Maths-
General
Easy

Question

left parenthesis x minus 1 right parenthesis squared equals 3 x minus 5
What is one possible solution to the equation above?

hintHint:

Hint:
We are given an equation in one variable. To find one solution of the equation means to find a value of x which satisfies this equation. First we expand the square in the left and side and then simplify the equation. We need to use this formula to expand the square term
<left parenthesis a plus b right parenthesis squared equals a squared plus 2 a b plus b squared
We may also need to use the quadratic formula, which is
x equals fraction numerator negative b plus-or-minus square root of b squared minus 4 a c end root over denominator 2 a end fraction

The correct answer is: 2 or 3


    The given equation is
    left parenthesis x minus 1 right parenthesis squared equals 3 x minus 5
    Expanding the square terms by the formula
    left parenthesis a plus b right parenthesis squared equals a squared plus 2 a b plus b squared
    We get,
    x squared minus 2 x plus 1 equals 3 x minus 5
    Taking all the terms to the left hand side, we have
    x squared minus 2 x plus 1 minus 3 x plus 5 equals 0
    Taking the co-efficient common which have the same power of x with them,
    x squared plus left parenthesis negative 2 minus 3 right parenthesis x plus left parenthesis 1 plus 5 right parenthesis equals 0
    Simplifying, we get,
    x squared minus 5 x plus 6 equals 0
    This is a quadratic equation of the form a x squared plus b x plus c equals 0
    The value of x is given by
    x equals fraction numerator negative b plus-or-minus square root of b squared minus 4 a c end root over denominator 2 a end fraction
    Comparing the above equation with the quadratic equation, we have
    a equals 1 semicolon b equals negative 5 semicolon c equals 6
    So the value of x is given by
    x equals fraction numerator negative left parenthesis negative 5 right parenthesis plus-or-minus square root of left parenthesis negative 5 right parenthesis squared minus 4.1.6 end root over denominator 2.1 end fraction
    Thus, we get
    x equals fraction numerator 5 plus-or-minus square root of 25 minus 24 end root over denominator 2 end fraction
    not stretchy rightwards double arrow x equals fraction numerator 5 plus-or-minus square root of 1 over denominator 2 end fraction
    not stretchy rightwards double arrow x equals fraction numerator 5 plus-or-minus 1 over denominator 2 end fraction
    Thus, we get two values of x
    By taking the ‘+’ sign, we have
    x equals fraction numerator 5 plus 1 over denominator 2 end fraction
    x equals 6 over 2
    x equals 3
    By taking the ‘-‘ sign, we get
    x equals fraction numerator 5 minus 1 over denominator 2 end fraction
    x equals 4 over 2
    x equals 2
    Hence, value of x = 2 or 3
    One possible solution to the equation  left parenthesis x minus 1 right parenthesis squared equals 3 x minus 5 text  is  end text 2
    or
    One possible solution to the equation  left parenthesis x minus 1 right parenthesis squared equals 3 x minus 5 text  is  end text 3

    Note:
    Instead of using the quadratic formula, we could also use middle term factorization to solve the quadratic equation x squared minus 5 x plus 6 equals 0
    in the following way:
    x squared minus 5 x plus 6 equals 0
    left parenthesis x minus 2 right parenthesis left parenthesis x minus 3 right parenthesis equals 0
    Thus we get values of x = 2 or 3 .

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