## What you need to know

There are 3 ways to solve a quadratic: factorisation (solving-quadratics-factoring), completing the square (completing square revision), and **the quadratic formula**. The quadratic formula is exactly what it sounds like – a formula that you can substitute a bunch of values into in order to get the solutions to any quadratic equation. So, suppose you have the general quadratic equation

ax^2+bx+c=0

Then, the solutions to this quadratic equation are given by

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

Notice that there is a “plus or minus” symbol in there (\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=\dfrac{-b+\sqrt{b^2-4ac}}{2a}\,\,\text{ and }\,\,x=\dfrac{-b-\sqrt{b^2-4ac}}{2a}

You are **not given **this formula in an exam so you do have to **memorise it**. The more you practice using it, the more it’ll stick in your head. Let’s see how this works with an example.

**Example: **Use the quadratic formula to find the solutions to x^2+2x-35=0.

Firstly, we have to identify what a,b, and c are in this case. There is no number in front of x^2, so its coefficient is 1, thus: a=1. The coefficient of x is 2, so b=2. The constant term is -35, so c=-35. Putting them into the formula, we get

x=\dfrac{-2\pm\sqrt{2^2-4\times1\times(-35)}}{2}

Then, the portion inside the square root becomes

2^2-4\times1\times(-35)=4-(-140)=144

Therefore, we get

x=\dfrac{-2+\sqrt{144}}{2}\,\,\text{ and }\,\,x=\dfrac{-2-\sqrt{144}}{2}

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

x=\dfrac{-2+12}{2}=5\,\,\text{ and }\,\,x=\dfrac{-2-12}{2}=-7

This particular quadratic could’ve been relatively easily factorised, so you can check that answers are correct. However, this also makes it not that useful in this particular case. That said, the quadratic formula *is* useful when the solutions are unpleasant decimals, so factorising is near impossible.

**Example: **Use the quadratic formula to find the solutions to 2x^2-6x+3=0 to 2dp.

You can try to factorise this is you’d like, but you’ll be there for a while. So, in this quadratic, we have that the coefficient of x^2 is 2, so a=2. The coefficient of x is -6, so b=-6. The constant term is 3, so c=3. Putting these into the formula, we get

x=\dfrac{-(-6)\pm\sqrt{(-6)^2-4\times2\times3}}{2\times2}

Inside the square root, we have

(-6)^2-4\times2\times3=36-24=12,

so, the solutions are

x=\dfrac{6+\sqrt{24}}{4}\,\,\text{ and }\,\,x=\dfrac{6-\sqrt{24}}{4}

18 is not a square number, which is how we know this won’t give us a nice answer. The question asks for 2dp, so putting these into the calculator, we get

x=2.366...=2.37\text{ (2dp), and }\,x=0.6339...=0.63\text{ (2dp)}

Once you have this formula memorised, and you remember what a,b, and c must be, it isn’t too difficult to use. Get practising, it’s one of the best ways to make sure you remember the formula.

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### Example Questions

1) Use the quadratic formula to find the solutions to x^2+11x+16=0 to 3sf.

Firstly, the quadratic formula is

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

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

x=\dfrac{-11\pm\sqrt{11^2-4\times1\times16}}{2}

The part inside the square root is

11^2-4\times1\times16=121-64=57

So, the solutions become

x=\dfrac{-11\pm\sqrt{57}}{2}

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

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

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2) Use the quadratic formula to find the solutions to 4x^2+7x-1=0 to 2dp.

Firstly, the quadratic formula is

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

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

x=\dfrac{-7\pm\sqrt{7^2-4\times4\times(-1)}}{2\times4}

The part inside the square root is

7^2-4\times4\times(-1)=49+16=65

So, the solutions become

x=\dfrac{-7\pm\sqrt{65}}{8}

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

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

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