Functions and algebra: Solve algebraic equations and inequalities

# Unit 1: Solve quadratic equations by factorisation

Dylan Busa

### Unit outcomes

By the end of this unit you will be able to:

• Apply the zero-product law.
• Solve rational quadratic equations which have fractions with the unknown in the denominator.
• Factorise quadratic expressions to solve for the unknown variable.

## What you should know

Before you start this unit, make sure you can:

## Introduction

In this unit we expand on what you learnt about quadratic equations in level 2 by looking at some slightly more complex quadratic equations.

Remember a quadratic equation is an equation which can be written as $\scriptsize a{{x}^{2}}+bx+c=0$ where $\scriptsize a$, $\scriptsize b$ and $\scriptsize c$ are constants and the highest power on $\scriptsize x$ is $\scriptsize 2$. $\scriptsize a{{x}^{2}}+bx+c=0$ is called the standard form of the quadratic equation.

Here are some examples of quadratic equations:

• $\scriptsize 2{{x}^{2}}+5x+2=0\quad \text{This is in standard form}$
• $\scriptsize a(a-3)=10\quad \text{This needs to be expanded to get the equation into standard form}$
• $\scriptsize \displaystyle \frac{4}{{x+1}}-1=\displaystyle \frac{{3x}}{{x+2}}\quad \text{ We must multiply through by the LCD and simplify into standard form}$

In level 2 we learnt that to solve a quadratic equation we need to follow some basic steps.

### Take note!

Steps to solve a quadratic equation:

1. Get one side of the equation equal to zero (write the quadratic in standard form).
2. Factorise the quadratic expression on the one side of the equation.
3. Apply the zero product property to set each factor equal to zero.
4. Solve for $\scriptsize x$ in each factor.
5. Check your solutions.

No matter how complicated the quadratic equation seems to be, these are always these steps to follow.

### Note

Sometimes we refer to the solution of an equation as the root of the equation. Because quadratic equations always give two solutions, quadratic equations have at most two roots.

## Solve quadratic equations by factorisation

We know that to solve a quadratic equation we need to apply the zero product property. The zero product property says that if $\scriptsize A\times B=0$ then $\scriptsize A=0$ or $\scriptsize B=0$ or both $\scriptsize A=0$ and $\scriptsize B=0$. To get a quadratic equation into the form $\scriptsize A\times B=0$, we need to find the factors of the quadratic expression on the left-hand side of the equation. Often, these can be found by factoristing the quadratic expression. This is always the first thing you should try.

The zero product property:

if $\scriptsize A\times B=0$ then $\scriptsize A=0$ or $\scriptsize B=0$ or both $\scriptsize A=0$ and $\scriptsize B=0$

### Example 1.1

Solve for $\scriptsize x$: $\scriptsize 2{{x}^{2}}+5x+2=0$.

Solution

Step 1: Get equation into standard form
The equation is already in standard form – $\scriptsize a{{x}^{2}}+bx+c=0$.

Step 2: Factorise the quadratic expression
\scriptsize \begin{align*}2{{x}^{2}}+5x+2 & =0\\\therefore (2x+1)(x+2) & =0\end{align*}

Step 3: Apply the zero product property
\scriptsize \begin{align*}(2x+1)(x+2) & =0\\\therefore (2x+1)=0\ & \text{ or }\ (x+2)=0\end{align*}

Step 4: Solve for $\scriptsize x$ in each factor
\scriptsize \begin{align*}(2x+1)=0\ & \text{ or }\ (x+2)=0\\\therefore 2x=-1\ & \text{ or }\ x=-2\\\therefore x=-\displaystyle \frac{1}{2}\ & \text{ or }\ x=-2\end{align*}

Step 5: Check solutions
It is always a good idea to check the solutions you get. This is not presented as part of your solution but should be done as an aside.

