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Quadratic Equations

GCSE Maths Algebra Quadratic Equations

Quadratic equations are equations of the form $ax^{2}+bx+c=0$ . The main ways to solve quadratic equations are:

Factorising

Using the quadratic formula

Completing the square

Let’s look at some examples of the different methods to solve quadratic equations.

Factorising Quadratic Equations

One method to solve quadratic equations is factorising. This method involves breaking down the quadratic expression $ax^{2}+bx+c$ into two factors.

if the product of two factors is equal to zero, then at least one of the factors must be equal to zero. We can use this property to find the solutions for the given quadratic equation.

Factorising is a useful method for solving quadratic equations when the given expression can be easily factorised. However, not all quadratic equations can be solved by factorisation

Example 1:

Solve the equations:

i) $(2x-1)(3x+2)=0$

ii) $2x^{2}+x-3=0$

i) $(2x-1)(3x+2)=0$

Either $2x-1=0$

$x = \frac{1}{2}$

Or, $3x+2=0$

$x = \frac{-2}{3}$

ii) $2x^{2}+x-3=0$

$2x^{2}+3x-2x-3=0$

$x(2x+3)-1(2x+3)=0$

$(x-1)(2x+3)=0$

So, either $x-1=0$

$x=1$

Or, $2x+3=0$

$x = \frac{-3}{2}$

Example 2:

(a) Find the value of: $\dfrac{a+\sqrt{a^{2}+b^{2}}}{a^{2}-2ab}$ When $a=-4$ and $b=-3$ , give your answer as a fraction in its lowest term.

(b) Expand the brackets and simplify:

$= 6x^{3}+9x^{2}-2x-3-9x^{2}+2x$

(c) Factorise and solve the equation:

$9x^{2}+5x-4=0$

(d) Find three consecutive even numbers with a sum of 78.

(a) $\dfrac{-4+\sqrt{(-4)^{2}+(-3)^{2}}}{(-4)^{2}-2(-4)(-3)}$

$= \dfrac{-4+\sqrt{25}}{-8}$

$= -\dfrac{1}{8}$

$(3x^{2}-1)(2x+3)-x(9x-2)$

(b) $= 6x^{3}+9x^{2}-2x-3-9x^{2}+2x = 6x^{3}-3$

(c) Factorisation: $(9x - 4)(x + 1)$

Solutions: $x = -1$ or $x = \dfrac{4}{9}$

(d) Let the numbers be $2x$ , $2x+2$ , and $2x+4$ . Then:

$2x + 2x+2 + 2x+4 = 78$

$6x+6=78$

$x=12$

The three even numbers are: 24, 26, 28.

Solving Quadratic Equations by the Quadratic Formula

We can also use the quadratic formula to find solutions to a quadratic equation. The quadratic formula is:

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Example 1:

Solve the quadratic equation:

$2x^{2}-7x+1=0$

For the equation, $a = 2$ , $b = -7$ and $c = 1$ .

We substitute these numbers into the formula above:

$x=\dfrac{7\pm \sqrt{49-8}}{4}$

We obtain $x = \frac{7- \sqrt{41}}{4}$

$= 0.49$ (3 significant figures)

Or, $x = \frac{7 + \sqrt{41}}{4}$

$= 3.35$ (3 significant figures)

Example 2:

Solve the equation: $\frac{x-5}{x-2}+1=\frac{3x}{x-1}$

1. Simplify the left-hand side:

$\frac{x-5}{x-2}+1=\frac{x-5+x-2}{x-2}$

$=\frac{2x-7}{x-2}$

2. Cross-multiply:

$\frac{2x-7}{x-2}=\frac{3x}{x-1}$

$(2x-7)(x-1)=3x(x-2)$

3. Expand and transpose:

$2x^2-9x+7=3x^2-6x$

$x^2-3x-7=0$

4. Solve using the quadratic formula:

$x=\dfrac{-3\pm \sqrt{9-4\left( -7\right) }}{2}$

$x=\dfrac{-3\pm \sqrt{37}}{2}$

The solutions are $x=\dfrac{-3 - \sqrt{37}}{2} \approx -4.54$ and $x=\dfrac{-3 + \sqrt{37}}{2} \approx 1.54$ , both to three significant figures.

Solving Quadratic Equations by Completing the Squares

Solving a quadratic equation by completing the square is a technique that helps us find the solutions by rewriting the equation in a special form. This method makes it easier to see the relationship between the parts of the equation and identify the solutions.

Example 1:

Solve the equation $x^2 - 6x + 7 = 0$ using the completing the square method.

