[Music] [Music] welcome to our 10th power episode i am teacher andrew your soul mate allow me to guide you in learning the different concepts and skills in mathematics 9. in the previous mathaton we learned that quadratic inequality in one variable is an inequality that contains a polynomial of degree 2 and can be expressed in one of the following standard forms where a b and c are real numbers and a is not equal to zero a x squared plus b x plus c is greater than zero a x squared plus bx plus c is less than zero ax squared plus bx plus c is less than or equal to zero and ax squared plus bx plus c is greater than or equal to zero you will see a set of quadratic inequalities that we must write in standard form 1 negative two x squared is less than or equal to negative eight the second one zero point one x squared plus two point three x is less than negative and the third one negative three x minus five is less than or equal to two x squared item number one the inequality negative two x squared is less than or equal to negative eight some of you might suggest that we increase both sides by eight yes that is correct negative two x squared plus eight less than or equal to 0 is the standard form of this inequality but we can also do the following we may make the coefficient of x squared positive also negative 2 and negative 8 are both even integers hence we divide both sides of the inequality by negative 2. hence the inequality can now be written as x squared is greater than or equal to four yes you heard me right the symbol has been changed from is less than or equal to 2 is greater than or equal to you know why because we divided both sides of the inequality by a negative number remember whenever you multiply or divide an inequality by a negative number you must flip the inequality sign the next step is to decrease both sides by 4 and there the standard form is achieved so negative 2x squared is less than or equal to negative 8 can also be written as x squared minus 4 greater than or equal to 0. let's record this in the table item number two zero point one x squared plus two point three x is less than negative four i could sense it you are planning to increase both sides by four and that's right and you must get that the standard form is 0.1 x squared plus two point three x plus four is less than zero but if you are a little anxious that the coefficients of x squared and x are not integers well relief is here we can multiply both sides of the inequality by 10. since 0.1 and 2.3 both have one decimal place the resulting inequality is x squared plus 23x plus 40 is less than zero and you might have raised one strand of your eyebrow while the symbol has not been changed there is no need to do this since we multiplied by a positive number let's record the results in the table so two down one to go hagama let's do this so item number three negative three x minus five is less than or equal to two x squared are we thinking of the same thing and that is to decrease both sides by 2x squared so what must we get then ah of course negative 2x squared minus 3x minus 5 is less than or equal to zero i already arranged the terms in descending powers but if your premium feels itchy because of the negative coefficient of x squared say no more we just have to divide both sides by negative 1 yes by negative 1 only since negative 3 and negative 5 are not even and you must remember that we need to change the symbol since we divide both sides by a negative number hence the inequality can now be expressed as 2x squared plus 3x plus 5 is greater than or equal to 0 so let's update the table warm-up is done we are now ready to take on more challenges and since we are mathematician we will survive solving word problems involving quadratic inequalities in one variable problem number one kevin shot an arrow straight upward with a velocity of 96 meters per second from an altitude of 20 meters after how many seconds will this arrow be more than 100 meters high use the position equation s equals negative 16 t squared plus v sub zero times t plus s sub zero where s is the altitude after t seconds s sub 0 is the initial height and b sub 0 is the initial velocity straight upward what do we have to solve for it is the value of the variable t which is the amount of time in seconds that the arrow will take to reach an altitude higher than 100 meters this altitude is the variable s we express this given information as the inequality s is greater than 100 according to the position equation s is equal to negative 16 t squared plus b sub zero t plus s sub zero we must express s in terms of t by substituting the values of v sub zero and s sub zero in this position equation according to the problem the initial velocity v sub 0 is equal to 96 while the initial height s sub 0 is 20. hence s must be equal to negative 16 t squared plus 96 t plus 20. so we may rewrite the inequality s is greater than 100 as negative 16 t squared plus 96 t plus 20 is greater than 100 we should rewrite this inequality in standard form so let's decrease both sides of the inequality by 100 in order to set it to zero the resulting inequality is negative 16 t squared plus 96 t minus 80 is greater than zero but if we want the numerical coefficient of t squared to be positive and if we want to deal with smaller numbers we must divide both sides of the inequality by negative 16 by a negative number because we want to make the coefficient of t squared positive since negative divided by negative is positive and y negative of 16 simply because negative 16 96 and negative 80 are all divisible by 16 and this will result to the inequality t squared minus six t plus