Solving Quadratic Equations by Taking Square Roots

Solving Quadratic Equations by Taking Square Roots exercise

• Explain how to factor quadratic equations which are a difference of two squares.

  Hints

  The quadratic equation $x^2-4=0$ is a difference of two squares.

  To solve $x^2-4=0$, first write this equation as follows:

  $(\sqrt{~~~}+\sqrt{~~~})(\sqrt{~~~}-\sqrt{~~~})=0$.

  Next, put $x^2$ as well as $4$ under the square root sign:

  $(\sqrt{x^2}+\sqrt{4})(\sqrt{x^2}-\sqrt{4})=0$.

  Simplifying, we get:

  $(x+2)(x-2)=0$.

  Solution

  In order for such a quadratic equation to be a difference of two squares, it must meet the following requirements:

  1. The coefficient $b$ must be zero.
  2. The coefficient $a$ and $c$ must have different signs.
  3. The exponent must be even.

  Let's check the example $16x^2-81=0$:

  1. $b=0$ $~~~~~$ ✓
  2. $a=16$ and $c=-81$ $~~~~~$ ✓
  3. The exponent is $2$ $~~~~~$ ✓

  To factor a difference of two squares, write the quadratic term on as follows:

  $(\sqrt{~~~}+\sqrt{~~~})(\sqrt{~~~}-\sqrt{~~~})$.

  Next, put $16x^2$ as well as $81$ under the square root sign:

  $(\sqrt{16x^2}+\sqrt{81})(\sqrt{16x^2}-\sqrt{81})=0$.

  Now, take the square root of the terms above to get

  $(4x+9)(4x-9)=0$.

  Lastly, solve this equation using the zero product property; either we have that $4x+9=0$ or $4x-9=0$.

• Factor the quadratic equation.

  Hints

  The goal is to get an equation which looks like

  $(e+f)(e-f)=0$.

  Using the FOIL method of multiplication, you get

  $e^2-f^2=0$.

  Backwards, you have to do the following: first assign

  • $e^2=x^2$
  • $f^2=64$

  Then take the square root to get $e$ as well as $f$.

  Solution

  To factor Dr. Evil's equation, we first have to check to see if it is a difference of two squares. Keeping the standard form $ax^2+bx+c=0$ of a quadratic equation in mind, we check the following:

  1. $b=0$? $~~~~~$✓
  2. $a$ and $c$ must have different signs: $a=1$ and $c=-64$ $~~~~~$✓
  3. The exponent has to be even. And, sure, it is, namely $2$. $~~~~~$✓

  Now we can proceed as follows:

  First, rewrite the left side of the equation as the product of two binomials:

  $(\sqrt{~~~}+\sqrt{~~~})(\sqrt{~~~}-\sqrt{~~~})=0$.

  Next, put $x^2$ and $64$ under the square root sign:

  $(\sqrt{x^2}+\sqrt{64})(\sqrt{x^2}-\sqrt{64})=0$.

  Lastly, take the square root to get:

  $(x+8)(x-8)=0$.

  To solve the equation, we use the zero product property as well as opposite operations:

  Either we have

  $\begin{array}{rcl} x+8 & = & ~0\\ \color{#669900}{-8} & &\color{#669900}{-8}\\ x & = & ~-8 \end{array}$

  or

  $\begin{array}{rcl} x-8 & = & ~0\\ \color{#669900}{+8} & &\color{#669900}{+8}\\ x & = & ~8 \end{array}$

• Find the errors in Rap-Punzel's calculations.

  Hints

  Write the difference as the product of two binomials

  $(\sqrt{(~~~)^2}+\sqrt{(~~~)^2})(\sqrt{(~~~)^2}-\sqrt{(~~~)^2})=0$.

  Take the square root. Let's have a look at a few examples:

  • $\sqrt{36}=6$
  • $\sqrt{x^2}=x$
  • $\sqrt{x^6}=x^3$
  • ...

  If you have to take the square root of a product, you can multiply the square roots of the factors.

  Solution

  If you have the difference of two squares, you can factor as follows:

  $e^2-f^2=(e+f)(e-f)$.

  Let's consider a few examples:

  $4x^2-9=0$.

  Write the left side as the product of two binomials. We write

  $(\sqrt{4x^2}+\sqrt{9})(\sqrt{4x^2}-\sqrt{9})=0$.

  Then you can take the square root to get:

  $(4x+3)(4x-3)=0$.

  In a similar manner you can factor the following examples:

  • $25-x^4=0$ $\rightarrow$ $(\sqrt{25}+ \sqrt{x^4})(\sqrt{25}-\sqrt{x^4})=0$, and by taking the square root we get $(5+x^2)(5-x^2)=0$.
  • $25x^2-36=0$ $\rightarrow$ $(\sqrt{25x^2}+ \sqrt{36})(\sqrt{25x^2}-\sqrt{36})=0$, and by taking the square root we get $(5x+6)(5x-6)=0$.
  • $49x^6-121=0$ $\rightarrow$ $(\sqrt{49x^6}+ \sqrt{121})(\sqrt{49x^6}-\sqrt{121})=0$, and by taking the square root we get $(7x^3+11)(7x^3-11)=0$.

