Quadratic expressions and equations

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Quadratic expressions and equations

Question 1

a)

Factorise the equation

$\simplify{{a*c}x^2+{a*d+b*c}x+{b*d}=0}\text{.}$

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b)

Write down the roots of the equation above.

Input your answer as $x_1$ and $x_2$, where $x_1<x_2$.

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Advice

a)

As this question involves a number greater than $1$ before the $x^2$ value it has a factorised form $(ax+b)(cx+d)$.

To find $a$ and $c$, we need to consider the factors of $\var{a*c}$.

We are already given that one of them is $\var{a}$, so we know that the other one must be $\var{c}$.

This means our factorised equation must take the form

\[(\var{a}x+b)(\var{c}x+d)=0\text{.}\]

This expands to

\[ \simplify{ {a*c}x^2 + ({a}*d+{c}*b)x + a*b} \]

So we must find two numbers which add together to make $\var{a*d+b*c}$, and multiply together to make $\var{b*d}$.

Therefore $b$ and $d$ must satisfy

\begin{align}
b \times d &=\var{b*d}\\
\simplify{{a}d+{c}b} &= \var{a*d+b*c}\text{.}
\end{align}

$b = \var{b}$ and $d = \var{d}$ satisfy these equations:

\begin{align}
\var{b} \times \var{d} &=\var{b*d}\\
\simplify[]{ {a}*{d} + {b}*{c} } &= \var{a*d+b*c}
\end{align}

So the factorised form of the equation is

\[ \simplify{({a}x+{b})({c}x+{d}) = 0} \text{.}\]

b)

$\simplify{({a}x+{b})({c}x+{d}) = 0}$ when either $\var{a}x+\var{b} = 0$ or $\var{c}x+ \var{d} = 0$.

So the roots of the equation are $\var[fractionnumbers]{-b/a}$ and $\var[fractionnumbers]{-d/c}$.

Question 2

We can rewrite quadratic equations given in the form $ax^2+bx+c$ as a square plus another term - this is called "completing the square".

This can be useful when it isn't obvious how to fully factorise a quadratic equation.

a)

Write the following expression in the form $a(x+b)^2-c$.

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b)

Now solve the quadratic equation

\[ \simplify {x^2+{sml}x+{big}} = 0\text{.} \]

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Advice

Completing the square works by noticing that

\[ (x+a)^2 = x^2 + 2ax + a^2 \]

So when we see an expression of the form $x^2 + 2ax$, we can rewrite it as $(x+a)^2-a^2$.

a)

Rewrite $x^2+\var{sml}x$ as $\simplify[basic]{ (x+{sml/2})^2 - {sml/2}^2}$.

\begin{align}
\simplify[basic]{x^2+{sml}x+{big}} &= \simplify[basic]{(x+{sml/2})^2-{(sml/2)}^2+{big}} \\
&= \simplify[basic]{(x+{sml/2})^2+{-(sml/2)^2+big}} \text{.}
\end{align}

b)

We showed above that

\[ \simplify[basic]{x^2+{sml}x+{big}} = 0 \]

is equivalent to

\[ \simplify[basic]{(x+{bits[0]})^2-{bits[1]^2}} = 0 \text{.} \]

We can then rearrange this equation to solve for $x$.

\begin{align}
\simplify{(x+{bits[0]})^2-{(bits[1])^2} } &= 0 \\
(x+\var{bits[0]})^2 &= \var{bits[1]^2} \\
x+\var{bits[0]} &= \pm \var{bits[1]} \\
x &= -\var{bits[0]} \pm \var{bits[1]} \\[2em]

x_1 &= \var{-bits[0]-bits[1]} \text{,}\\
x_2 &= \var{-bits[0]+bits[1]} \text{.}
\end{align}

Question 3

We can rewrite quadratic equations given in the form $ax^2+bx+c$ as a square plus another term - this is called "completing the square".

This can be useful when it isn't obvious how to fully factorise a quadratic equation.

Rewrite the following expressions in the form \[(x+b)^2-c\] or \[a(x+b)^2-c\]

a)

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b)

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c)

$\simplify {x^2+ {evens2}x +{evens2-evens1}} =$ $\displaystyle{}$ question.score feedback.none Expected answer:

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d)

$\simplify {{all} x^2+{multiall}x+{odds3-evens2}} =$ $\displaystyle{}$ question.score feedback.none Expected answer:

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Advice

Completing the square works by noticing that

\[ (x+a)^2 = x^2 + 2ax + a^2 \]

So when we see an expression of the form $x^2 + 2ax$, we can rewrite it as $(x+a)^2-a^2$.

a)

We have $x^2+ \var{evens1}x$, so we can replace it with $(x+\var{evens1/2})^2-\var{evens1/2}^2 = (x+\var{evens1/2})^2 - \var{evens1^2/4}$.