$\scriptsize x=-\displaystyle \frac{1}{2}$:
\scriptsize \begin{align*}\text{LHS}&=2{{\left( {-\displaystyle \frac{1}{2}} \right)}^{2}}+5\left( {-\displaystyle \frac{1}{2}} \right)+2\\&=2\times \displaystyle \frac{1}{4}-\displaystyle \frac{5}{2}+2\\&=\displaystyle \frac{1}{2}-\displaystyle \frac{5}{2}+\displaystyle \frac{4}{2}\\&=\displaystyle \frac{0}{2}\\&=0\\&=\text{RHS}\end{align*}

$\scriptsize x=-2$:
\scriptsize \begin{align*}\text{LHS}&=2{{\left( {-2} \right)}^{2}}+5\left( {-2} \right)+2\\&=2\times 4-10+2\\&=8-10+2\\&=0\\&=\text{RHS}\end{align*}
Both solutions are valid. Therefore, $\scriptsize x=-\displaystyle \frac{1}{2} \text{ or }\ x=-2$.

Note: It is always best to test your solutions with the original equation.

### Example 1.2

Solve for $\scriptsize a$: $\scriptsize a(a-3)=10$

Solution

Step 1: Get equation into standard form
\scriptsize \begin{align*}a(a-3) & =10\\\therefore {{a}^{2}}-3a & =10\\\therefore {{a}^{2}}-3a-10 & =0\end{align*}

Step 2: Factorise the quadratic expression
\scriptsize \begin{align*}{{a}^{2}}-3a-10 & =0\\\therefore (a-5)(a+2) & =0\end{align*}

Step 3: Apply the zero product property
\scriptsize \begin{align*}(a-5)(a+2) & =0\\\therefore (a-5)=0\ & \text{ or }\ (a+2)=0\end{align*}

Step 4: Solve for $\scriptsize a$ in each factor
\scriptsize \begin{align*}(a-5)=0\ & \text{ or }\ (x+2)=0\\\therefore a=5\ & \text{ or }\ a=-2\end{align*}

Step 5: Check solutions
$\scriptsize a=5$:
\scriptsize \begin{align*}\text{LHS}&=5(5-3)\\&=10\\&=\text{RHS}\end{align*}

$\scriptsize x=-2$:
\scriptsize \begin{align*}\text{LHS}&=(-2)\left( {(-2)-3} \right)\\&=(-2)(-5)\\&=10\\&=\text{RHS}\end{align*}

Both solutions are valid. Therefore, $\scriptsize a=5 \text{ or }\ a=-2$.

Note: Remember, it is always best to test your solutions with the original equation.

### Example 1.3

Solve for $\scriptsize x$: $\scriptsize \displaystyle \frac{4}{{x+1}}-1=\displaystyle \frac{{3x}}{{x+2}}$

Solution

Step 1: Get equation into standard form
\scriptsize \begin{align*}\displaystyle \frac{4}{{x+1}}-1 & =\displaystyle \frac{{3x}}{{x+2}}\text{ LCD: }(x+1)(x+2) & \\\therefore 4(x+2)-(x+1)(x+2) & =3x(x+1)\\\therefore 4x+8-({{x}^{2}}+3x+2) & =3{{x}^{2}}+3x\\\therefore 4x+8-{{x}^{2}}-3x-2-3{{x}^{2}}-3x & =0\\\therefore -4{{x}^{2}}-2x+6 & =0\\\therefore 4{{x}^{2}}+2x-6 & =0\\\therefore 2(2{{x}^{2}}+x-3) & =0\end{align*}
Because there are unknowns in the denominator, we need to place restrictions on the solutions that we can accept. We are never permitted to have zero in the denominator. Therefore:
\scriptsize \begin{align*}(x+1)\ne 0\text{ } & \text{and }(x+2)\ne 0\\\therefore x\ne -1\text{ } & \text{and }x\ne -2\end{align*}

Step 2: Factorise the quadratic expression
\scriptsize \begin{align*}2(2{{x}^{2}}+x-3) & =0\\\therefore 2{{x}^{2}}+x-3 & =0\\\therefore (2x+3)(x-1) & =0\end{align*}

Step 3: Apply the zero product property
\scriptsize \begin{align*}(2x+3)(x-1) & =0\\\therefore (2x+3)=0\ & \text{ or }\ (x-1)=0\end{align*}

Step 4: Solve for $\scriptsize x$ in each factor
\scriptsize \begin{align*}(2x+3)=0\ & \text{ or }\ (x-1)=0\\\therefore x=-\displaystyle \frac{3}{2}\ & \text{ or }\ x=1\end{align*}

Step 5: Check solutions
Neither solution contradicts our restrictions.

