Quadratic Formula

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The Quadratic Formula:

The quadratic formula is a formula that you can substitute values into in order to find the solutions to any quadratic equation. This is just one method of solving quadratics, you will encounter more throughout the course.

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The Quadratic Formula

The quadratic formula is a formula that you can substitute values into in order to get the solutions to any quadratic equation.

x=−b±b2−4ac2ax=dfrac{-textcolor{red}{b}pmsqrt{textcolor{red}{b}^2-4textcolor{blue}{a}textcolor{Orange}{c}}}{2textcolor{blue}{a}}

The atextcolor{blue}{a}, btextcolor{red}{b} and ctextcolor{Orange}{c} values correspond to values in a quadratic equation, shown below

ax2+bx+c=0textcolor{blue}{a}x^2+textcolor{red}{b}x+textcolor{Orange}{c}=0

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Take note:

Notice that there is a “plus or minus” symbol in there (±){textcolor{blue}{(pm)}}.

This is because a quadratic has up to two real solutions – putting a plus sign there will give you one solution and putting a minus sign there will give you the other. In other words, the two solutions are

x=−b+b2−4ac2a       x=dfrac{-btextcolor{blue}{+}sqrt{b^2-4ac}}{2a} ,,,,,,, and        x=−b−b2−4ac2a,,,,,,, x=dfrac{-btextcolor{blue}{-}sqrt{b^2-4ac}}{2a}

You are not given this formula in an exam so you do have to memorise it.

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Example 1: Quadratic where a=1a=1

Use the quadratic formula to solve the following quadratic equation:

x2+2x−35=0x^2+2x-35=0

[2 marks]

Firstly, we have to identify what a,b,a,b, and cc are:

a=1a=1, b=2b=2, c=−35c=-35

Next we need to substitute these into the formula:

x=−2±22−4×1×(−35)2x=dfrac{-2pmsqrt{2^2-4times1times(-35)}}{2}

Simplifying this we get

x=−2+1442x=dfrac{-2+sqrt{144}}{2}, x=−2−1442x=dfrac{-2-sqrt{144}}{2}

We know 144=12sqrt{144}=12, so the two final solutions are

x1=−2+122=5x_1=dfrac{-2+12}{2}=5, x2=−2−122=−7x_2=dfrac{-2-12}{2}=-7

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Example 2: Quadratic where a>1a>1

Use the quadratic formula to solve the following quadratic equation:

2x2−6x+3=02x^2-6x+3=0

Give your answer to 22 decimal places.

[2 marks]

a=2a=2, b=−6b=-6, c=3c=3

Putting these into the formula, we get

x=−(−6)±(−6)2−4×2×32×2x=dfrac{-(-6)pmsqrt{(-6)^2-4times2times3}}{2times2}

so, the solutions are

x=6+124x=dfrac{6+sqrt{12}}{4} and x=6−124x=dfrac{6-sqrt{12}}{4}

1212 is not a square number, which is how we know this won’t give us a nice answer.

The question asks for 22dp, so putting these into the calculator, we get

x1=2.366…=2.37x_1=2.366…=2.37 (2dp)(2dp)  and x2=0.6339…=0.63x_2=0.6339…=0.63 (2dp)(2dp)

Note: You can put the first quadratic formula straight into the calculator without any simplifying and use the + and – to get your two answers.

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Quadratic Formula Example Questions

Question 1: Use the quadratic formula to find the solutions to x2+11x+16=0x^2+11x+16=0 to 3 significant figures.

[2 marks]

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Firstly, the quadratic formula is

 

x=−b±b2−4ac2ax=dfrac{-bpmsqrt{b^2-4ac}}{2a}

 

Then, we can identify that here, a=1,b=11,a=1, b=11, and c=16c=16. Putting these values into the formula, we get

 

x=−11±112−4×1×162x=dfrac{-11pmsqrt{11^2-4times1times16}}{2}

 

The part inside the square root is

 

112−4×1×16=121−64=5711^2-4times1times16=121-64=57

 

So, the solutions become

 

x=−11±572x=dfrac{-11pmsqrt{57}}{2}

 

Putting these into a calculator (one with +, one with -), we get the final solutions

 

x=−1.7250…=−1.73 (3sf), and  x=−9.2749…=−9.27 (3sf)x=-1.7250…= -1.73text{ (3sf), and },x=-9.2749…=-9.27text{ (3sf)}

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Question 2: Use the quadratic formula to find the solutions to x2−2x=44x^2-2x=44 to 3 significant figures.