Check that this is equivalent to the original expression by expanding the brackets:

\begin{align}
(x+\var{evens1/2})^2 - \var{evens1^2/4} &= \simplify[basic]{ x^2 + 2*{evens1/2}*x + {evens1/2}^2 - {evens1^2/4} } \\
&= x^2 + \var{evens1}x \text{.}
\end{align}

b)

Replace $x^2 + \var{odds}x$ by $\simplify[basic]{(x+{odds}/2)^2-({odds}/2)^2}$, to obtain

\begin{align}
x^2 + \var{odds}x &= \simplify[basic]{(x+{odds}/2)^2-({odds}/2)^2} \\[0.5em]
&= \simplify[basic]{ (x+{odds}/2)^2 - {odds^2}/4} \text{.}
\end{align}

c)

Replace $x^2+\var{evens2}x$ with $(x+\var{evens2/2})^2 - \var{evens2/2}^2$. Remember to keep the $\var{evens2-evens1}$ term on the end!

\begin{align}
\simplify[basic]{ x^2 + {evens2}x + {evens2-evens1}} &= \simplify[basic]{ (x+{evens2/2})^2 - {evens2/2}^2 + {evens2-evens1} } \\
&= \simplify[basic]{ (x+{evens2/2})^2 + {evens2-evens1 - evens2^2/4} }
\end{align}

d)

First, notice that $\simplify[basic]{ {all}x^2 + {multiall}x } = \simplify[basic]{ {all}*( x^2 + {multiall/all} x)}$.

Then, we can replace $x^2 + \var{multiall/all}x$ with $(x+\var{multiall/all/2})^2 - \var{multiall/all/2}^2$.

\begin{align}
\simplify[basic]{ {all}x^2 + {multiall}x + {odds3-evens2}} &= \simplify[basic]{ {all}*( x^2 + {multiall/all} x) + {odds3-evens2}} & \text{Extract the common factor of } \var{all} \\
&= \simplify[basic]{ {all}*( (x+{multiall/all/2})^2 - {multiall/all/2}^2) + {odds3-evens2} } & \text{Complete the square}\\
&= \simplify[basic]{ {all}*(x+{multiall/all/2})^2 - {all}*{(multiall/all/2)^2} + {odds3-evens2} } & \text{Expand the constant term}\\
&= \simplify[basic]{ {all}*(x+{multiall/all/2})^2 + {odds3-evens2 - (multiall/2)^2/all}} & \text{Collect constants}
\end{align}

Question 4

When quadratic equations can't be factorised, or if equations are difficult to factorise (perhaps if the coefficients are large), we need to use the quadratic formula to solve the equations.

Use the quadratic formula to calculate values for $x$ in these equations. Input the possible values as $x_1$ and $x_2$, where $x_1<x_2$.

a)

$\simplify{x^2+{a+m}x+{a*m}=0}$

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b)

$\simplify{{a1}x^2+{a2}x+{a3}={a4}}$

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c)

$\simplify{{b1}x^2+{b2}x+{b3}={b4}x}$

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Advice

The quadratic formula is

\[x={\frac {-b\pm\sqrt{b^2-4\times a\times c}}{2a}}\text{.}\]

a)

From the equation, we can read off values for $a$, $b$ and $c$:

\[\begin{align}
a&=1\text{,}\\
b&=\var{a+m}\text{,}\\
c&=\var{a*m} \text{.}
\end{align}\]

Substituting these values into the quadratic formula,

\[x = \frac {-\var{a+m}\pm\sqrt{\var{a+m}^2-4\times \var{a*m}}}{2}\text{.}\]

Note the $\pm$ symbol in the formula. This means there are two solutions: one using $+$, the other using $-$.

The two solutions are

\[\begin{align}
x_1&=\var{m}\text{,}\\
x_2&=\var{a}\text{.}
\end{align}\]

b)

Note that the right-hand side of the given equation is not zero. We need to rewrite it in the form $ax^2+bx+c=0$:

\[\begin{align}
\simplify{{a1}x^2+{a2}x+{a3}}&=\var{a4}\\
\simplify{{a1}x^2+{a2}x+{a3-a4}}&=0\text{.}
\end{align}\]

Then we can read off values for $a$, $b$ and $c$:

\[\begin{align}
a&=\var{a1}\\
b&=\var{a2}\\
c&=\var{a3-a4} \text{.}
\end{align}\]

We can now substitute these values into the quadratic formula:

\[x = {\frac {-\var{a2}\pm\sqrt{\var{a2}^2-4\times \var{a1}\times \var{a3-a4}}}{2\times\var{a1}}}\text{.}\]

So the two solutions are

\[\begin{align}
x_1&=\var{dpformat(x1,2)}\\
x_2&=\var{dpformat(x2,2)}\text{.}
\end{align}\]

c)

We first rearrange our equation into the form $ax^2+bx+c=0$:

\[\begin{align}
\simplify{{b1}x^2+{b2}x+{b3}}&=0=\var{b4}x\\
\simplify{{b1}x^2+{b2-b4}x+{b3}}&=0\text{.}
\end{align}\]

We can then read off the values for $a, b$ and $c$, which are

\[\begin{align}
a&=\var{b1}\text{,}\\
b&=\var{b2-b4}\text{,}\\
c&=\var{b3}\text{.}
\end{align}\]

Substituting these values into the quadratic formula,

\[x = {\frac {-\var{b2-b4}\pm\sqrt{\var{b2-b4}^2-4\times \var{b1}\times \var{b3}}}{2\times\var{b1}}},\]

we obtain solutions

\[\begin{align}
x_1&=\var{dpformat(p1,2)}\text{,}\\
x_2&=\var{dpformat(p2,2)}\text{.}
\end{align}\]

Question 5

The coefficients of the following equation involve the unknown value $k$. We can use the quadratic formula to find expressions for the values of $x$ in terms of $k$.