$\scriptsize x=1$:
\scriptsize \begin{align*}\text{LHS}&=\displaystyle \frac{4}{{x+1}}-1\\&=\displaystyle \frac{4}{{1+1}}-1\\&=1\end{align*}
\scriptsize \begin{align*}\text{RHS}&=\displaystyle \frac{{3x}}{{x+2}}\\&=\displaystyle \frac{{3(1)}}{{1+2}}\\&=1\\&=\text{LHS}\end{align*}

$\scriptsize x=-\displaystyle \frac{3}{2}$:
\scriptsize \begin{align*}\text{LHS}&=\displaystyle \frac{4}{{\left( {-\displaystyle \frac{3}{2}} \right)+1}}-1\\&=\displaystyle \frac{4}{{\left( {-\displaystyle \frac{1}{2}} \right)}}-1\\&=-8-1\\&=-9\end{align*}
\scriptsize \begin{align*}\text{RHS}&=\displaystyle \frac{{3x}}{{x+2}}\\&=\displaystyle \frac{{3\left( {-\displaystyle \frac{3}{2}} \right)}}{{\left( {-\displaystyle \frac{3}{2}} \right)+2}}\\&=\displaystyle \frac{{-\displaystyle \frac{9}{2}}}{{\displaystyle \frac{1}{2}}}\\&=-\displaystyle \frac{9}{2}\times \displaystyle \frac{2}{1}\\&=-9\\&=\text{LHS}\end{align*}

Both solutions are valid. Therefore, $\scriptsize x=1 \text{ or }\ x=-\displaystyle \frac{3}{2}$.

### Exercise 1.1

Solve for the unknown in each case:

1. $\scriptsize 6{{g}^{2}}+12g=0$
2. $\scriptsize 13{{y}^{2}}-26y=-13$
3. $\scriptsize a(2a+1)=15$
4. $\scriptsize 13{{t}^{2}}=325$
5. $\scriptsize \displaystyle \frac{{d-2}}{{d+1}}=\displaystyle \frac{{2d+1}}{{d-7}}$
6. $\scriptsize \displaystyle \frac{{5x}}{{x-2}}+\displaystyle \frac{6}{{{{y}^{2}}-2y}}=-\displaystyle \frac{3}{y}-2$

The full solutions are at the end of the unit.

## Solve more complex quadratic equations

Let’s now work through some examples of how to solve more complex quadratic equations.

### Example 1.4

Solve for the unknown: $\scriptsize {{x}^{2}}=13$

Solution

Step 1: Get equation into standard form
\scriptsize \begin{align*}{{x}^{2}} & =13\\\therefore {{x}^{2}}-13 & =0\end{align*}
Note: In this case the coefficient of the $\scriptsize x$ term in $\scriptsize {{x}^{2}}+bx+c$ is zero.

Step 2: Factorise the quadratic expression
We can factorise the quadratic expression using the difference of two squares by recognising that $\scriptsize {{\left( {\sqrt{{13}}} \right)}^{2}}=13$.
\scriptsize \begin{align*}{{x}^{2}}-13 & =0\\\therefore \left( {x+\sqrt{{13}}} \right)\left( {x-\sqrt{{13}})} \right) & =0\end{align*}

Step 3: Apply the zero product property
\scriptsize \begin{align*}\left( {x+\sqrt{{13}}} \right)\left( {x-\sqrt{{13}})} \right) & =0\\\therefore \left( {x+\sqrt{{13}}} \right)=0\ & \text{ or }\ \left( {x-\sqrt{{13}}} \right)=0\end{align*}

Step 4: Solve for $\scriptsize x$ in each factor
\scriptsize \begin{align*}\left( {x+\sqrt{{13}}} \right)=0\ & \text{ or }\ \left( {x-\sqrt{{13}}} \right)=0\\\therefore x=-\sqrt{{13}}\ & \text{ or }\ x=\sqrt{{13}}\end{align*}

Step 5: Check solutions
$\scriptsize x=-\sqrt{{13}}$:
\scriptsize \begin{align*}\text{LHS}&={{\left( {-\sqrt{{13}}} \right)}^{2}}\\&=13\\&=\text{RHS}\end{align*}

$\scriptsize x=\sqrt{{13}}$:
\scriptsize \begin{align*}\text{LHS}&={{\left( {\sqrt{{13}}} \right)}^{2}}\\&=13\\&=\text{RHS}\end{align*}
Both solutions are valid. Therefore, $\scriptsize x=-\sqrt{{13}} \text{ or }\ x=\sqrt{{13}}$.

