Solving Math Equations but they keep getting HARDER | jensenmath.ca

watch me solve 11 equations where each one gets more difficult than the previous one let me know in the comments how far you could get through these 11 equations here's the first equation when solving any of these equations we're really just trying to find out what value of x makes the left side of the equation equal to the right side we're starting off with a really simple linear equation where we can just use inverse operations to isolate the X and then that'll be the answer to the equation let me start by rewriting the question I'll try and isolate the term that has the X by getting rid of the minus 7 by doing the inverse of subtracting seven which is adding seven I'll make sure I add seven to both sides of the equation I do that because on the left - 7 + 7 cancels out I'm left with 3x = 18 and then to isolate the X I'll do the inverse of multiplying by 3 which is dividing by 3 I make sure I do that to both sides so it stays balanced three is on the left left cancel and I get my final answer of x = 18 3 which is 6 here's question number two this is another linear equation so not too difficult yet you might be tempted to try and distribute this 1/3 but I think it'll be easier if what I do is try and get rid of that fraction right away I'm going to multiply both sides of this equation by three the left by three and the right by three the reason I do that is because on the left I have a 3 ID 3 which cancels so now I have no fractions left on the left side of the equation I just have 1 * x + 9 which is x + 9 on the right I have 3 * x + 1 I'll distribute on the left side to get 3x + 3 now I want to get both of the variable terms on one side of the equation and both of the constant terms on the other side of the equation I'll bring the constant terms to the left so I have 9 - 3 on the left the variable terms on the right I have 3x - 1X 9 - 3 is 6 3x - 1X is 2x I'll isolate the X by doing the inverse of multiplying by two which is dividing by two so I have 6 / 2 = x so X is 3 here's our third equation this equation is quadratic so we may get up to two answers to this equation when solving any quadratic it's usually good to set it to zero first so I'll start by moving that 12 to the other side so I have x^2 + 4x - 12 = 0 I'll now solve this quadratic by factoring I notice that the a value is one so to factor it I just need to find numbers that multiply to the C value of -12 and add to the B value of four the numbers that satisfy that product and sum are six and and -2 so that quadratic factors to x + 6 * x - 2 now that it's in factored form I can use the zero product rule that tells me the product of two factors is zero if either of the two factors are zero so if x + 6 was Zero the whole product would be zero or if x- 2 was 0 the whole product would be zero x + 6 would be zero if x was -6 and x minus 2 would be zero if x was 2 so we have two solutions to this equation -6 and two here's equation number four once again this is a quadratic equation so I'll start by setting it equal to zero now I'm going to want to get this into factored form if possible I'll start by taking out a common factor of two when I divide all three terms by two I get 5 y^2 - 8 y + 3 now the quadratic in the brackets the a value is not one it's five so if this is factorable I'm going to have to factor it the Long Way by decomposition I need to find numbers that multiply to a * C 5 * 3 which is 15 and add to the B value of8 the numbers that satisfy that product and sum are -5 and -3 so what I can do is split the middle term 8 y split it into -5 Yus 3 Y and now I continue factoring by grouping I take a common factor from the first two terms I could common factor a 5 Y and then have y -1 and then from the last two terms I could common factor a -3 and then have y - 1 again notice the common binomial of y - 1 I could common factor out that common binomial and once I remove that common binomial as a factor I'm left with 5 Yus 3 that's now fully factored using the zero product rule I know the product would be zero if any of the factors would be zero y- - 1 would be Z if y was 1 and 5 y - 3 would be Z if 5 y was 3 and if y was then 3 over5 there's my second answer here's the fifth equation this is our first exponential equation notice I have two Powers equal to each other that have different bases but I can rewrite the base of the power on the right eight so that it has the same base as the power on the left so I can rewrite eight as a 2 Cubed on the right I now have a power of a power when you have two exponents on top of each other you can just find their product so it's 2 ^ of 3 * x - 2 make sure you distribute the 3 to the X and the -2 to get 3x - 6 now that I have power equals power where the bases of those powers are equal I know the only way this equation could be true is if their exponents were equal to each other so I can just set the exponents equal and solve move the 3x to the other side and I'm left with x = -6 there's my solution here's equation six we're at the Midway point we once again have an exponential equation but if we look at the bases it would be very difficult to rewrite those as Powers with the same base so we'll use a different strategy I'm going to take the common logarithm of both sides of this equation I do that so that I can take the power of the argument of the logarithm use the power rule which tells me I can write it as the coefficient of the log this gets my unknown out of the exponent I'll expand both sides of that equation by doing 3x * log 4 and -1 * log 4 on the right x * Log 5 and 2 * Log 5 so I have 3x log 4 - log 4 = x Log 5 + 2 Log 5 I now want to get both terms that have an x on the same side of the equation and both terms that that don't have an x on the other side of the equation I'll bring the terms with the X to the left so on the left I have 3x log 4 - x Log 5 on the right I have 2 Log 5 but remember I could take that coefficient of two and write it as the exponent on the argument of the log so that' be equal to log of 5^2 which is log 25 so I'll just say log 25 bring over the negative log 4 becomes plus log 4 now what I want to do I know both terms on the left have an X so I'll common factor out an X which will give me 3 log 4 minus Log 5 on the right I have two logs being added I can use the product rule of logs which tells me I can write it as a single log by multiplying their arguments 25 * 4 is 100 oh and log 100 what exponent goes on 10 to get 100 that's just two on the left side of the equation I'm missing a bracket there I want to combine those two logarithmic Expressions before I can do that I would have to take that coefficient of three and make it the exponent on the four so 4 cubed is 64 when two logs are being subtracted we write it as a single log by dividing the arguments so it would become log of 64 / 5 I can isolate X just by dividing that logarithm to the other side so I have xal 2 / log of 64 over 5 that's the exact answer to this equation if we want an approximate answer it's about 1.86 here's equation s our first trig equation trig functions are periodic so there's going to be an infinite number of answers to this equation but we'll start just by finding the two answers that are between 0 and 2 pi I need to isolate that sinx first so I'll have sinx equals 1 over2 I see that I have a negative s ratio so I'm going to draw a cartisian grid and figure out where my answers are going to be because my sign ratio is negative I know sign is negative in quadrant 3 and four based on the cast rule so I know I'll get an answer in both of those quadrants I'll draw a terminal arm in both of those quadrants and I know those terminal arms will have the exact same reference angle and if you know your special triangles remember the half equilateral special triangle with side lengths 2 1 and < tk3 and angles pi over 3 pi over 6 notice that s from pi/ 6 right s is opposite over hypotenuse is 1 / 2 so if I place that reference angle pi over 6 in quadrant 3 and 4 I will get a sign ratio of - 1 over2 so I can label this reference angle in both of those quadrants as pi/ 6 and then calculate the principal angle to get to each of those terminal arms my first answer would be P Pi by Pi / 6 and my second answer would be pi/ 6 less than going a full 2 pi so let's do those two calculations Answer 1 pi plus piun / 6 so answer 1 is 7 piun over 6 s of 7 piun over 6 isga half and my second answer was piun 6 less than 2 pi so 2 piun minus < / 6 which is 11 piun over 6 so 7 Pi 6 and 11 Pi 6 are the two answers between 0 and 2 pi like I said there's an infinite number of answers we could find co-terminal angles to both of those by just adding 2 pi to them as many times as we want and we would get more answers to the equation so for both of them I could just say we could add 2 pi K times as long as K is an integer value and I would get more answers there we go that's the seventh one done you can see the difficulty is really starting to ramp up here's the eighth equation this is a polinomial equation to figure out when a polinomial is zero we're going to want to get it into factored form so I have to start by testing for zeros of this function so I can figure out what a factor of this is possible zeros of a polinomial function are factors of the constant term at