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5 Effective Methods to Solve Quadratic Equations
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Jun 28, 2025
In this video, I’ll walk you through five effective methods for solving quadratic equations. Whether you’re a student looking to sharpen your math skills or simply curious about different approaches, this video is designed to help you master the art of solving quadratic equations. #maths #mathematics #mathstricks #equation #quadraticequation
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in this video I'll show you five
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different ways to solve a quadratic
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equation be sure to watch until the end
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and let me know in the comments which
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method you think is the best the problem
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is to solve the quadratic equation 3 *
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x^2 + 7 * x - 10 = 0 Let's solve the
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equation using the first method which is
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the classic approach to solving a
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quadratic equation using the
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discriminant recall that the coefficient
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of the highest degree of the variable is
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denoted by the letter A the coefficient
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of the first degree of the variable is
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denoted by B and the constant term is
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denoted by C to solve the quadratic
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equation we need to find the
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discriminant using the formula b^ 2 - 4
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* a * C let's substitute the values and
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calculate it we substitute 7 for B and
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get 7^ 2 - 4 * a we substitute 3 for a
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and multi mly by C where we substitute
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-10 for C this results in 7^ s which
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equals 49 we multiply -4 by 3 to get
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-12 then we multiply -12 by -10
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resulting in a positive
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120 adding 120 gives us 169 the
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discriminant turns out to be positive
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this means the quadratic equation has
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exactly two roots we find the roots
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using the formula x sub 1 and x sub 2
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our two Roots will have the form minus B
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plus or minus the square root of the
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discriminant divided 2 * a substituting
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7 for B and adding the minus sign we get
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- 7 plus or minus the square root of the
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discriminant the discriminant we found
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is 169 which gives us the square root of
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169 as 13 dividing by 2 * a where a = 3
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we multiply by 3 rewriting we have - 7 +
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or -3 all divided 6 now let's find the
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first
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root to find the first root we take the
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minus sign giving us Min 7 -3 which
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equals
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-2 this is then divided 6 for the second
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root we take the plus sign giving us - 7
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+ 13 which is the same as 13 - 7
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resulting in six this gives us 6 / 6 we
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simplify the first fraction by dividing
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by two yielding the first root as - 10 /
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3 the second fraction is simplified by
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dividing by 6 resulting in one thus our
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quadratic equation has two Roots - 10 /
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3 and 1 the second method of solving
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involves finding the solution to our
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quadratic equation by using the sum of
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the coefficients specifically let's
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notice that the sum of the coefficients
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equals z that is if we now discard the
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variable X we end up with 3 + 7 - 10
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which equals z if the sum of the
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coefficients equals z then one of the
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roots must be equal to one indeed if we
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now substitute one in place of the
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variable X we obtain a valid equation to
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find the second root we need to divide
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both sides of our equation by the
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coefficient of the highest degree term
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so we divide both sides by 3 and we get
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x^2 + 7/3 * x - 10/3 and all of this
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equals 0 we divided both sides of the
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equation by 3 so that the coefficient of
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the highest degree term becomes one now
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if one of the roots is one then the
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other root will be equal to the constant
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term which is -
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10/3 using the third method let's solve
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our equation using vieta's theorem or as
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some clarify the theorem that is the
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converse of vieta's theorem to do this
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by dividing by three we make the
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coefficient of the highest degree equal
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to 1 as a result we obtain an equation
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of the
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form x^ 2 + 7/3 x - 10/3 and all of this
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equals zero in other words we have
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obtained what is called a reduced
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quadratic equation of the form x^2 + BX
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+ q and all of this equals z according
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to vieta's theorem the product of the
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roots X1 * X2 is equal to q and the sum
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of the roots X1 + X2 is equal to the
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coefficient of the first degree term
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with the opposite sign meaning it is
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equal to P typically in school these two
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equations are written in the reverse
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order but applying V theorem is more
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convenient in the order you see on the
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screen so first we select the product
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which equals
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-1/3 the simplest case of such Roots is
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the constant term and one so we take the
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constant term which is -1/3 and take one
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and we verify that these two Roots
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satisfy our equation and we again obtain
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the same
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Roots fourth method of solution the
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transposition method
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first the coefficient of the highest
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degree term needs to be transposed to
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the constant term this means multiplying
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-10 by
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3 as a result you get what is called an
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auxiliary equation of the form x^2 + 7 x
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- 10 * 3 after transposing we obtain a
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quadratic equation of the form x^2 + 7 x
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- 30 = 0 now we can find the root of the
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quadratic equation using vieta's theorem
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we start by selecting the product of the
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roots which is -10 and 3 the product
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equals -30 and the sum equals the number
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seven with the opposite sign meaning -7
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now that we found the roots of the
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auxiliary equation we must remember to
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divide the obtained Roots by the number
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that was transposed which is three as a
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result we find that the first root equal
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=
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-1/3 and the second root after dividing
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3 by 3 equal 1 thus we have again
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obtained the same Roots fifth method of
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solution this method is based on
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completing the
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square first we need to make the
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coefficient of the highest degree term
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equal to
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1 to do this we divide both sides of the
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equation by three as a result we obtain
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an equivalent equation of the form x^2 +
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7/3 x - 10/3 = 0 now let's recall the
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formula for the square of a sum
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specifically if we have a^ 2 + 2 * a * b
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+ b^ 2 it is equivalent to the square of
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the sum of these numbers A + B all
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squared we will now try try to complete
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the square on the left side of the
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equation keeping in mind that the
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coefficient of the first deegree term is
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positive so we will complete the square
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of the sum if the coefficient were
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negative we would complete the square of
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the difference instead we already have
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the square of the first term which is
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x^2 so we rewrite it next we have the
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term that is 2 * the first term
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multiplied by some unknown second term
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we have the first term but we don't have
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the two we obtain this two artificially
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by multiplying the second term by two
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however to keep the equation unchanged
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we must also divide the second term by
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two as a result the second term can be
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written as 2 * 76 multiplied X we now
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have x^2 + 2 * 76 x let's compare this
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with our formula x^2 is equivalent to a^
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2 we have the 2 in the X so B is 76 this
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means we are missing the square of 76 to
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complete the square we add 76 squar to
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the equation since this term was not
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originally in the equation we must
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subtract it to keep the equation
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balanced so we subtract 76 SAR and add
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what was left which was -1/3 now the
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first three terms form a perfect square
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according to our formula which is x + 76
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all 2ar - 76 squared we Square 7 to get
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49 and square 6 to get 36 so we have -
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49 over 36 we also bring the second
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fraction to a common denominator of 36
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which requires multiplying the numerator
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and denominator by 12 this gives us -10
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over 36 adding these together we get -69
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over 36 this equation now equals 0 we
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move the constant term to the other side
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with the opposite sign and get x + 76
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all 2ar = 169 36 we take the square root
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of both sides x + 76 = plus orus the <
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TK of 169 /
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36 from this we find that X = -76 plus
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orus the < tk69 over
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36 the square < TK of 169 is 13 and the
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square < TK of 36 is 6 so we have plus
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orus 136 We Now find the
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roots for the root Corr responding to
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the minus sign we subtract 136 from
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-76 which gives us
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-26 or -1/3 for the root corresponding
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to the plus sign we add 136 to -76
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giving us 66 or 1 in the end we get the
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same two Roots as before -1/3 and 1 the
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problem is solved if you found the
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solution helpful please like the video
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leave a comment and let me know which
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method you liked best thank you for
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watching and see you in the next video
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goodbye
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