
• $x=\sqrt{5}$
• $x=-\sqrt{5}$

Is there another way to solve for x?

In math, there's always another way!

We can use the approach mentioned in hint 1 in order to transform this equation as shown below:

$\left(x^{2}-7\right)^{2}+2 x^{2}-14=0$

$\left(x^{4}-14 x^{2}+49\right)+2 x^{2}-14=0$

$x^{4}-12 x^{2}+35=0$

Now if we use structure in a slightly different way, we can let $a=x^{2}$ and rewrite this equation in a form that allows us to solve it by factoring:

$\left(x^{2}\right)^{2}-12\left(x^{2}\right)+35=0$

$a^{2}-12 a+35=0$

$(a-7)(a-5)=0$

We can conclude that $a = 7$ or $a = 5$. Since $a=x^{2}$, we can write $x^{2}=7$ and $x^{2}=5$, which yield the same solutions for $x$ that we already found.

Once again, math is consistent! However, it is often easier (and more fun) to look for structural features when solving quadratic equations.

2. C. $m^{2}−7m+6=0$

We are asked to rewrite the equation in terms of $m$, where $m = 2x + 3$. In order to do this, we need to find all of the places where the expression $2x+3$ shows up in the equation, and then substitute $m$ wherever we see them!

For instance, note that $-14 x-21=-7(2 x+3)$. This means that we can rewrite the equation as:

$(2 x+3)^{2}-14 x-21=-6$

$(2 x+3)^{2}-7(2 x+3)=-6$

What if I don't see this factorization?

Since $m=2x + 3$ we can see that $m^{2}=(2 x+3)^{2}$. But what about the slightly trickier $-14x -21$?

We know that we must replace all $x$ terms with $m$-terms. If we solve the equation $m=2x+3$ for $x$ and substitute this result into the trickier expression, we can rewrite it in terms of $m$.

Indeed, we can see that $x=\frac{m-3}{2}$. So

$\begin{aligned} -14 x-21 &=-14\left(\frac{m-3}{2}\right)-21 \\ &=-7(m-3)-21 \\ &=-7 m+21-21 \\ &=-7 m \end{aligned}$

While this approach is longer, we can use it to always find the correct factorization in case we do not see that $−14x−21=−7(2x+3)$.

Now we can substitute $m=2x+3$:

$(m)^{2}-7(m)=-6$

Finally, let's manipulate this expression so that it shares the same form as the answer choices:

$m^{2}-7 m+6=0$

In conclusion, $m^{2}-7 m+6=0$ is equivalent to the given equation when $m=2x+3$.

3. $x=2$ and $x= -2$

We could solve for $x$ by expanding $\left(x^{2}+4\right)^{2} \text { and }-11\left(x^{2}+4\right)$, combining terms that are alike, and using the quadratic formula or factoring to solve for $x$. However there is a more elegant way to approach this problem. Let's use structural features to rewrite the equation in a simpler form.

Note that if we let $p=x^{2}+4$, we can rewrite the equation:

$\left(x^{2}+4\right)^{2}-11\left(x^{2}+4\right)+24=0$

In particular, we can express it in the form:

$p^{2}-11 p+24=0$

Let's solve this equation in terms of $p$:

$p^{2}-11 p+24=0$

$(p-8)(p-3)=0$

$p = 8$ or $p = 3$

Since $p=x^{2}+4$, let's substitute this value back into our two solutions in order to solve for $x$.

$x^{2}+4=8$ or $x^{2}+4=3$

When we solve $x^{2}+4=8$, we find that $x=\pm 2$.

Note that there are no real solutions to the equation $x^{2}+4=3$.

Why not?

When we try to solve this equation, we immediately get stuck:

$x^{2}+4=3$

$x^{2}=-1$

For any positive or negative value of $x$, the value of $x^{2}$, must be positive. We can conclude that there are no real values of $x$ that satisfy this equation.

Later on, we will learn that this equation can be solved within the domain of imaginary numbers. Imaginary numbers are of the form $a+b i(\text { where } i=\sqrt{-1})$.

In conclusion, the two solutions of the equation $\left(x^{2}+4\right)^{2}-11\left(x^{2}+4\right)+24=0$ are

$x=2$ and $x= -2$:

Is there another way to solve for x?

In math, there's always another way!

We can use the approach mentioned in hint 1 in order to transform this equation as shown below:

$\begin{array}{r} \left(x^{2}+4\right)^{2}-11\left(x^{2}+4\right)+24=0 \\ \left(x^{4}+8 x^{2}+16\right)-11 x^{2}-44+24=0 \\ x^{4}-3 x^{2}-4=0 \end{array}$

Now if we use structure in a slightly different way, we can let $a=x^{2}$ and rewrite this equation in a form that allows us to solve it by factoring:

$\begin{array}{r} \left(x^{2}\right)^{2}-3\left(x^{2}\right)-4=0 \\ a^{2}-3 a-4=0 \\ (a-4)(a+1)=0 \end{array}$

We can conclude that $a = 4$ and $a = -1$. Since $a=x^{2}$, we can write $x^{2}=4$ and $x^{2}=-1$, which yield the same solutions for $x$ that we already found.

Once again, math is consistent! However, it is often easier (and more fun) to look for structural features when solving quadratic equations.

4. $x=-\frac{3}{2}$ and $x=0$

We could solve for $x$ by expanding $(2 x+3)^{2}$, combining terms that are alike, and using the quadratic formula or factoring to solve for $x$. However there is a more elegant way to approach this problem. Let's use structural features to rewrite the equation in a simpler form.

Note that $-6 x-9=-3(2 x+3)$. This means that we can rewrite the equation as:

$(2 x+3)^{2}-3(2 x+3)=0$

If we let $p= 2x + 3$, we can see that this equation is in the form:

$p^{2}-3 p=0$

Let's solve this equation in terms of $p$:

$p^{2}-3 p=0$

$p(p-3)=0$

$p=0$ or $p=3$

Since $p= 2x + 3$, let's substitute this value back into our two solutions in order to solve for $x$:

$2x+3=0$ or $2x+3=3$

When we solve $2x+3=0$, we find that $x=-\frac{3}{2}$.

When we solve $2x+3=3$, we find that $x=0$.

In conclusion, the two solutions of the equation $(2 x+3)^{2}-6 x-9=0$ are $x=-\frac{3}{2}$ and $x=0$.

Is there another way to solve for x?

In math, there's always another way!

We can use the approach mentioned in hint 1 in order to transform this equation as shown below:

$\begin{aligned} (2 x+3)^{2}-6 x-9 &=0 \\ \left(4 x^{2}+12 x+9\right)-6 x-9 &=0 \\ 4 x^{2}+6 x &=0 \\ x(4 x+6) &=0 \end{aligned}$

We can conclude that $x=0$ and $x=-\frac{3}{2}$ are both solutions to this equation. Once again, math is consistent! However, it is often easier (and more fun) to look for structural features when solving quadratic equations.