the end so factors of 18 are plus or minus 1 2 3 6 9 or 18 I just need to find one value that makes that polinomial b0o and then I can divide by its corresponding factor to start getting it into factored form so I'll do a test let me test out1 if I evaluate this I would get 1 + 1 - 11 - 9 + 18 and that equals z so because negative 1 makes that polinomial be zero I know that's one answer to my equation but to get the rest of the answers I should divide that polinomial by the factor that has a z of1 x + 1 has a z of1 so x + 1 is a factor of the polinomial so what I'm going to do is divide that polinomial by x + 1 and I'll do that by synthetic division the zero of the divisor is NE 1 the coefficients of the polinomial are 1 -11 9 and 18 if I run the synthetic division bring down the one multiply by1 add in that column multiply by1 add multiply by 1 add multiply by 1 and I get a remainder of zero and the rest of the numbers are the coefficients of each term in my quotient so I could rewrite the original equation as a product of x + one which was the divisor and the quotient which was X Cub - 2x^2 - 9x + 18 and now I not notice that degree 3 polinomial I have as my second factor I could factor that by grouping I could take a common factor from the first two terms and a common factor from the last two terms and I'll end up with a common binomial I'll take out an X2 from the first two terms and have x - 2 take it to9 from the last two terms and again have x - 2 I have that common binomial of x - 2 when I factor that out I'm left with x^2 - 9 and that x^2 - 9 is a difference of squares it's an x^2 - A 3^ 2 so I can factor that to x - 3 * X +3 now that the polinomial is fully factored I know any number that would make any of those factors be zero is a solution to the equation so looking at my factors from left to right the zeros would be - 1 2 3 and-3 so all of those are answers to the original equation here's the ninth equation this is the first logarithmic equation our unknown is in the argument of a logarithm I notice that I have two logarithms with the same base that are being added together I could combine them together by multiplying their arguments x * x - 4 is x^2 - 4x now this logarithm is the common logarithm if we don't see a base it's 10 and a logarithmic expression is a way of getting the exponent one that goes on the base of the logarithm 10 to make it equal to the argument so I could rewrite this in exponential form as 10 ^ 1 = x^2 - 4x this is now a quadratic equation to solve I'll set it to Zero by moving the 10 over x^2 - 4x - 10 now this quadratic it's not factorable nothing multiplies to -10 and adds to4 so that means if there are zeros there will be irrational zeros and we'll have to get them using quadratic formula quadratic formula uses the a b and c values of the standard form quadratic it' be xal B so 4 plus or minus the Square t of b^ 2 so4 2 - 4 * a * C all divid 2 * a I'll simplify underneath the square root underneath the square root I have 16 + 40 That's 56 this radical expression can be simplified there is a perfect square factor that goes into 56 4 goes into 56 14 times so I can split this into < TK4 * < TK4 but the square root of 4 is just two so I'll rewrite that as 2un 14 now both terms in the numerator being divided by two so I could write this as 4 / 2 is 2 plus or minus 2un / 2 is justun 14 so I have two answers to this equation x = 2 +un 14 and also x = 2 minus < tk14 but when solving a logarithmic equation you have to understand that the argument of a logarithm is restricted to being greater than zero so if either of the answers we got make either of the logarithmic Expressions undefined by making the arguments less than or equal to zero then we call it an extraneous root notice that 2 - < TK4 is going to give us a negative answer so of course that's going to make both of these log Expressions undefined so 2 - < tk14 is an extraneous root so the only real answer to this equation is 2 +un 14 here's the 10th equation only two equations left this is another trig equation but this one is especially difficult because we have two different trig functions in this equation I'll start by using an identity that tells me that secant 2 x is equal to tan^ 2 x + 1 I'll distribute the two to get 2 tan ^ 2 x + 2 - 3 + Tan x = 0 I'll rewrite this in standard form and collect the 2us 3 together I might at this point actually also use a let statement I'll say let Tan x equal K that'll make this equation look easier it's just 2 k^ 2 + K - 1 = 0 so that looks like easy quadratic to solve where I would want numbers that have a product of a * C so a product of -2 and a sum of the B value which is one the