in this video we're looking at modulus inequalities we're going to look at four examples the first two are using an equal signs that are a little bit easy in nature and then they get a little bit harder in example three and four just a little bit of revision of what modulus is so these parallel lines are standing for a modulus so if I'm looking for the modulus of say um five it basically turns your value into its absolute value so it's positive value so the modulus of five is five now that could have also been the case the modulus of negative five which basically means you turn your value into a positive value so five so it's usually got to do with maybe areas of triangles you might have seen the modulus or the values for lengths of um coordinate geometry so that's what the modulus is doing for us so if we had another example here the modulus of X is equal to 3 technically the values for X there are negative 3 because the modulus of that is three but I could also use x as positive 3 because the modulus of positive 3 obviously has to be tree so there's two values there for X now the easiest way to find the values for X here would be to square both sides so what I mean by that is if we keep our basic example here of the modulus of X is equal to 3. so in order to find those two values of X the easiest way is to just come along and square both sides because when we Square the X we get x squared and we square root a 3 and we get nine and the square root of nine is plus or minus 3. so that's how we're going to solve these one two three four examples of modulus we're going to square both sides and solve for x basically so let's have a look now at example one so it's asking us to solve the inequality for X an element of the real number so real numbers are positive and negative uh whole or rational number so fractions and decimals so the method of doing this is to square both sides so I'm going to go x minus 2 all to be squared is equal to 4 to be squared multiplying x minus 2 by itself is my next step because the square means you multiply it by itself and 4 squared there will give you 16. so first term by my second bracket and second term by my second bracket is equal to 16. and multiplying in my bracket stand so first term by second term and so on is going to give me x squared minus 2x minus 2X and negative 2 by negative 2 will give me a plus 4 is equal to 16. if I move that 16 over or subtract 16 from both sides I'm getting x squared minus 4X um plus 4 minus 16 is equal to zero so my quadratic is x squared minus 4X minus 12 is equal to 0. so I'm trying to find the values of X so I need to solve my quadratic here so you can use your guide number trial and error using your two brackets or your minus B formula your quadratic formula I'm going to just do the two brackets here I think it should be straightforward enough um so it's equal to zero so the multiples of minus 12 to get to minus 4 are going to be minus six and a plus 2. so there are my factors and when I solve those I get X is equal to 6 and X is equal to minus 2. so therefore my two values for X are positive six and negative two and you can always check these by subbing them back in to your modules so that's example one let's have a look there at example two so solve the following inequality for X an element of r x minus five modulus is equal to the modulus of X plus one so the process once again is to square both sides so it's x minus five to be squared is equal to X plus one two B squared squaring something means you multiplied by itself so x minus five multiplied by x minus five is equal to X plus one times X plus one so multiplying out your bracket so first term by the second bracket and second term by the second bracket on both sides of the equal so it's a little bit Ultra heavy here already so just take your time as you multiply out your brackets so multiplying out the first two you get x squared minus five x minus five x minus five by minus five is a positive 25 and on the right hand side of the equals we get One X Plus One X plus one by one which is one tidying it up a little bit x squared minus 10x plus 25 is equal to x squared plus 2X plus 1. so to solve a quadratic we need to move everything to one side so I'm basically just going to tidy this up a little bit and move everything to the left of my equals so I have x squared minus 10x plus 25 and then I'm moving over the x squared to become a negative x squared negative 2x plus 1 will change to a negative one and that is now going to equal to zero on the right hand side hopefully you can spot here that the x squared minus x squared will cancel out so that's leaving me with minus 10 x minus 2x is minus 12x plus 25 minus 1 is plus 24 is equal to zero I need to solve that linear equation now so let me just rewrite it up here so I'm going to have x's on one side and numbers on the other so subtract 24 from both sides we'll leave him with minus 12x is equal to negative 24 and dividing across by negative 12 will give me a value of x as positive 2. so there's only one value for x on example two and that's X is equal to 2. looking at example tree next solved following inequality for X an element of or so it's now getting into a greater than symbol rather than the equals but we carry these out in the same fashion we Square both sides so it's x minus four all to be squared greater than 3 to be squared squaring something once again means you multiply it by itself so x minus 4 by x minus 4 is greater than 3 trees which is nine so first term by my second bracket second term by my second bracket greater than 9 which is giving me x squared minus four x minus 4X plus 16 is greater than 9. so just tidying up this now