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AN AMAZING METHOD for Solving a 4th Degree Equation
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Jun 28, 2025
In this video, I show you a method for solving a complex fourth-degree equation. I start with a challenging equation and walk you through each step, simplifying the problem and finding the solution. By using a combination of the method of undetermined coefficients and another approach that you might not have seen before, I explain the entire process in a clear and accessible way. #algebra #mathematics #maths #equation #howtosolvemathproblem #uniquemaths
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0:00
hello there welcome to we learn
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daily in this video we will tackle the
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solution of a complex fourth deegree
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equation using a method you've likely
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never seen
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before to gauge the difficulty try
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solving this equation on your own right
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now let's start if we substitute the
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root it won't help much so instead we'll
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immediately Square both sides to do this
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we need to ensure that the left side of
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the equation is non- negative we start
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by setting the condition that x^2 - 5 is
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greater than or equal to0 now we can
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square both sides of the equation
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squaring both sides we get x^2 - 5 all
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squar equals the expression under the
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square root which is x + 5 should we
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require the non- negativity of the
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expression under the square root there's
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no need because this automatically
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follows from the equation we derived
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since the left side is non- negative due
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to the squaring x + 5 will also be non-
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negative therefore we don't need to
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write an additional condition for the
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non- negativity of the expression under
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the square root we won't solve the first
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inequality here's the plan we'll solve
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the given equation then substitute the
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roots into our condition and select the
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ones that satisfy
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it expanding the brackets and moving
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everything to the left side we get X
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the^ of 4 - 10 * x^ 2 + 25 - x - 5 equal
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0 this gives us a rather complex fourth
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degree equation spoiler alert this
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equation has four irrational Roots we'll
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solve this equation using two methods
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the first method is one we are already
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familiar with which is based on the
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method of undetermined
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coefficients we'll start with that the
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second method is something you've likely
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never seen before let's start with the
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first method the method of undetermined
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coefficients we rewrite the left side
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and combine like terms the equation
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becomes x ^ 4 - 10 * x^2 - x + 20 now
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let's try to factor the left side using
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the method of undetermined
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coefficients we'll attempt to factor the
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left side into the product of two
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quadratic trinomials of the form x^2 + a
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* x + B multiplied x^2 + C * x + D next
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we need to find these coefficients a b c
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and d and then substitute them into our
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factorization we'll determine these
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coefficients using the method of
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undetermined coefficients let's write
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this out immediately to avoid rewriting
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it later we expand the brackets and
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combine like terms the equation becomes
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x^ of 4 + a + C * x cubed + B + D + a c
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* x^ 2 + a d + b c * x + B * D now the
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left side equals the right side which
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means we need to equate the coefficients
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for corresponding powers of X for X to
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the^ of 4 the coefficients are the same
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in the right side the coefficient for X
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cubed is a + C but there is no X cubed
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term in the left side so we set the
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coefficient on the right side to zero
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this gives us the first equation for
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finding the coefficient
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a + c equal 0 the next step is to equate
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the coefficient for X2 with the
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corresponding term on the left side
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which gives us B+ D plus a c = -10 for
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the linear term we get a d + BC = -1
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finally the constant term gives us B * D
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= 20 we'll try to solve the system of
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equations with integer Solutions let's
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start with the last equation B * d = 20
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you can choose pairs like four and five
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but if we choose four and five
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substituting into the second equation
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will give us something inconvenient like
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AC = 9 so let's choose
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bal4 and D = -5 substituting these into
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the equations we find AC = -1 we also
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keep a + Cal 0 and for a d + BC we
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substitute and get ne
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1 now we have simpler pairs to work with
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for example if a is 1 B = -4 then C will
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be -1 and D equal -5 the coefficients
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match we substitute these coefficients
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back into our factorization the left
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side can now be written as the product
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of two quadratic trinomials x^2 + x - 4
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* x^2 - x - 5
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all set to zero to find the roots we
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solve each quadratic equation for the
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first quadratic equation X2 + x - 4 = 0
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or x^2 - x - 5 = 0 we find The Roots the
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Roots are then X first and 2 for the
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first equation we have minus B so min-1
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plus or minus the square root of the
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discriminant we calculate the
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discriminant it turns out to be 17 / 2
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and then we immediately find the roots 3
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and
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four for the second pair of roots we
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have 1 plus or minus the square < TK of
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21 / 2 now each of the
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roots we have four rational roots are
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substituted back into the equation we
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Square them subtract five and compare
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them to zero it's not difficult try
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doing it
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yourself I've already done it and there
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are two Roots left and let's do this
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before moving on to an incredible method
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of solving this fourth degree equation
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let's finish the first method completely
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write down the answer and then we'll
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move on to the method you've never heard
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of The Final Answer after filtering out
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leaves us with two Roots out of the four
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we get min-1 - the < TK of 17 divid 2
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and the second root is 1 + theare < TK
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of 21 / by two now let's move on to the
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second method so in order to avoid
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solving this complex fourth degree
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equation and it's now clear why it's
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complex because it has four irrational
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Roots we will instead solve a second
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deegree equation that is a quadratic
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equation it's not clear how to obtain a
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quadratic equation from a fourth degree
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equation well it turns out we can solve
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a quadratic equation with respect to the
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number five here look we have the number
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five and over here we have 5^ squ
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we will solve the quadratic equation
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with respect to this five but to make it
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less intimidating let's replace the
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number five with t now this becomes t^2
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and most importantly don't forget to
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also replace the 10 if you forget to
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replace the 10 nothing will work you can
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try it yourself so we replace the 10 as
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well writing it as 2 * 5 which becomes 2
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* T now we proceed to solve after this
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substitution we get a Quadra atic
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equation of the following form T ^2 -
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2x^2 + 1 * T and what's left is+ X to
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4th power - x and all of this equals 0
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so we have obtained a quadratic equation
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with respect to T let's solve it as
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usual we find the discriminant the
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discriminant equal B ^2 - 4 * a * C
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which gives us 2 x^2 + 1^ 2 - 4x 4 power
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- - x we combine like terms 4X 4th power
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- 4x 4th power cancels each other out
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leaving us with 4 x^2 + 4 x + 1 which is
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equal to 2 x + 1^ 2 this simplifies
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nicely we proceed to the roots we find
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the roots T first and second for the
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first equation we have minus B which is
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minus we remove the minus sign so we get
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2 x^2 + 1 plus orus the square root of
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the discriminant
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when we find this root we get the
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absolute value of this sum the absolute
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value simplifies with a plus minus sign
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and since we already have a plus minus
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here we no longer need the absolute
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value we simply write this expression
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after taking the square root 2x + 1 and
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divide everything by two this gives us
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two Roots let's write them separately
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either T equals let's immediately divide
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by two then we get x^2 + x + 1 and the
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second root is T = x^2 - x let's check
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yes everything is correct now we perform
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the reverse substitution since t equal 5
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we substitute 5 in place of T and we get
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x^2 + x we move this 5 to the right side
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with a minus sign so we get x^2 + x - 4
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= 0 the second equation is x^2 - x and
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we move the five to the right side with
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a minus sign so it becomes - 5 = 0 thus
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we OB contained exactly the same two
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quadratic equations that we got in the
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first method naturally the solution is
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the same we write the answer and
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naturally we get the same Roots after
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filtering let me remind you that after
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filtering through this inequality we get
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minus1 minus the < TK of 17 / 2 and 1 +
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the < TK of 21 / 2 the problem is solved
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if the solution is clear don't forget to
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like the video subscribe and write a
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comment about which method you found
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better thank you for watching goodbye
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