Step 1: Write the equation in the form $(x - a)^2 = b$

$x^2 - 6x + 7 = 0$

To complete the square, we need to find the square of half the coefficient of the x term. In this case, half the coefficient of x is -3, and the square of -3 is 9. Therefore, we add and subtract 9:

$x^2 - 6x + 9 - 9 + 7 = 0$

Now, we can rewrite the left side as a perfect square:

$(x - 3)^2 - 2 = 0$

Step 2: Solve for x

Add 2 to both sides:

$(x - 3)^2 = 2$

Take the square root of both sides:

$x - 3 = \pm \sqrt{2}$

Add 3 to both sides:

$x = 3 \pm \sqrt{2}$

So, the solutions for x are:

$x = 3 + \sqrt{2}$ and $x = 3 - \sqrt{2}$

Example 2:

Solve the equation $2x^2 - 8x + 5 = 0$ using the completing the square method.

Step 1: Divide the entire equation by the coefficient of the $x^2$ term, in this case, 2.

$\frac{2x^2}{2} - \frac{8x}{2} + \frac{5}{2} = 0$

$x^2 - 4x + \frac{5}{2} = 0$

Step 2: Write the equation in the form $(x - a)^2 = b$

To complete the square, we need to find the square of half the coefficient of the x term. In this case, half the coefficient of x is -2, and the square of -2 is 4. Therefore, we add and subtract 4:

$x^2 - 4x + 4 - 4 + \frac{5}{2} = 0$

Remember that $\frac{5}{2} = 2.5$

Now, we can rewrite the left side as a perfect square:

$(x - 2)^2 - \frac{3}{2} = 0$

Step 3: Solve for x

Add $\frac{3}{2}$ to both sides:

$(x - 2)^2 = \frac{3}{2}$

Take the square root of both sides:

$x - 2 = \pm \sqrt{\frac{3}{2}}$

Add 2 to both sides:

$x = 2 \pm \sqrt{\frac{3}{2}}$

So, the solutions for x are:

$x = 2 + \sqrt{\frac{3}{2}}$ and $x = 2 - \sqrt{\frac{3}{2}}$

Applications of Quadratic Equations

Quadratic equations can be used to solve real-world problems. Let’s look at two examples of this:

Example 1:

A solid cuboid of length x metres has a square cross-section of side $(x-3)$ metres.

Write down, in terms of x, expressions for:

i) The surface area of the square cross-section.

ii) The surface area of one of the longer faces of the cuboid.

Given that the total surface area of the cuboid is 39 square metres, write down an equation in x and show that it reduces to:

$2x^{2}-8x-7=0$

Find the two values of x which satisfy this equation, giving your answers correct to three significant figures. Hence, write down the dimensions of the cuboid.

i) Since the cross-section is a square with side $(x-3)$ meters, its surface area will be:

$A_1 = (x-3)^2$

You can also simplify: $(x-3)^2 = x^2 - 6x + 9$

ii) The surface area of one of the longer faces of the cuboid:

A longer face of the cuboid has a length of x meters and a width of $(x-3)$ meters. Therefore, the surface area of one of the longer faces will be:

$A_2 = x(x-3)$

You can also simplify: $x(x-3) = x^{2}-3x$

Now, we are given that the total surface area of the cuboid is 39 square metres. A cuboid has 6 faces: 2 square cross-sectional faces and 4 longer faces. Using the expressions we found for $A_1$ and $A_2$ , we can write down the equation for the total surface area:

$2A_1 + 4A_2 = 39$

Substitute the expressions for $A_1$ and $A_2$ :

$2(x-3)^2 + 4x(x-3) = 39$

Now, let’s expand and simplify this equation:

$2(x^2 - 6x + 9) + 4x^2 - 12x = 39$

$2x^2 - 12x + 18 + 4x^2 - 12x = 39$

Combine like terms:

$6x^2 - 24x + 18 = 39$

Now, subtract 39 from both sides to set the equation to 0:

$6x^2 - 24x - 21 = 0$

To get the equation in the form of $2x^{2}-8x-7=0$ , we can divide both sides by 3:

$2x^2 - 8x - 7 = 0$

Now, we will solve for x. This quadratic equation does not easily factor, so we will use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 2$ , $b = -8$ , and $c = -7$ . Plugging these values into the formula, we get:

$x = \frac{8 \pm \sqrt{(-8)^2 - 4(2)(-7)}}{2(2)}$

$x = \frac{8 \pm \sqrt{64 + 56}}{4}$

$x = \frac{8 \pm \sqrt{120}}{4}$

Now, we have two possible values for x:

$x_1 = \frac{8 + \sqrt{120}}{4}$

$x_2 = \frac{8 - \sqrt{120}}{4}$

Now, let’s simplify further:

$x_1 \approx \frac{8 + 10.954}{4} \approx \frac{18.954}{4} \approx 4.738$

$x_2 \approx \frac{8 - 10.954}{4} \approx \frac{-2.954}{4} \approx -0.739$

So, the two values of x which satisfy the equation $2x^2 - 8x - 7 = 0$ are approximately $x_1 \approx 4.738$ and $x_2 \approx -0.739$ .

