The Zeroes of a Quadratic Function

The zero, or |x|-intercept, of a function |f| is a value of |x| for which |f(x)=0.| A function can have multiple zeroes.

Finding the zeroes of a quadratic function amounts to finding the point or points of intersection between the parabola and the |x|-axis. There may be 1, 2, or none.

Note that |x| cannot be directly isolated in the equation |0=ax^2+bx+c|, because variable |x| is found in 2 terms that are not like terms.

The zero product rule is as follows.

A product of factors is zero if and only if at least one of its factors is zero.

||a \times b = 0\\ \Updownarrow\\a=0\ \ \text{or}\ \ b=0||

1. Replace |f(x)| with |0.|

2. Factor the polynomial.

3. Apply the zero product rule.

Determine the zeroes of the function |f(x)=4x^2+12x+9.|

1. Replace |f(x)| with |0|
   ||\begin{align}f(x)&=4x^2+12x+9\\0&=4x^2+12x+9\end{align}||

2. Factor the polynomial
   This polynomial is a perfect square trinomial. This gives the following equation.
   ||\begin{align} 0&=4x^2+12x+9\\ &=(2x+3)^2\end{align}||So, we have |0=(2x+3)^2| or |0=(2x+3)(2x+3).|

3. Apply the zero product rule
   Since the 2 factors are identical, the function has only one zero.
   ||\begin{align}2x+3&=0\qquad\\ 2x&=-3\\x&=-\dfrac{3}{2}\end{align}||

Answer: The zero of the function is |-\dfrac{3}{2}.|

Determine the zeroes of the function |f(x)=x^2-0.8x-3.84.|

1. Replace |f(x)| with |0|
   ||\begin{align}f(x)&=x^2-0.8x-3.84\\0&=x^2-0.8x-3.84\end{align}||

2. Factor the polynomial
   ||\begin{align}&x^2-0.8x-3.84\\=\ &(x^2-0.8x\color{#3a9a38}{+0.16})-3.84\color{#3a9a38}{-0.16}\\=\ &(x-0.4)^2-4\\=\ &\big((x-0.4)+2\big)\big((x-0.4)-2 \big)\\=\ &(x+1.6)(x-2.4) \end{align}||Therefore, |0=(x+1.6)(x-2.4).|

3. Apply the zero product rule
   ||\begin{aligned}x+1.6&=0\\ x_1&=-1.6\end{aligned}\qquad \begin{aligned}x-2.4&=0\\ x_2&=2.4\end{aligned}||

Answer: The 2 zeroes of the function are |-1.6| and |2.4.|

Determine the zeroes of the function |f(x)=x^2-3x-10.|

1. Replace |f(x)| with |0|
   ||\begin{align}f(x)&=x^2-3x-10\\0&=x^2-3x-10\end{align}||

2. Factor the polynomial
   The polynomial is factored with the product-sum technique. We are looking for 2 numbers, |m| and |n,| where the product |m \times n| must be equal to |-10| and the sum |m+n| must be equal to |-3.| Looking at the different factors of |-10,| there is |\color{#3b87cd}m=\color{#3b87cd}{-5}| and |\color{#3b87cd}n=\color{#3b87cd}{2}.|
   Next, complete the factorization. ||\begin{align}&x^2-3x-10\\=x^2\color{#3b87cd}{-5}x+\color{#3b87cd}2x-10 \\=x(x-5) + 2(x-5)\\=(x-5)(x+2)\end{align}||So, the result is |0=(x-5)(x +2).|

3. Apply the zero product rule
   Check for which values of |x| each factor is |0.| ||\begin{aligned}x-5&=0\\x_1&=5\end{aligned}\qquad \begin{aligned}x+2&=0\\x_2&=-2\end {aligned}||

Answer: The 2 zeroes of the function are |-2| and |5.|

1. Replace |f(x)| with |0.|

2. Determine the value of |a,| |b,| and |c.|

3. Apply the quadratic formula.

Determine the zeroes of the function |f(x)=x^2-3x-10.|

1. Replace |f(x)| with |0|
   ||\begin{align}f(x)&=x^2-3x-10\\0&=\color{#ec0000}1x^2\color{#3b87cd}{-3}x\color{#3a9a38}{-10}\end{align}||

2. Determine the value of |a,| |b,| and |c|
   ||\color{#ec0000}a=\color{#ec0000}1,\ \color{#3b87cd}b=\color{#3b87cd}{-3}, \ \color{#3a9a38}c=\color{#3a9a38}{-10}||