[2 marks]

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Firstly, the quadratic formula is

 

x=−b±b2−4ac2ax=dfrac{-bpmsqrt{b^2-4ac}}{2a}

 

Then, we can identify that here, a=1,b=−2,a=1, b=-2, and c=−44c=-44. Putting these values into the formula, we get

 

x=−(−2)±(−2)2−4×1×(−44)2x=dfrac{-(-2)pmsqrt{(-2)^2-4times1times(-44)}}{2}

 

The part inside the square root is

 

(−2)2−4×1×(−44)=4+176=180(-2)^2-4times1times(-44)=4+176=180

 

So, the solutions become

 

x=2±1802x=dfrac{2pmsqrt{180}}{2}

 

Putting these into a calculator (one with +, one with -), we get the final solutions

 

x=7.7082…=7.71 (3sf), and  x=−5.7082…=−5.71 (3sf)x=7.7082…=7.71text{ (3sf), and },x=-5.7082…=-5.71text{ (3sf)}

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Question 3: Use the quadratic formula to find the solutions to 4x2+7x−1=04x^2+7x-1=0 to 2 decimal places.

[2 marks]

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Firstly, the quadratic formula is

 

x=−b±b2−4ac2ax=dfrac{-bpmsqrt{b^2-4ac}}{2a}

 

Then, we can identify that here, a=4,b=7,a=4, b=7, and c=−1c=-1. Putting these values into the formula, we get

 

x=−7±72−4×4×(−1)2×4x=dfrac{-7pmsqrt{7^2-4times4times(-1)}}{2times4}

 

The part inside the square root is

 

72−4×4×(−1)=49+16=657^2-4times4times(-1)=49+16=65

 

So, the solutions become

 

x=−7±658x=dfrac{-7pmsqrt{65}}{8}

 

Putting these into a calculator (one with +, one with -), we get the final solutions

 

x=0.1327…=0.13 (2dp), and  x=−1.8827…=−1.88 (2dp)x=0.1327…=0.13text{ (2dp), and },x=-1.8827…=-1.88text{ (2dp)}

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Question 4: Use the quadratic formula to find the solutions to x2+8x=−13x^2+8x=-13 to 2 decimal places.

[2 marks]

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Firstly, the quadratic formula is

 

x=−b±b2−4ac2ax=dfrac{-bpmsqrt{b^2-4ac}}{2a}

 

Then, we can identify that here, a=1,b=8,a=1, b=8, and c=13c=13. Putting these values into the formula, we get

 

x=−8±82−4×1×132×1x=dfrac{-8pmsqrt{8^2-4times1times13}}{2times1}

 

The part inside the square root is

 

82−4×8×13=64−52=128^2-4times8times13=64-52=12

 

So, the solutions become

 

x=−8±122x=dfrac{-8pmsqrt{12}}{2}

 

Putting these into a calculator (one with +, one with -), we get the final solutions

 

x=−2.2679…=−2.27 (2dp), and  x=−5.73205…=−5.73 (2dp)x=-2.2679…=-2.27text{ (2dp), and },x=-5.73205…=-5.73text{ (2dp)}

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Question 5: Use the quadratic formula to find the solutions to 25x2−30x+7=025x^2-30x+7=0, giving your answer in surd form.

[3 marks]

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Firstly, the quadratic formula is

 

x=−b±b2−4ac2ax=dfrac{-bpmsqrt{b^2-4ac}}{2a}

 

Then, we can identify that here, a=25,b=−30,a=25, b=-30, and c=7c=7. Putting these values into the formula, we get

 

x=−(−30)±(−30)2−4×25×72×25x=dfrac{-(-30)pmsqrt{(-30)^2-4times25times7}}{2times25}

 

The part inside the square root is

 

302−4×25×7=900−700=20030^2-4times25times7=900-700=200

 

So, the solutions become

 

x=30±20050x=dfrac{30pmsqrt{200}}{50}

 

Putting these into a calculator (one with +, one with -), we get the final solutions,

 

x=3±25x=dfrac{3pmsqrt{2}}{5}

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Specification Points Covered

Algebra – 18. solve quadratic equations (including those that require rearrangement) algebraically by factorising, by completing the square and by using the quadratic formula; find approximate solutions using a graph

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