However, since x represents the length of the cuboid, a negative value is not physically meaningful in this context. Therefore, we can disregard the negative value and conclude that the valid value for x is approximately 4.738 metres.

The length of the cuboid is given by x:

Length ≈ 4.738 meters

The square cross-section has a side length of $(x-3)$ metres:

Width = Height ≈ (4.738 – 3) metres ≈ 1.738 metres

So, the dimensions of the cuboid are approximately:

Length: 4.738 metres

Width: 1.738 metres

Height: 1.738 metres

Example 2:

A train takes x hours to travel between London and York, which are 320 km apart.

(a) Write an expression, in terms of x, for the average speed of the train in km/h.

(b) A car takes $2 \frac{1}{2}$ hours longer than a train to travel between London and York. The average speed of the train is 80 km/h greater than the average speed of the car. Form an equation in terms of x and show that it simplifies to $2x^{2}+5x-20=0$ .

(c) Solve the equation $2x^2+5x-20=0$ , giving your answer correct to two decimal places.

(d) Find the average speed of the car correct to the nearest km/h.

(a) To find the average speed of the train, we can use the formula:

Average Speed = Distance / Time

The distance between London and York is given as 320 km, and the time taken by the train is given as x hours. Therefore, the average speed of the train in km/h is:

$= \frac{320}{x}$ km/h

(b) Now, let’s consider the car. The car takes $2 \frac{1}{2}$ hours longer than the train to travel between London and York:

Time (car) = x + 2.5 hours

The average speed of the train is 80 km/h greater than the average speed of the car. Let’s denote the average speed of the car as S (car). Then, the average speed of the train can be expressed as:

Average Speed (train) = S (car) + 80

Using the formula for average speed, we can write the expressions for both the car and the train:

S (car) = $\frac{320}{x + 2.5}$

S (train) = $\frac{320}{x}$

Now, we can write the relationship between the average speeds of the train and the car:

$\frac{320}{x} = \frac{320}{x + 2.5} + 80$

To solve for x, we’ll first get rid of the fractions by multiplying both sides of the equation by x(x + 2.5):

$x(x+2.5) \left(\frac{320}{x}\right) = x(x+2.5) \left[\frac{320}{x+2.5} + 80\right]$

Then we simplify:

$320(x + 2.5) = 320x + 80x(x + 2.5)$

Now, let’s expand and simplify the equation:

$320x + 800 = 320x + 80x^2 + 200x$

We can subtract 320x from both sides:

$800 = 80x^2 + 200x$

Now, let’s divide the entire equation by 40 to simplify it further:

$20 = 2x^2 + 5x$

Finally, subtract 20 from both sides to set the equation to 0:

$2x^2 + 5x - 20 = 0$

(c) To find the values of x that satisfy this equation, you can either use factoring (if possible), the quadratic formula, or numerical methods. In this case, the equation does not factor easily, so we will use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, a = 2, b = 5, and c = -20. Plugging these values, we get:

$x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-20)}}{2(2)}$

$x = \frac{-5 \pm \sqrt{25 + 160}}{4}$

$x = \frac{-5 \pm \sqrt{185}}{4}$

Now, we have two possible values for x:

$x_1 = \frac{-5 + \sqrt{185}}{4}$

$x_2 = \frac{-5 - \sqrt{185}}{4}$

These are the two values of x that satisfy the equation $2x^2 + 5x - 20 = 0$ . However, since x represents the time taken by the train in hours, a negative value is not physically meaningful in this context. Therefore, we can disregard the negative value and conclude that the valid value for x is:

$x \approx \frac{-5 + \sqrt{185}}{4}$

$x \approx \frac{-5 + 13.60}{4}$

$x \approx \frac{8.60}{4}$

$x \approx 2.15$ hours

So, the train takes approximately 2.15 hours to travel between London and York.

(d) We have already found the time taken by the train (x) to be approximately 2.15 hours. The car takes $2 \frac{1}{2}$ hours longer than the train to travel between London and York. So, the time taken by the car can be calculated as follows:

Time (car) = x + 2.5 hours ≈ 2.15 + 2.5 ≈ 4.65 hours

Now, we can find the average speed of the car using the formula:

Average Speed = Distance / Time

The distance between London and York is 320 km. Therefore, the average speed of the car in km/h is:

Average Speed (car) = 320 km / 4.65 hours ≈ 68.82 km/h

Rounded to the nearest km/h, the average speed of the car is approximately 69 km/h.

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