a little bit is giving me minus eight X plus 16 and if I subtract the 9 from both sides or move over that nine that's leaving me with negative 9 greater than zero because we can only solve our quadratic equations when it is um greater than or less than or equal to zero so that's my quadratic equation now I'm going to come up and solve my quadratic so again guide number minus B formula two brackets whichever way you want so what we're trying to do first of all now is find the roots of the equation so I'm going to let it equal to zero here just to solve the roots because I'm looking for where it crosses the x-axis so when that's equal to zero so my two brackets x and x and equal to zero and the multiples of seven are seven and one and a minus and a minus because minus by minus gives me plus so the roots are X is equal to seven and X is equal to positive one so what does that mean so if we come back if we come down here just do a little sketch of this and it's crossing the x-axis at number one and number seven and it's a positive quadratic so maybe it looks something like that so we're looking for where it's greater than zero so greater than zero is above the x-axis so it's this portion of my graph here and this portion of my graph here so the values for X here are going to be X greater than seven so I could use seven eight nine ten eleven all the way to Infinity so anything basically in this yellow portion here but I could also go down to anything below one so X less than one because I can include zero minus one minus two minus three all these values here the only values I can't use are anything between 1 and 7. um so that's example three Dawn let's have a look now at example four solved following inequality for X an element of the real numbers so again we have an inequality here modulus on both sides so we need to square both sides um to deal with the positive and negative of that inequality so that's going to give me X Plus 1 on the left hand side and I'm going to square it out and that's going to be greater than uh 2 times X plus 3 all to be squared so what you're doing here is you'll have to square the 2 and then you'll have to square the bracket so we can write that out if it makes it easier so it's X Plus 1 to be squared greater than at 2 to be squared and the X plus 3 to be squared so I'm going to do them separately so on the left hand side that'll give me X Plus 1 times X Plus 1 and on the right hand side two twos are 4 and then I need to multiply out my two brackets so I'm just going to use square brackets here just to keep them separately and I need to times them by each other as well so first term by my second bracket and then second term by my second bracket greater than 4 times the x times the second bracket and then the plus 3 by the X Plus 3. so just take your time with your multiplication here now I'm just going to multiply in my bracket here so uh use the arrows if it helps so that's going to give me X by X is x squared plus 1X Plus 1X Plus 1. greater than 4 times x by X is x squared 3x 3 by X is 3x and 3 by 3 is 9. just timesing in that 4 now into this bracket multiplying it in is going to give me uh greater than or equal to 4 x squared well that 3 plus 3 is 6 and 6 by 4 is 24. X and four nines is 36. and on the left I have my x squared plus 2X Plus 1. tidying that up now I'm going to move everything to the left hand side so uh minus all of this quadratic in other words so that's going to give me x squared plus 2X plus 1 and as I move them over let me just change the colors here so that'll be minus 4x squared minus 24x minus 36 and that will all be greater than zero so grouping terms I have x squared minus 4x squared will give me a negative 3x squared I have a 2X minus a 24 is a minus 22x and then I have my 1 minus 35 which or 36 is minus 35 greater than or equal to zero so make it a little bit easier I'm going to multiply across by negative 1 to change the signs so that's going to give me positive 3x squared positive 22x and a positive 35 but remember when you multiply across or divide by a minus you have to change the direction of your inequality sign so my inequality sign there is flipping so now I'm looking to factorize this so using your minus B formula if it's a little bit easier or you can use your um your two brackets uh either way uh let's see what we get so the factors of 35 that are multiplying to 35 and adding to 22 we could have a 5 and a 7 and positive and positive so that would give me 15 plus 7 is 22. yeah so there are my Solutions so they're my factors solving these now I get 3x is equal to minus seven X is equal to minus seven over three or two and one-thirds and my second root is going to be X is equal to negative five so there are my roots where it crosses the axis so technically there if when I'm solving that I'm going to have that equal to a zero so let's see the question again wants us to solve the inequality so there's a range of values here so I'm just going to sketch this out once again and see how it looks so minus 5 would be down here minus two and a third probably around here so minus seven over three and it's a positive quadratic so just sketching it looks something like this and if I come back up to the question here the quadratic that I was solving was less than zero so I'm looking at this portion here where it's below my x-axis so the values of X below the x axis would be X um greater than minus five and less than minus seven over three so a combined inequality there would be between minus seven over three and negative five so that's my final solution there so that's example for them so that's four examples there of the modulus inequality ranging in difficulty