### Example 1.5

Solve for the unknown: $\scriptsize n+2=\sqrt{{7+2n}}$.

Solution

Step 1: Get equation into standard form
We need to ‘get rid’ of the square root on the RHS. Therefore, we need to square both sides of the equation. However, before we do this, it is important to recognise that the value inside the square root sign must be greater than or equal to zero. In other words:
\scriptsize \begin{align*}7+2n & \ge 0\\\therefore 2n & \ge -7\\\therefore n & \ge -\displaystyle \frac{7}{2}\end{align*}
\scriptsize \begin{align*}n+2 & =\sqrt{{7+2n}}\quad \quad \text{Restriction: }\ge -\displaystyle \frac{7}{2}\\\therefore {{(n+2)}^{2}} & ={{\left( {\sqrt{{7+2n}}} \right)}^{2}}\quad \text{Make sure you square the whole of the LHS}\\\therefore {{n}^{2}}+4n+4 & =7+2n\\\therefore {{n}^{2}}+2n-3 & =0\end{align*}

Step 2: Factorise the quadratic expression
\scriptsize \begin{align*}{{n}^{2}}+2n-3 & =0\\\therefore (n+3)(n-1) & =0\end{align*}

Step 3: Apply the zero product property
\scriptsize \begin{align*}(n+3)(n-1) & =0\\\therefore (n+3)=0\text{ } & \text{or (}n-1)=0\end{align*}

Step 4: Solve for $\scriptsize x$ in each factor
\scriptsize \begin{align*}(n+3)=0\text{ } & \text{or (}n-1)=0\\\therefore n=-3\text{ } & \text{or }n=1\end{align*}

Step 5: Check solutions
You can either check your solutions by testing to see if they make the $\scriptsize \text{LHS}=\text{RHS}$ or you can test them against the original restriction that $\scriptsize n\ge -\displaystyle \frac{7}{2}$.

$\scriptsize n=-3$:
\scriptsize \begin{align*}\text{LHS}&=\text{ }n+2\\&=-3+2\\&=-1\end{align*}
\scriptsize \begin{align*}\text{RHS}&=\sqrt{{7+2n}}\\&=\sqrt{{7+2(-3)}}\\&=1\ne \text{LHS}\end{align*}
$\scriptsize n=-3$ conflicts with the restriction that $\scriptsize n\ge -\displaystyle \frac{7}{2}$.

$\scriptsize n=1$:
\scriptsize \begin{align*}\text{LHS}&=\text{ }n+2\\&=1+2\\&=3\end{align*}
\scriptsize \begin{align*}\text{RHS}&=\sqrt{{7+2n}}\\&=\sqrt{{7+2(1)}}\\&=3\\&=\text{LHS}\end{align*}
$\scriptsize n=1$ does not conflict with the restriction that $\scriptsize n\ge -\displaystyle \frac{7}{2}$.

Only one solution is valid. Therefore, $\scriptsize n=1$.

### Example 1.6

Solve for $\scriptsize y$: $\scriptsize 3\sqrt{{2y+1}}-5=-2(2y+1)$

Solution

The first thing we need to recognise is that there are restrictions because of the square root.
\scriptsize \begin{align*}2y+1&\ge 0\\\therefore y&\ge -\displaystyle \frac{1}{2}\end{align*}
$\scriptsize 3\sqrt{{2y+1}}-5=-2(2y+1)\quad \quad \text{Restriction: }y\ge -\displaystyle \frac{1}{2}$

While we could square both sides to ‘get rid’ of the square root, a better technique would be to recognise that if $\scriptsize A=\sqrt{{2y+1}}$ then $\scriptsize (2y+1)={{A}^{2}}$. Therefore, we can start by doing a substitution to make the equation simpler.

