Solving quadratic equations

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Quadratic equations

A quadratic equation involves $x^2$ and when we look to solve an equation of this sort we need to find the value of the unknown x. Because the unknown could really be any value, trial and error is not a very good method to use as it will take a huge amount of time. Therefore, we need a method to use in order to solve an equation of the type $x^2+bx+c=0$ .

By realising that two unknowns when multiplied together equals 0, one of the unknowns must be 0 itself. This is because the only way for $a \times b=0$ to work is for either a or b to be 0 (or both of a and b to be 0). Therefore, by factorising the equation into two different brackets multiplied together, we can see that one or both of these brackets must equal 0.

Example

Solve the following equations:

1) $(x+4)(x-2)=0$

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

1) This equation has two distinct parts which are multiplied to get 0. This means that one of the two brackets MUST be equal to zero. So we can now just work out the value of x in either case. $(x+4)=0$ will tell us that tells us that $x=-4(x-2)=0$ . So the answer for the unknown is $x=2$ or $-4$ .

2) For this equation we must first factorise into double brackets using the method that we have already seen in this lesson. To get a product of 7 we must have the pair of numbers 7 and 1, of which one must be negative since we really have $-7$ . Then for the two numbers to add to 6 the 7 must be positive and the 1 negative, this give us $(x+7)(x-1)=0$

Now we can solve by setting either bracket to equal 0.

$(x+7)=0$ means that $x=-7$

$(x-1)=0$ means that $x=1$

So $x=-7$ or $1$

Whenever we get an equation that contains $x^2$ we should always look to make it into the same form as $x^2+bx+c$ .

This means that if we have something like $x^2+10$ this is really the same as $x^2+0x+10$ when in the correct form but we just avoid writing the 0 term. This will of course help us when we come to solving the equation as we can use the same method as before except the pair of numbers needed must add to 0.

The same can be done for something like $x^2+5x$ . We can put this as $x^2+5x+0$ and see that this must mean the pair of numbers needed must multiply to 0, so from what we have already stated earlier, one of them must be equal to 0 itself! This would be factorised into $(x+0)(x+5)$ which is the same as $x(x+5)$ .

Example

Solve the following quadratic equations for the unknown value of x.

1) $x^2-16=0$

2) $x^2+19x=0$

3) $2x^2-50=0$

1) Here we are missing a value for in the equation so the must be 0. This leaves us with $x^2+0x-16=0$ . So we need a pair of numbers that add to 0 and multiply to –16. This has to be 4 and –4 as there are no other options. So we are left with $(x+4)(x-4)=0$

This then gives us two values for when either bracket is equal to zero. So we have an answer of $x=4$ or $-4$

2) Putting this into the equation we have $x^2+19x+0=0$ So we need a pair of numbers which add to 19 but multiply to 0. Clearly one of the pair must be 0, leaving the other one as 19. This gives $(x+0)(x+19)=0$ and means that the unknown value must be $x=0$ or $-19$

3) With this equation we can immediately divide the entire thing by 2 so that we have $x^2$ instead of $2x^2$ . This then gives us $x^2-25=0$ which is the same as $x^2+0x-25=0$ . So we require a pair of numbers to add to 0 and multiply to 25. This means we must use 5 and $-5$ . Leaving us with $(x+5)(x-5)=0$ and then by solving the two individual brackets equal to 0 we are left with $x=5$ or $-5$ .