numbers that satisfy that product and sum are two and Nega 1 I'll factor that quadratic by decomposition I'll split that K into a 2K minus 1K and then Factor by grouping take it a 2K from the first two terms take out a Nega 1 from the last two terms take out that common binomial of k + 1 and see we actually get two answers for K we get K = -1 and k equal a half but remember those Ks are actually Tan x so we're about halfway done this question now let me make some more room like I said those Ks are actually Tan x so I actually have Tan x = -1 and tan xal a half let me draw a carian grid for both of those if I focus on the equation Tan x = -1 I have a NE tan ratio which happens in quadrants 4 and two so I'll draw a terminal arm in both of those quadrants and if you remember your special triangles I can get a tan ratio opposite over adjacent of 1 over 1 when the reference angle is pi over 4 I know the reference angle in both of those quadrants is going to be the same so the reference angles are both pi over 4 and I know the tan ratio is going to change to negative because it's in two quadrants where tan is negative so my first answer goes from the initial arm to that first terminal arm which is piun / 4 Less Than Pi so I'll say X1 = < - < 4 which is 3 pi over 4 and X2 is pi over 4 less than 2 pi so I'll say X2 2 piun - < over 4 which is 7 piun over4 notice any answers to an equation that involves tan are going to be separated by a difference of Pi right to get from 3 piun 4 to 7 piun 4 you can just add Pi that's because the period of a tan function is pi so I could get infinitely many more answers by continuing to add Pi so I could summarize both of those answers together I could just say xal 3 pi over 4 plus add pi as many times as you want as long as K is an integer value how about the equation tan xal a half that's a positive tan ratio tan is positive in quadrant 3 and 1 I'll draw a terminal alarm in both of those quadrants I know those principal angles are going to have the same reference angle and I can't make a tan ratio of a half from a special triangle so I would use my calculator to find an approximate answer I would do inverse tan of a half and it's about 0.46 so I know that's the reference angle for both of those principal angles so my third answer X3 is 0.46 my fourth answer which goes all the way to here goes past Pi by .46 so Pi + 0.46 it's about 3.60 notice once again the relationship between these answers is that I just add the period of tan which is pi so I could summarize that full set of Answers by saying x equals my first answer 46 plus pi K * where K is an integer here's the 11th equation the very last one the most difficult one this is an exponential equation where you may not know where to start there's actually a hidden quadratic here and you'll only be able to see it if we multiply both sides of this equation by 2 to the X to start with on the left side of the equation let me distribute that 2 to the^ of x to both terms and you'll have to remember your exponent rule that says when multiplying powers of the same base you add the exponents together and keep the base the same so when I do 2 the x * 2 the X I keep the base the same and add the exponents x + x is 2x when I distribute to the second term in the brackets I keep the factor of 12 and 2 x * 2 the X those Powers have the same base so I can keep that base of two the same add the exponents X plusx is zero on the next line I'm going to do a couple things this 2^ 2x I'm going to use the power of a power rule and reite write that as 2 x^ 2 12 * 2 0 well I know 2 ^ of 0 is 1 so 12 * 1 is just 12 now I'm going to bring this term to the left side of this equation and now I'm going to use a let statement so you can see the hidden quadratic let's let 2 ^ x equal K so what I have is k^2 - 7 k + 12 = 0 that quadratic I could Factor the numbers that multiply to 12 and add to -7 are -3 and4 so the solutions to this equation are K = 3 and K = 4 but remember those Ks are actually 2 to the^ of X so I have 2 ^ x = 3 and 2 ^ x = 4 well 2 the x = 4 that's easy to solve what exponent goes on two to get four that's two 2^ s is four but how do I find what exponent goes on two to get three I'll have to use a logarithm I can say the exponent equals log base 2 of three that'll find me what exponent goes on two to get three as an approximate answer that's about 1.58 there you have it through all 11 difficulties of high school math level equations let me know how far you got and which ones you found the most [Music] difficult and

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Id: a_b-AWfIrr8

Length: 23min 28sec (1408 seconds)

Published: Thu Feb 15 2024