Let $\scriptsize A=\sqrt{{2y+1}}$
\scriptsize \begin{align*}3A-5 & =-2{{A}^{2}}\\\therefore 2{{A}^{2}}+3A-5 & =0\\\therefore (2A+5)(A-1) & =0\\\therefore A=-\displaystyle \frac{5}{2}\text{ } & \text{or }A=1\end{align*}

Now substitute $\scriptsize A$ back for $\scriptsize \sqrt{{2y+1}}$:
$\scriptsize \sqrt{{2y+1}}=-\displaystyle \frac{5}{2}\ \text{or }\sqrt{{2y+1}}=1$
\scriptsize \begin{align*}\sqrt{{2y-1}} & =-\displaystyle \frac{5}{2}\\\therefore 2y-1 & =\displaystyle \frac{{25}}{4}\\\therefore 2y & =\displaystyle \frac{{25+4}}{4}=\displaystyle \frac{{29}}{4}\\\therefore y & =\displaystyle \frac{{29}}{8}\end{align*}

This does not conflict with the resistriction that $\scriptsize y\ge -\displaystyle \frac{1}{2}$ but we need to check the solution.
\scriptsize \begin{align*}\text{LHS}&=3\sqrt{{2y+1}}-5\\&=3\sqrt{{2\left( {\displaystyle \frac{{29}}{8}} \right)+1}}-5\\&=3\sqrt{{\displaystyle \frac{{29}}{4}+\displaystyle \frac{4}{4}}}-5\\&=3\sqrt{{\displaystyle \frac{{33}}{4}}}-5\\&=3\displaystyle \frac{{\sqrt{{33}}}}{2}-5\\&=\displaystyle \frac{{3\sqrt{{33}}-10}}{2}\end{align*}
\scriptsize \begin{align*}\text{RHS}&=-2(2y+1)\\&=-2\left( {2\left( {\displaystyle \frac{{29}}{8}} \right)+1} \right)\\&=-2\left( {\displaystyle \frac{{29}}{4}+\displaystyle \frac{4}{4}} \right)\\&=-2\left( {\displaystyle \frac{{33}}{4}} \right)\\&=-\displaystyle \frac{{66}}{4}\end{align*}
$\scriptsize \text{LHS}\ne \text{RHS}$ so we cannot accept the solution.

\scriptsize \begin{align*}\sqrt{{2y+1}} & =1\\\therefore 2y+1 & =1\\\therefore y & =0\end{align*}

Check the solution:
\scriptsize \begin{align*}\text{LHS}&=3\sqrt{{2y+1}}-5\\&=3\sqrt{1}-5\\&=-2\end{align*}
\scriptsize \begin{align*}\text{RHS}&=-2(2y+1)\\&=-2\\&=\text{LHS}\end{align*}

Therefore, $\scriptsize y=0$ only.

### Exercise 1.2

Solve for the unknowns in the following:

1. $\scriptsize 17={{a}^{2}}$
2. $\scriptsize 4+a=\sqrt{{a+6}}$
3. $\scriptsize 2k=\sqrt{{5-12k}}$
4. $\scriptsize 5\sqrt{{5y+1}}-4=5y+1$

The full solutions are at the end of the unit.

## Summary

In this unit you have learnt the following:

• How to solve quadratic equations using factorisation.
• How to apply the zero product property.
• How to solve quadratic equations involving square roots.
• How to find and apply restrictions to quadratic equations.

# Unit 1: Assessment

#### Suggested time to complete: 35 minutes

1. Solve for the unknown in each of the following:
1. $\scriptsize 21+5a=6{{a}^{2}}$
2. $\scriptsize \displaystyle \frac{x}{{x+3}}+\displaystyle \frac{{x+9}}{{{{x}^{2}}-9}} =\displaystyle \frac{2}{{x-3}}$
3. $\scriptsize 2{{x}^{3}}=10x$
4. $\scriptsize \displaystyle \frac{1}{2}p=\sqrt{{-p-5}}-4$
5. $\scriptsize \displaystyle \frac{{z-1}}{{{{z}^{2}}-z-6}}-\displaystyle \frac{4}{{{{z}^{2}}-4}} =\displaystyle \frac{4}{{z+2}}$
2. Solve for $\scriptsize y$ in terms of $\scriptsize k$: $\scriptsize {{y}^{2}}-6yk=7{{k}^{2}}$

The full solutions are at the end of the unit.

