Quadratic,Formula,Lucid,Explan education Quadratic Formula : Lucid Explanation of Its Derivation and

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Derivation of Quadratic formula:To solve ax^2 + bx + c = 0 where a ( ` 0 ), b, c are constants which can take real number values.ax^2 + bx + c = 0or ax^2 + bx = -cDividing by 'a' on both sides, we getx^2 + (bDa)x = -cDaor x^2 + 2x(bD2a) = -cDa .........(i)The L.H.S. of equation(i) has (first term)^2 and 2(first term)(second term) terms where fist term = x and second term = (bD2a).If we add (second term)^2 {= (bD2a)^2}, the L.H.S. of equation(i) becomes a perfect square.Adding (bD2a)^2 to both sides of equation(i), we getx^2 + 2x(bD2a) + (bD2a)^2 = -cDa + (bD2a)^2or (x + bD2a)^2 = b^2D4a^2 - cDa = ( b^2 - 4ac)D(4a^2)or (x + bD2a) = ±{( b^2 - 4ac)D(4a^2)} = ±( b^2 - 4ac)D2aor x = -bD2a ± (b^2 - 4ac)D2aor x = {-b ± (b^2 - 4ac)}D2aThis is the Quadratic Formula. (Derived.)I Applying Quadratic Formula in Finding the roots :Example I(1) : Solve x^2 + x - 42 = 0 using Quadratic Formula.Comparing this equation with ax^2 + bx + c = 0, we geta = 1, b = 1 and c = -42Applying Quadratic Formula here, we getx = {-b ± (b^2 - 4ac)}D2a= [ (-1) ± {(1)^2 - 4(1)(-42)}]D2(1)= [ (-1) ± {1 + 168}]D2(1) = [ (-1) ± {169}]D2(1) = [(-1) ± 13]D2(1)= (-1 + 13)D2, (-1 - 13)D2 = 12D2, -14D2 = 6, -7 Ans.Example I(2) :Solve 8 - 5x^2 - 6x = 0 using Quadratic FormulaMultiplying the given equation by -1, we get5x^2 + 6x - 8 = 0(-1) = 0Comparing this equation with ax^2 + bx + c = 0, we geta = 5, b = 6 and c = -8Applying Quadratic Formula here, we getx = {(-b) ± (b^2 - 4ac)}D2a= [ (-6) ± {(6)^2 - 4(5)(-8)}]D2(5)= [ (-6) ± {36 + 160}]D10 = [ (-6) ± {196}]D10 = [(-6) ± 14]D10= (-6 + 14)D10, (-6 - 14)D10 = 8D10, -20D10 = 4D5, -2 Ans.Example I(3) :Solve 2x^2 + 3x - 3 = 0 using Quadratic FormulaComparing this equation with ax^2 + bx + c = 0, we geta = 2, b = 3 and c = -3Applying Quadratic Formula here, we getx = {(-b) ± (b^2 - 4ac)}D2a= [(-3) ± {(3)^2 - 4(2)(-3)}]D2(2)= [(-3) ± {9 + 24}]D4 = [-3 ± (33)]D4 Ans.II To find the nature of the roots :By Quadratic Formula, the roots of ax^2 + bx + c = 0 are± = {-b + (b^2 - 4ac)}D2a and ² = {-b - (b^2 - 4ac)}D2a.Let (b^2 - 4ac) be denoted by ” (called Delta).Then ± = (-b + ”)D2a and ² = (-b - ”)D2a.The nature of the roots (± and ²) depends on ”.” ( = b^2 - 4ac) is called the DISCRIMINANT of ax^2 + bx + c = 0.Three cases arise depending on the value of” (= b^2 - 4ac) is zero or positive or negative.(i) If ” ( = b^2 - 4ac) = 0, then ± = -bD2a and ² = -bD2ai.e. the two roots are real and equal.Thus ax^2 + bx + c = 0 has real and equal roots, if ” = 0.(ii) If ” ( = b^2 - 4ac) > 0, the roots are real and distinct.(ii) (a) if ” is a perfect square, the roots are rational.(ii) (b) if ” is not a perfect square, the roots are irrational.(iii) If ” ( = b^2 - 4ac) < 0, ” is not real.It is called an imaginary number.i.e. ±, ² are imaginary when ” is negative.When ” is negative, the roots are imaginary.Example II(1) :Find the nature of the roots of the equation, 5x^2 - 2x - 7 = 0.Solution : The given equation is 5x^2 - 2x - 7 = 0.Comparing this equation with ax^2 + bx + c = 0, we geta = 5, b = -2 and c = -7.Discriminant = ” = b^2 - 4ac = (-2)2 - 4(5)(-7) = 4 + 140 = 144 = 12^2Since the Discriminant is positive and a perfect square,the roots of the given equation are real, distinct and rational. Ans.Example II(2) :Find the nature of the roots of the equation, 9x^2 + 24x + 16 = 0.Solution : The given equation is 9x^2 + 24x + 16 = 0.Comparing this equation with ax^2 + bx + c = 0, we geta = 9, b = 24 and c = 16Discriminant = ” = b^2 - 4ac = (24)^2 - 4(9)(16) = 576 - 576 = 0.Since the Discriminant is zero,the roots of the given equation are real and equal. Ans.Example II(3) : Find the nature of the roots of the equation, x^2 + 6x - 5 = 0.Solution : The given equation is x^2 + 6x - 5 = 0.Comparing this equation with ax^2 + bx + c = 0, we geta = 1, b = 6 and c = -5.Discriminant = ” = b^2 - 4ac = (6)^2 - 4(1)(-5) = 36 + 20 = 56Since the Discriminant is positive and is not a perfect square,the roots of the given equation are real, distinct and irrational. Ans.Example II(4) : Find the nature of the roots of the equation, x^2 - x + 5 = 0.Solution : The given equation is x^2 - x + 5 = 0.Comparing this equation with ax^2 + bx + c = 0, we geta = 1, b = -1 and c = 5.Discriminant = ” = b^2 - 4ac = (-1)^2 - 4(1)(5) = 1 - 20 = -19.Since the Discriminant is negative,the roots of the given equation are imaginary. Ans.III To find the relation between the roots and the coefficients :Let the roots of ax^2 + bx + c = 0 be± (called alpha) and ² (called beta).Then By Quadratic Formula± = {-b + (b^2 - 4ac)}D2a and ² = {-b - (b^2 - 4ac)}D2aSum of the roots = ± + ²= {-b + (b^2 - 4ac)}D2a + {-b - (b^2 - 4ac)}D2a= {-b + (b^2 - 4ac) -b - (b^2 - 4ac)}D2a= {-2b}D2a = -bDa = -{(coefficient of x)D(coefficient of x^2)}.Product of the roots = (±)(²)= [{-b + (b^2 - 4ac)}D2a][{-b - (b^2 - 4ac)}D2a]= [{-b + (b^2 - 4ac)}][{-b - (b^2 - 4ac)}]D(4a^2)The Numerator is product of sum and difference of two terms whichwe know is equal to the difference of the squares of the two terms.Thus, Product of the roots = ±²= [(-b)^2 - {(b^2 - 4ac)}^2]D(4a^2)= [b^2 - (b^2 - 4ac)]D(4a^2) = [b^2 - b^2 + 4ac)]D(4a^2) = (4ac)D(4a^2)= cDa = (constant term)D(coefficient of x^2)Example III(1) : Find the sum and product of the roots of the equation 3x^2 + 2x + 1 = 0.Solution : The given equation is 3x^2 + 2x + 1 = 0.Comparing this equation with ax^2 + bx + c = 0, we geta = 3, b = 2 and c = 1.Sum of the roots = -bDa = -2D3. Product of the roots = cDa = 1D3.Example III(2) : Find the sum and product of the roots of the equation x^2 - px + pq = 0.Solution : The given equation is x^2 - px + pq = 0.Comparing this equation with ax^2 + bx + c = 0, we geta = 1, b = -p and c = pq.Sum of the roots = -bDa = -(-p)D1 = p.Product of the roots = cDa = pq D1 = pq.Example III(3) : Find the sum and product of the roots of the equation lx^2 + lmx + lmn = 0.Solution : The given equation is lx^2 + lmx + lmn = 0.Comparing this equation with ax^2 + bx + c = 0, we geta = l, b = lm and c = lmn.Sum of the roots = -bDa = -(lm)D l = -m Product of the roots = cDa = lmnDl = mn.IV To find the Quadratic Equation whose roots are given :Let ± and ² be the roots of the Quadratic Equation.Then, we know (x - ±)(x - ²) = 0.or x^2 - (± + ²)x + ±² = 0.But, (± + ²) = sum of the roots and ±² = Product of the roots.The required equation isx^2 - (sum of the roots)x + (product of the roots) = 0.Thus, The Quadratic Equation with roots ± and ² isx^2 - (± + ²)x + ±² = 0.Example IV(1) :Find the quadratic equation whose roots are 3, -2.Solution: The given roots are 3, -2.Sum of the roots = 3 + (-2) = 3 - 2 = 1;Product of the roots = 3 x (-2) = -6.We know the Quadratic Equation whose roots are given isx^2 - (sum of the roots)x + (product of the roots) = 0.So, The required equation is x^2 - (1)x + (-6) = 0.i.e. x^2 - x - 6 = 0 Ans.Example IV(2) :Find the quadratic equation whose roots are lm, mn.Solution: The given roots are lm, mn.Sum of the roots = lm + mn = m(l + n);Product of the roots = (lm)(mn) = l(m^2)n.We know the Quadratic Equation whose roots are given isx^2 - (sum of the roots)x + (product of the roots) = 0.So, The required equation is x^2 - m(l + n)x + l(m^2)n = 0. Ans.Example IV(3) :Find the quadratic equation whose roots are (5 + 7), (5 - 7).Solution: The given roots are 5 + 7, 5 - 7.Sum of the roots = (5 + 7) + (5 - 7) = 10;Product of the roots = (5 + 7)(5 - 7) = 5^2 - (7)^2 = 25 - 7 = 18.We know the Quadratic Equation whose roots are given isx^2 - (sum of the roots)x + (product of the roots) = 0.So, The required equation is x^2 - (10)x + (18) = 0.i.e. x^2 - 10x + 18 = 0 Ans.For more about Quadratic Formula, go to,http://www.math-help-ace.com/Quadratic-Formula.html