The Quadratic Formula

1. Solve for \(x\). Enter the solutions from least to greatest.

\(x2−8x+7=0\)

2. Solve for \(x\).

\(x^{2}+16 x+64=0\)

3. Solve for \(x\). Enter the solutions from least to greatest.

\(x^{2}+12 x+27=0\)

4.Solve for \(x\).

\(x^{2}+10 x+25=0\)

1.

\( \text { lesser x} = 1\)

\( \text { greater x} = 7\)

To factor \(x^{2}-8 x+7 \text { as }(x+a)(x+b)\), we need to find numbers \(a\) and \(b\) such that \(a+b=−8\) and \(ab = 7\).

\(a=−7\) and \(b=-1\) satisfy both conditions, so our equation can be re-written:

\((x−7)(x−1)=0\)

According to the zero-product property, we know that

\(x−7=0\) or \(x - 1 = 0\)

which means

\(x=7x\) or \(x=1\)

In conclusion,

\( \text { lesser x} = 1\)

\( \text { greater x} = 7\)

2. \(x = -8\)

Both \(x^{2}\) and \(64\) are perfect squares, since \(x^{2}=(x)^{2}\) and \(64=(8)^{2}\).

Additionally, \(16x\) is twice the product of the roots of \(x^{2}\) and \(64\), since \(16 x=2(x)(8)\).

\(x^{2}+16 x+64=(x)^{2}+2(x)(8)+(8)^{2}\)

So we can use the square of a sum pattern to factor:

\(a^{2}+2(a)(b)+b^{2}=(a+b)^{2}\)

In this case, \(a = x\) and \(b = 8\):

\((x)^{2}+2(x)(8)+(8)^{2}=(x+8)^{2}\)

So our equation can be re-written:

\((x+8)^{2}=0\)

The only possible solution is when \(x+8 = 0\), which is

\(x = -8\)

3.

\( \text { lesser x} = -9\)

\( \text { greater x} = -3\)

To factor \(x^{2}+12 x+27\) as \((x+a)(x+b)\), we need to find numbers \(a\) and \(b\) such that

\(a+b=12\) and \(a b=27\).

\(a = 3\) and \(b = 9\) satisfy both conditions, so our equation can be re-written:

\((x+3)(x+9)=0\)

According to the zero-product property, we know that

\(x+3=0\) or \(x+9 = 0\)

which means

\(x=−3\) or \(x=-9\)

In conclusion,

\( \text { lesser x} = -9\)

\( \text { greater x} = -3\)

4. \(x = -5\)

Both \(x^{2}\) and \(25\) are perfect squares, since \(x^{2}=(x)^{2} \text { and } 25=(5)^{2}\).

Additionally, \(10x\) is twice the product of the roots of \(x^{2} \text { and } 25, \text { since } 10 x=2(x)(5)\).

\(x^{2}+10 x+25=(x)^{2}+2(x)(5)+(5)^{2}\)

So we can use the square of a sum pattern to factor:

\(a^{2}+2(a)(b)+b^{2}=(a+b)^{2}\)

In this case, \(a = x\) and \(b = 5\):

\((x)^{2}+2(x)(5)+(5)^{2}=(x+5)^{2}\)

So our equation can be re-written:

\((x+5)^{2}=0\)

The only possible solution is when \(x + 5 = 0\), which is

\(x = -5\)

1. Solve for \(x\). Enter the solutions from least to greatest.

\(4 x^{2}+40 x+84=0\)

2. Solve for \(x\).

\(4 x^{2}-48 x+144=0\)

3. Solve for \(x\). Enter the solutions from least to greatest.

\(5 x^{2}-20 x+15=0\)

4. Solve for \(x\).

\(3 x^{2}-54 x+243=0\)

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.

1. Find one value of \(x\) that is a solution to the equation:

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

2. Let \(m=2x+3\).

Which equation is equivalent to \((2 x+3)^{2}-14 x-21=-6\) in terms of \(m\)?

Choose 1 answer:

A. \(m^{2}+7m+6=0\)

B. \(m^{2}−7m−15=0\)

C. \(m^{2}−7m+6=0\)

D. \(m^{2}+7m-15=0\)

3. Find one value of \(x\) that is a solution to the equation:

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

4. Find one value of \(x\) that is a solution to the equation:

\((2 x+3)^{2}-6 x-9=0\)

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.