# Unit 1: Solutions

### Exercise 1.1

1. .
\scriptsize \begin{align*} 6g^2+12g&=0\\ \therefore 6(g^2+2g)&=0\\ \therefore g^2+2g&=0\\ \therefore g(g+2)&=0\\ \therefore g=0 &\text{ or }g=-2 \end{align*}
2. .
\scriptsize \begin{align*} 13y^2-26y&=-13\\ \therefore 13y^2-26y+13&=0\\ \therefore 13(y^2-2y+1)&=0\\ \therefore y^2-2y+1&=0\\ \therefore (y-1)(y-1)=0\\ \therefore y=1 \end{align*}
Note: It may look like there is only one solution or root. However, there are still two roots. They just happen to be the same.
3. .
\scriptsize \begin{align*} 2(2a+1)&=15\\ \therefore 2a^2+a-15&=0\\ \therefore (2a-5)(a+3)&=0\\ \therefore a=\displaystyle \frac{5}{2}&\text{ or }a=-3 \end{align*}
4. .
\scriptsize \begin{align*} 13t^2&=325\\ \therefore 13t^2-325&=0\\ \therefore 13(t^2-25)&=0\\ \therefore t^2-25&=0\\ \therefore (t+5)(t-5)&=0\\ \therefore t=-5&\text{ or }t=5 \end{align*}
5. .
\scriptsize \begin{align*}\displaystyle \frac{{d-2}}{{d+1}} & =\displaystyle \frac{{2d+1}}{{d-7}}\quad \quad \text{Restriction: }d\ne -1,d\ne 7\\\therefore (d-2)(d-7) & =(2d+1)(d+1)\\\therefore {{d}^{2}}-9d+14 & =2{{d}^{2}}+3d+1\\\therefore -{{d}^{2}}-12d+13 & =0\\\therefore {{d}^{2}}+12d-13 & =0\\\therefore (d-1)(d+13) & =0\\\therefore d=1\text{ } & \text{or }d=-13\end{align*}
6. .
\scriptsize \begin{align*}\displaystyle \frac{{5x}}{{x-2}}+\displaystyle \frac{6}{{{{x}^{2}}-2x}} & =-\displaystyle \frac{3}{x}-2\\\therefore \displaystyle \frac{{5x}}{{x-2}}+\displaystyle \frac{6}{{x(x-2)}} & =-\displaystyle \frac{3}{x}-2\quad \quad \text{Restriction: }x\ne 2,x\ne 0\\\therefore 5x(x)+6 & =-3(x-2)-2x(x-2)\\\therefore 5{{x}^{2}}+6 & =-3x+6-2{{x}^{2}}+4x\\\therefore 7{{x}^{2}}-x & =0\\\therefore x(7x-1) & =0\\\therefore \xcancel{{x=0}}\text{ } & \text{or }x=\displaystyle \frac{1}{7}\\\therefore x & =\displaystyle \frac{1}{7}\end{align*}

Back to Exercise 1.1

### Exercise 1.2

1. .
\scriptsize \begin{align*}17&=a^2\\ \therefore a^2-17&=0\\ \therefore \left(a+\sqrt{17}\right)\left(a-\sqrt{17}\right)&=0\\ \therefore a=-\sqrt{17}&\text{ or }a=\sqrt{17} \end{align*}
2. .
\scriptsize \begin{align*}4+a & =\sqrt{{a+6}}\quad \quad \text{Restriction: }a\ge -6\\\therefore {{(4+a)}^{2}} & =a+6\\\therefore 16+8a+{{a}^{2}} & =a+6\\\therefore {{a}^{2}}+7a+10 & =0\\\therefore (a+2)(a+5) & =0\\\therefore a=-2\text{ } & \text{or }\xcancel{{a=-5}}\\\therefore a & =-2\end{align*}
3. .
\scriptsize \begin{align*}2k & =\sqrt{{5-8k}}\quad \quad \text{Restriction: }k\le \displaystyle \frac{5}{8}\\\therefore 4{{k}^{2}} & =5-8k\\\therefore 4{{k}^{2}}+8k-5 & =0\\\therefore (2k-1)(2k+5) & =0\\\therefore k=\displaystyle \frac{1}{2}\text{ } & \text{or }\xcancel{{k=-\displaystyle \frac{5}{2}}}\\\therefore k & =\displaystyle \frac{1}{2}\end{align*}
4. .
\scriptsize \begin{align*}5\sqrt{5y+1}-4&=5y+1\\ \text{Let }\sqrt{5y+1}&=A\\ \therefore 5A-4&=A^2\\ \therefore A^2-5A+4&=0\\ \therefore (A-4)(A-1)&=0\\ \therefore A=4&\text{ or }A=1\\ \therefore \sqrt{5y+1}=4&\text{ or }\sqrt{5y+1}=1\\ \therefore 5y+1=16&\text{ or }5y+1=1\\ \therefore 5y=15&\text{ or }5y=0\\ \therefore y=3&\text{ or }y=0 \end{align*}