Solve the equation $\simplify {x^2+{b_3}*k*x+{c_2}k^2=0}$. Give your answer in terms of $k$. Assuming $k$ is positive, enter the lowest root first.

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Advice

The quadratic formula is

\[{\frac {-b\pm\sqrt{b^2-4\times a\times c}}{2a}}\text{.}\]

We can list our values for $a, b$ and $c$.

\[\begin{align}
a&=1\\
b&=\var{b_3}k\\
c&=\var{c_2}k^2
\end{align}\]

Then by substituting them into the quadratic formula, we obtain

\[x=\frac {-\var{b_3}k\pm\sqrt{\var{b_3}^2k^2-4\times \var{c_2}k^2}}{2}\]

We can then simplify this equation to

\[\begin{align}
x&=\frac {-\var{b_3}k\pm k\sqrt{\var{b_3}^2-\var{4c_2}}}{2}\\
\end{align}\]
\[\begin{align}
&=k\left(\frac{-\var{b_3}}{2}\pm \frac {\sqrt{\var{b_3}^2-\var{4c_2}}}{2}\right)\\
\end{align}\]
\[\begin{align}
&=k\left(-\frac{\var{b_3}}{2} \pm\frac{\sqrt{\var{(b_3^2-4c_2)}}}{2}\right)\\
\end{align}\]

This means our possible values for $x$ in terms of $k$ are,

\begin{align}
x_1 &= \left( \simplify[all,!noleadingminus,!collectnumbers,!simplifyfractions]{-{b_3}/2 - {sqrt(b_3^2-4c_2)}/2} \right) k = \var[fractionnumbers]{-b_3/2 - sqrt(b_3^2-4*c_2)/2}k \\
x_2 &= \left( \simplify[all,!noleadingminus,!collectnumbers,!simplifyfractions]{-{b_3}/2 + {sqrt(b_3^2-4c_2)}/2} \right) k = \var[fractionnumbers]{-b_3/2 + sqrt(b_3^2-4*c_2)/2}k
\end{align}

Question 6

Factorise the following quadratic equations.

a)

$\simplify{x^2+{v1+v2}x+{v1*v2}=0}$

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b)

$\simplify{x^2+{v3+v4}x+{v3*v4}}=0$

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c)

$\simplify{x^2+{v5*v6}}=0$

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Advice

Quadratic equations of the form

\[x^2+bx+c=0\]

can be factorised to create an equation of the form

\[(x+m)(x+n)=0\text{.}\]

When we expand a factorised quadratic expression we obtain

\[(x+m)(x+n)=x^2+(m+n)x+(m \times n)\text{.}\]

To factorise an equation of the form $x^2+bx+c$, we need to find two numbers which add together to make $b$, and multiply together to make $c$.

a)

\[\simplify{x^2+{v1+v2}x+{v1*v2}=0}\]

We need to find two values that add together to make $\var{v1+v2}$ and multiply together to make $\var{v1*v2}$.

\[\begin{align}
\var{v1} \times \var{v2}&=\var{v1*v2}\\
\var{v1}+\var{v2}&=\var{v1+v2}\\
\end{align} \]

So the factorised form of the equation is

\[\simplify{(x+{v1})(x+{v2})}=0\text{.}\]

b)

We can begin factorising by finding factors of $\var{v3*v4}$ that add together to give $\var{v3+v4}$.

\[\begin{align}
\var{v3} \times \var{v4}&=\var{v3*v4}\\
\var{v3}+\var{v4}&=\var{v3+v4}\\
\end{align} \]

So the factorised form of the equation is

\[\simplify{(x+{v3})(x+{v4})}=0\text{.}\]

c)

When factorising the quadratic expression

\[\simplify{x^2+{v5*v6}=0}\]

we need to find two values that add together to make $0$ and multiply together to make $\var{v5*v6}$.

\begin{align}
\var{v5} \times \var{v6}& = \var{v5*v6}\\
\simplify[]{ {v5} + {v6}} &= 0 \\
\end{align}

So the factorised form of the equation is

\[\simplify{(x+{v5})(x+{v6})}=0\text{.}\]

Numbas has failed

Quadratic expressions and equations

Quadratic expressions and equations

Question 1

Question 1

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Question 2

Question 2

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Question 3

Question 3

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Question 4

Question 4

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Question 5

Question 5

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Question 6

Question 6

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