3. Apply the quadratic formula
   ||\begin{align}x_{1,2} &= \dfrac{-\color{#3b87cd}b \pm \sqrt{\color{#3b87cd}b^2-4\color{#ec0000}a\color{#3a9a38}c}}{2\color{#ec0000}a}\\\\ &= \dfrac{-(\color{#3b87cd}{-3}) \pm \sqrt{(\color{#3b87cd}{-3})^2-4(\color{#ec0000}1)(\color{#3a9a38}{-10})}}{2(\color{#ec0000}1)} \\ &= \dfrac{3 \pm \sqrt{9 + 40}}{2}\\ &= \dfrac{3 \pm \sqrt{49}}{2}\\&= \dfrac{3 \pm 7}{2} \end{align}||
   Next, separate the formula into 2 parts, one using |+| and the other using |-.|
   ||\begin{aligned} x_1 &= \dfrac{3 + 7}{2}\\&=5 \end{aligned}\qquad\begin{aligned}x_2 &= \dfrac{3 - 7}{2}\\&=-2\end{aligned}||

Answer: The 2 zeroes of the function are |-2| and |5.| This is the same answer obtained before by factoring.

Determine the zeroes of the function |f(x)=2x^2+3x-4.|

1. Replace |f(x)| with |0|
   ||\begin{align}f(x)&=2x^2+3x-4\\0&=2x^2+3x-4\end{align}||

2. Determine the value of |a,| |b,| and |c|
   ||a=2,\ b=3,\ c=-4||

3. Apply the quadratic formula
   ||\begin{align}x_{1,2} &= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\\\\ &= \dfrac{-3 \pm \sqrt{3^2-4(2)(-4)}}{2(2)} \\ &= \dfrac{-3 \pm \sqrt{9+32}}{4}\\ &= \dfrac{-3 \pm \sqrt{41}}{4}\end{align}||
   Because |41| is not a square number, keep the square root and separate the formula into 2 parts, one using |+| and the other using |-.|
   ||\begin{aligned} x_1 &=\dfrac{-3 + \sqrt{41}}{4}\\&\approx0.85 \end{aligned}\qquad\begin{aligned}x_2 &=\dfrac{-3 - \sqrt{41}}{4}\\&\approx -2.35\end{aligned}||

Answer: The zeroes are |\approx 0.85| and |\approx -2.35.| For a more precise answer, keep the radical.

||\begin{align}x_1 &= \dfrac{-3 + \sqrt{41}}{4}\\ x_2&=\dfrac{-3 - \sqrt{41}}{4}\end{align}||

Determine the zeroes of the function |f(x)=-6x^2+2x-3.|

1. Replace |f(x)| with |0|
   ||\begin{align}f(x)&=-6x^2+2x-3\\0&=-6x^2+2x-3\end{align}||

2. Determine the value of |a,| |b,| and |c|
   ||a=-6,\ b=2,\ c=-3||

3. Apply the quadratic formula
   ||\begin{align}x_{1,2} &= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\\\\ &= \dfrac{-2 \pm \sqrt{(2)^2-4(-6)(-3)}}{2(-6)} \\ &= \dfrac{-2 \pm \sqrt{4-72}}{-12}\\ &= \dfrac{-2 \pm \sqrt{\color{#EC0000}{-68}}}{-12} \end{align}||
   We cannot continue solving the equation, because the number under the square root is negative. The function has no zero.

Answer: The function does not have any zeroes.

The number of zeroes in a quadratic function can be determined by analysing the discriminant in the quadratic formula.

• When |b^2-4ac>0,| the function has 2 zeroes.

• When |b^2-4ac=0,| the function has 1 zero.

• When |b^2-4ac<0,| the function has no zeroes.

1. Replace |f(x)| with |0.|

2. Isolate the brackets.

3. Take the square root of the 2 sides of the equality.

4. Solve the equations.

There are 2 answers when you take the square root of a number: a positive one and a negative.