five is less than zero hence we have to determine the set of values of t for which the polynomial t squared minus six t plus five is less than zero or negative and the first thing we do is to solve the quadratic equation that corresponds to the given inequality and i am referring to t squared minus six t plus five equals zero observe that the symbol in the inequality is just changed to equal sign the quadratic equation can be solved by factoring hence we rewrite it as the binomial t minus 1 times the binomial t minus five equals zero and equating each factor to zero we can conclude that the values of t are one and five one and five are the roots of the equation and are also called the critical points that will serve as boundaries when we divide the number line later into regions or intervals these values will make the polynomial and inequality equal to zero remember we are looking for values of the variable t for which the polynomial t squared minus sixty plus five is negative or less than zero we may rewrite the inequality in a way that shows the factored form of the polynomial of degree two now let's test one and five against the inequality to make this crystal clear c the computation let's plot the critical points now on the number line we will use hollow circles since these critical points are not part of the inequality's solution set then we will draw a broken line through each of them to divide the number line into regions or intervals why broken lines again it's because the critical points are not part of the solution set but we filipinos must always aim to be part of solution mustn't we the number line should look like this let's add the following details on the graph at t equals one the value of the factor t minus one is zero and at t equals five the value of the factor t minus five is zero we must identify which among the regions and the number line contains the set of values of t for which the product of the two factors t minus 1 and t minus 5 is negative their values then must be of different signs so we are looking at two cases here case one t minus one is less than zero and t minus five is greater than zero case 2 t minus 1 is greater than 0 and t minus 5 is less than 0. in the region to the left of 1 we may use 0 as the test value and when t is equal to zero t minus one should be equal to zero minus one and the result is negative one hence in this region where t is less than one the value of t minus 1 is negative in the region between 1 and 5 we may use 3 as the test value and when t is equal to 3 t minus 1 will be equal to three minus one or simply two hence in this region where one is less than t and t is less than five the value of t minus one is positive in the region to the right of 5 we may use 9 as the test value so when t is equal to 9 t minus 1 will be equal to 9 minus 1 or simply 8. hence in this region where t is greater than 5 the value of t minus 1 is also positive let's add these observations in the graph now let's use the same set of test values to determine the sign of the value of the factor t minus 5 at each interval in the region to the left of 1 we may use 0 as the test value so when t is equal to zero t minus five will be equal to zero minus five or negative five hence in this region where t is less than one the value of t minus five is negative in the region between 1 and 5 we may use 3 as the test value when t is equal to 3 t minus 5 is equal to 3 minus 5 and the result here is negative 2 hence in this region where 1 is less than t but t is less than 5 the value of t minus 5 is also negative in the region to the right of 5 we may use 9 as the test value and when t is equal to 9 t minus 5 must be equal to 9 minus 5 or simply 4. hence in this region where t is greater than 5 the value of t minus 5 is positive let's add these observations in the graph look at the number line closely recall that we were looking at two cases to determine the solution set of the inequality so where on the graph do we see that the value of the factor t minus one is negative and that of the other factor t minus five is positive nowhere hence the solution set for case 1 is an empty set where on the graph do we see that the value of the factor t minus 1 is positive and that of the other factor d minus 5 is negative it is in the region where 1 is less than t and t is less than 5. hence the solution set for case 2 can be expressed by the same double inequality 1 is less than t and t is less than 5. written in interval notation this inequality is expressed this way it is an open interval we use parenthesis since the boundaries 1 and 5 in this case are not part of the solution set therefore the solution set for the inequality t squared minus 6 t plus 5 is less than 0 is the union of the solution set for case 1 and of the solution set for case 2 which is just equal to the solution set for case 2. hence the values of t for which the inequality holds true must be greater than 1 but less than 5. let's plot this set on the number line this means that after more than a second but less than 5 seconds the arrow will reach an altitude that is higher than 100 meters let's solve another problem anna is instructed to make a garden plot which has an area less than 18 square feet the length should be three feet longer than the width so what are the possible dimensions of the garden plot the garden plot is presumably rectangular its area is less than 18 square feet recall that the area of a rectangle is obtained by getting the product of its length and its width hence we have the inequality a is less than 18 or l times w is less than 18. the length l must be three feet longer than the