• Identify which factorizations of the equations are correct.

  Hints

  Look at the example

  $16x^4-36=0$.

  First, write as

  $(\sqrt{16x^4}+\sqrt{36})(\sqrt{16x^4}-\sqrt{36})=0$.

  Then, take the square roots to get

  $(4x^2+6)(4x^2-6)=0$.

  If you have to take the square root of a product, you can take the square root of each factor and multiply those square roots after.

  You can also check the factorization using the FOIL method for multiplication:

  $(4x^2+6)(4x^2-6)=(4x^2)(4x^2)-(4x^2)(6)+(6)(4x^2)-(6)(6)=16x^4-36$ $~~~~~$✓

  Solution

  All the equations have in common, that the regarding left side is the difference of two squares. So you can factor them as follows:

  1. $9x^2-9=0$ $\rightarrow$ $(\sqrt{9x^2}+ \sqrt{9})(\sqrt{9x^2}-\sqrt{9})=0$ and by taking the square root we get $(3x+6)(3x-6)=0$; the answer was correct.
  2. $x^6-100=0$ $\rightarrow$ $(\sqrt{x^6}+ \sqrt{100})(\sqrt{x^6}-\sqrt{100})=0$ and by taking the square root we get $(x^3+10)(x^3-10)=0$; the answer was wrong, because $\sqrt{x^6}=x^3$ instead of $x^2$.
  3. $100-16x^2=0$ $\rightarrow$ $(\sqrt{100}+ \sqrt{16x^2})(\sqrt{100}-\sqrt{16x^2})=0$ and by taking the square root we get $(10+4x)(10-4x)=0$; the answer was wrong. Keep in mind to take the square root for both factors.
  4. $9x^6-144=0$ $\rightarrow$ $(\sqrt{9x^6}+ \sqrt{144})(\sqrt{9x^6}-\sqrt{144})=0$ and by taking the square root we get $(3x^3+12)(3x^3-12)=0$; the answer was correct.
  5. $49x^2-64=0$ $\rightarrow$ $(\sqrt{49x^2}+ \sqrt{64})(\sqrt{49x^2}-\sqrt{64})=0$ and by taking the square root we get $(7x+8)(7x-8)=0$; the answer was correct.

• Explain any differences between $16x^4 -81=0$ and $ax^2 +bx +c=0$.

  Hints

  Compare the terms step by step.

  $x$ is the variable.

  Another example for such an equation is

  $16x^4+81=0$.

  Solution

  Let's check the differences between the two equations:

  1. The degree, or largest exponent value of $x$ in a term, for $16x^4 -81=0$ is $4$ and for $ax^2 +bx +c=0$ is $2$.
  2. There is no term in $16x^4 -81=0$ of the form $bx$. This means that $b=0$.
  3. Normally $a$ and $b$ could have all possible values, except $a=0$, and also a positive as well as a negative sign. For $16x^4 -81=0$, $a=16$ and $c=-81$ have different signs.

• Determine the factored equation.

  Hints

  $4x^6-36=0$ is equivalent to

  $(2x^3+6)(2x^3-6)=0$.

  You still have to take the square root of the corresponding terms.

  The square root of a power with an even exponent is the power with the same base and the half the value of the exponent:

  $\sqrt{x^{2n}}=x^n$

  Solution

  Each time you have to factor an equation, where the left side is the difference of two squares, you can write the left side as follows:

  $(\sqrt{~~~}+\sqrt{~~~})(\sqrt{~~~}-\sqrt{~~~})$,

  and put the corresponding terms under the square root sign after.

  Let's try it with $169x^4-144=0$. We have

  $(\sqrt{169x^4}+\sqrt{144})(\sqrt{169x^4}-\sqrt{144})=0$.

  Now, take the square root to get

  $(13x^2+12)(13x^2-12)=0$.

  Let's try it again with $81-36x^2=0$. We have

  $(\sqrt{81}+\sqrt{36x^2})(\sqrt{81}-\sqrt{36x^2})=0$.

  Again, we take the square root to get

  $(9+6x)(9-6x)=0$.

  Just one last example: $121-16x^8=0$.

  Keeping the square root of a product in mind, $\sqrt{36x^2}=\sqrt{36}\times\sqrt{x^2}=6x$, we have

  $(\sqrt{121}+\sqrt{16x^8})(\sqrt{121}-\sqrt{16x^8})=0$.

  Now, we take the square root

  $(11+4x^4)(11-4x^4)=0$.