Determine the zeroes of the function |f(x)=-3(x+5)^2+12.|

1. Replace |f(x)| with |0|
   ||\begin{align}f(x)&=-3(x+5)^2+12\\ 0&=-3(x+5)^2+12\end{align}||

2. Isolate the brackets
   ||\begin{align}-12&=-3(x+5)^2\\4&=(x+5)^2\end{align}||

3. Take the square root
   ||\begin{align}\color{#ec0000}{\sqrt{\color{black}4}}&=\color{#ec0000}{\sqrt{\color{black}{(x+5)^2}}}\\ \pm\ 2&=x+5\end{align}||

4. Solve the equations
   ||\begin{aligned}-2&=x+5\\-7&=x_1 \end{aligned} \qquad \begin{aligned} 2&=x+5\\-3&=x_2\end{aligned}||

Answer: The 2 zeroes of the function are |-7| and |-3.|

Determine the zeroes of the function |f(x)=2(x-1)^2+6.|

1. Replace |f(x)| with |0|
   ||\begin{align}f(x)&=2(x-1)^2+6\\ 0&=2(x-1)^2+6\end{align}||

2. Isolate the brackets
   ||\begin{align}-6&=2(x-1)^2\\-3&=(x-1)^2\end{align}||

3. Take the square root
   ||\begin{align}\color{#ec0000}{\sqrt{\color{black}{-3}}}&=\color{#ec0000}{\sqrt{\color{black}{(x-1)^2}}}\end{align}||
   It is impossible to take the square root of a negative number. So, stop solving the problem because the function does not have a zero.

Answer: The function does not have any zeroes.

The number of zeroes possible in a quadratic function in standard form can be determined by analyzing parameters |a| and |k.| The sign of parameter |a| indicates the direction of the parabola’s opening (up or down), while parameter |k| indicates the vertical location of the vertex.

The function in the previous example has no zeroes, since |a>0| and |k>0.|

Here is the formula for finding the zeroes of a quadratic function in standard form.
||x_{1,2}= h \pm \sqrt{-\dfrac{k}{a}}||

1. Replace |f(x)| with |0.|

2. Determine the value of parameters |a,| |h,| and |k.|

3. Apply the zeroes of a function formula.

Determine the zeroes of the function |f(x)=2(x+1)^2-8.|

1. Replace |f(x)| with |0|
   ||\begin{align}f(x)&=2(x+1)^2-8\\ 0&=2(x+1)^2-8\end{align}||

2. Determine the value of parameters |a,| |h,| and |k|
   ||a=2,\ h=-1,\ k=-8||

3. Apply the zeroes of a function formula
   ||\begin{align}x_{1,2}&= h \pm\sqrt{-\dfrac{k}{a}}\\ &= -1 \pm \sqrt{-\dfrac{-8}{2}}\\&= -1 \pm \sqrt{4}\\&= -1 \pm 2\end{align}||
   Next, separate the formula into 2 parts, one using |+| and the other using |-.|
   ||\begin{aligned}x_1&=-1-2\qquad \\&=-3\end{aligned}\begin{aligned}x_2&=-1+2\\&=1\end{aligned}||

Answer: The 2 zeroes of the function are |-3| and |1.|

The number of zeroes in a quadratic function can be determined by analyzing the radicand (the expression under the root) in the zeroes formula.

• When |- \dfrac{k}{a} >0,| the function has 2 zeroes.

• When |- \dfrac{k}{a} =0,| the function has 1 zero.

• When |- \dfrac{k}{a} <0,| the function has no zeroes.

Find the zeroes of the function |f(x)=-0.5(x+2.7)(x-6.2).|

It is necessary to determine |x_1| and |x_2.| There are subtractions inside the brackets in the rule for factored form, which can be found in this example.

||\begin{align} f(x) &= -0.5(x+2.7)(x-6.2) \\ f(x) &= -0.5\big(x-(\color{#3a9a38}{-2.7})\big)\big(x-\color{#3a9a38}{6.2}\big) \end{align}||

Answer: The 2 zeroes of the function are |-2.7| and |6.2.|

Find the zeroes of the function |f(x)=-0.5(x+2.7)(x-6.2).|

1. Replace |f(x)| with |0|
   ||\begin{align}f(x)&=-0.5(x+2.7)(x-6.2) \\ 0&=-0.5(x+2.7)(x-6.2)\end{align}||

2. Factor the polynomial
   The polynomial is already factored.

3. Apply the zero product rule
   Verify the |x| values where the factors are |0.| Since the factor |-0.5| does not contain variable |x,| it is not taken into account.
   ||\begin{aligned}x+2.7&=0\\ x_1&=-2.7\end{aligned}\qquad\!\! \begin{aligned}x-6.2&=0\\ x_2&=6.2\end{aligned}||

Answer: The zeroes of the function are |-2.7| and |6.2.|