Back to Exercise 1.2

### Unit 1: Assessment

1. .
1. .
\scriptsize \begin{align*}21+5a&=6a^2\\ \therefore 6a^2-5a-21&=0\\ \therefore (3a-7)(2a+3)&=0\\ \therefore a=\displaystyle \frac{7}{3}&\text{ or }a=-\displaystyle \frac{3}{2} \end{align*}
2. .
\scriptsize \begin{align*}\displaystyle \frac{x}{{x+3}}+\displaystyle \frac{{x+9}}{{{{x}^{2}}-9}} & =\displaystyle \frac{2}{{x-3}}\\\therefore \displaystyle \frac{x}{{x+3}}+\displaystyle \frac{{x+9}}{{(x+3)(x-3)}} & =\displaystyle \frac{2}{{x-3}}\quad \quad \text{Restrictions: }x\ne -3,x\ne 3\\\therefore x(x-3)+(x+9) & =2(x+3)\\\therefore {{x}^{2}}-3x+x+9 & =2x+6\\\therefore {{x}^{2}}-4x+3 & =0\\\therefore (x-1)(x-3) & =0\\\therefore x=1\text{ } & \text{or }\xcancel{{x=3}}\\\therefore x & =1\end{align*}
3. .
\scriptsize \begin{align*}2x^3&=10x\\ \therefore 2x^3-10x&=0\\ \therefore 2x(x^2-5)&=0\\ \therefore 2x\left(x+\sqrt{5}\right)\left(x-\sqrt{5}\right)&=0\\ \therefore x-0&\text{ or }x=-\sqrt{5}\text{ or }x=\sqrt{5} \end{align*}
4. .
\scriptsize \begin{align*} \displaystyle \frac{1}{2}p&=\sqrt{-p-5}-4\\ \therefore \displaystyle \frac{1}{2}p+4&=\sqrt{-p-5}\\ \therefore \left(\displaystyle \frac{1}{2}p+4\right)^2&=-p-5\\ \therefore \displaystyle \frac{1}{4}p^2+4p+16&=-p-5\\ \therefore p^2+16p+64+4p+20&=0\\ \therefore p^2+20p+84&=0\\ \therefore (p+6)(p+14)&=0\\ \therefore p=-6&\text{ or }\xcancel{p=14} \end{align*}
5. .
\scriptsize \begin{align*}\displaystyle \frac{{z-1}}{{{{z}^{2}}-z-6}}-\displaystyle \frac{4}{{{{z}^{2}}-4}} & =\displaystyle \frac{4}{{z+2}}\\\therefore \displaystyle \frac{{z-1}}{{(z-3)(z+2)}}-\displaystyle \frac{4}{{(z+2)(z-2)}} & =\displaystyle \frac{4}{{(z+2)}}\quad \quad \text{Restrictions: }z\ne 3,z\ne \pm 2\\\therefore (z-1)(z-2)-4(z-3) & =4(z-3)(z-2)\\\therefore {{z}^{2}}-3z+2-4z+12 & =4{{z}^{2}}-20z+24\\\therefore 3{{z}^{2}}-13z+10 & =0\\\therefore (3z-10)(z-1) & =0\\\therefore z=\displaystyle \frac{{10}}{3}\ & \text{ or }\ z=1\end{align*}
2. .
\scriptsize \begin{align*} y^2-6yk&=7k^2\\ \therefore y^2-6yk-7k^2&=0\\ \therefore (y-7k)(y+k)&=0\\ \therefore y=7k&\text{ or }y=-k \end{align*}

Back to Unit 1: Assessment