If the difference of the roots of the quadratics equation x²-ax+bis 1
then
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⏩Given If the difference of the roots of the quadratic equation x2 - ax + b = 0 is 1. then prove that a2 = 4b + 1.
Given If the difference of the roots of the quadratic equation x2 - ax + b = 0 is 1. then prove that a2 = 4b + 1.Let α and β be the roots of the quadratic equation
Given If the difference of the roots of the quadratic equation x2 - ax + b = 0 is 1. then prove that a2 = 4b + 1.Let α and β be the roots of the quadratic equation So x^2 – ax + b = 0
Given If the difference of the roots of the quadratic equation x2 - ax + b = 0 is 1. then prove that a2 = 4b + 1.Let α and β be the roots of the quadratic equation So x^2 – ax + b = 0Comparing with ax^2 + bx + c we get a = 1, b = - a and
c = b
c = bNow sum of roots α + β = - b / a
c = bNow sum of roots α + β = - b / a = -(-a) / 1
c = bNow sum of roots α + β = - b / a = -(-a) / 1 = a
c = bNow sum of roots α + β = - b / a = -(-a) / 1 = aProduct of roots = c/a
c = bNow sum of roots α + β = - b / a = -(-a) / 1 = aProduct of roots = c/a = b / 1
c = bNow sum of roots α + β = - b / a = -(-a) / 1 = aProduct of roots = c/a = b / 1 = b
c = bNow sum of roots α + β = - b / a = -(-a) / 1 = aProduct of roots = c/a = b / 1 = bAccording to question α – β = 1
c = bNow sum of roots α + β = - b / a = -(-a) / 1 = aProduct of roots = c/a = b / 1 = bAccording to question α – β = 1 (α – β^2 = α + β)^2 - 4αβ
c = bNow sum of roots α + β = - b / a = -(-a) / 1 = aProduct of roots = c/a = b / 1 = bAccording to question α – β = 1 (α – β^2 = α + β)^2 - 4αβ 1^2 = a^2 – 4b
c = bNow sum of roots α + β = - b / a = -(-a) / 1 = aProduct of roots = c/a = b / 1 = bAccording to question α – β = 1 (α – β^2 = α + β)^2 - 4αβ 1^2 = a^2 – 4bOr we get a^2 = 4b + 1
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