width w so l must be equal to w plus 3 and therefore area will be expressed as the binomial w plus 3 times w and in terms of the variable w the inequality may now be written as the binomial w plus 3 times w is less than 18. the inequality must be written as w squared plus 3w is less than 18 after applying distributive property of multiplication on the left side then we decrease both sides by 18. this will result to w squared plus 3w minus 18 is less than zero and we have to determine the set of values of the variable w for which the value of the polynomial w squared plus 3w minus 18 is negative now that the inequality has been written in standard form we can now start taking steps in solving it first we solve the quadratic equation w squared plus 3w minus 18 equals 0 which corresponds to the inequality and it can be solved by factoring hence the equation may be written as the binomial w plus six times the binomial w minus three equals zero equating each factor to zero we can conclude that the values of w are negative 6 and 3. these are the critical points that will serve as boundaries in dividing the number line into regions or intervals also the inequality may now be expressed as the binomial w plus six times the binomial w minus three is less than zero let's test the root of the equation against this inequality by substituting each of them in the inequality we must be convinced that they are not part of the inequalities solution set now let's plot the critical points now on the number line we will use hollow circles since these critical points are not part of the inequality solution set then we will draw broken line through each of them to divide the number line into regions or intervals why broken lines again it's because the critical points are not part of the solution set now let's add the following details on the graph at w equals negative six the value of the factor w plus six is zero at w equals three the value of the factor w minus three is zero the inequality binomial w plus 6 times binomial w minus 3 is less than 0 implies that the values of the factor must be of different signs and we have two cases for this case one w plus six is less than zero and w minus three is greater than zero for case two w plus six is greater than zero and w minus 3 is less than 0. so we will use a test value to determine the sign of the value of each factor in each region or interval the region to the left of negative 6 we may use negative 7 as the test value when w equals negative 7 w plus 6 will be equal to negative 7 plus 6 or simply negative one hence in this region where w is less than negative six the value of w plus six is negative in the region between negative six and three we may use one as the test value so when w is equal to one w plus six will be equal to one plus six and the result is positive seven hence in this region where negative 6 is less than w and w is less than 3 the value of w plus 6 is positive in the region to the right of 3 we may use 5 as the test value and when w is 5 w plus 6 will be equal to 5 plus 6 in which the result is 11. hence in this region where w is greater than 3 the value of w plus 6 is also positive let's add these observations in the graph and let's use the same set of test values to determine the sign of the value of the factor w minus 3 at each interval in the region to the left of negative 6 we may use negative 7 as the test value and when w is negative 7 w minus three will be equal to negative seven minus three in which the result is negative ten hence in this region where w is less than negative six the value of w minus three is negative in the region between negative 6 and 3 we may use 1 as the test value when w is equal to 1 w minus 3 will be equal to 1 minus 3 in which the result is negative 2 hence in this region where negative 6 is less than w and w is less than 3 the value of w minus 3 is also negative in the region to the right of 3 we may use 5 as the test value and when w is 5 w minus 3 will be equal to 5 minus 3 in which the result is positive 2. hence in this region where w is greater than 3 the value of w minus 3 is positive let's add these observations in the graph now where on the graph do we see that the signs of the values of the two factors are different only in the region where negative 6 is less than w and w is less than three hence this will already be the solution set for the inequality w squared plus three w minus 18 is less than zero written in interval notation the solution can be expressed by the open interval containing the boundaries negative 6 and 3. however not all values in the solution set can be accepted as answers to the problem we are to identify the possible dimensions of the guardian plot anna is instructed to make and since length is expressed in terms of the width that is represented by the variable w w must be positive hence in the solution set negative 6 is less than w and w is less than 3 the accepted set of values of w can be expressed by the inequality 0 is less than w and w is less than 3. increasing all sides of the double inequality by 3 we shall obtain the range of values of the variable l which represents the length of the garden plot hence the length can be longer than 3 feet but it must be shorter than 6 feet considering only the integer values of w the table will show us the possible dimensions of the garden plot hence the garden plot could be one foot by four feet or two feet by five feet thank you for tuning in kagamath it has been my pleasure to be your guide in learning the different concepts and skills in mathematics night i am teacher andrew your soulmate bye [Music] [Music] [Music] you