Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

If there's a problem like 3x^2 - 4x - 7 = 0, can that be solved the same way as the quadratic equation formula?

Can it be solved the same way as the quadratic equation formula ax^2 + bx + c = 0? (Except obviously instead of the + there's - )

5 Answers

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  • 1 month ago
    Favourite answer

    Yes. You just use negative values for b and c:

    a = 3

    b = -4

    c = -7

    x = [ -b ± √(b² - 4ac)] / (2a)

    x = [ -(-4) ± √((-4)² - 4(3)(-7))] / (2 * 3)

    x = [ 4 ± √(16 + 84)] / 6

    x = [ 4 ± √(100)] / 6

    x = (4 ± 10) / 6

    x = -6/6 and 14/6

    x = -1 and 7/3

  • 1 month ago

    3x^2-4x-7=0

    =>

    (3x-7)(x+1)=0

    =>

    x=7/3 or x=-1

    Using the quadratic formula, get

    a=3

    b=-4

    c=-7

    =>

    x=[4+/-sqr((-4)^2-4*3(-7)]/(2*3)

    =>

    x=[4+/-sqr(16+84]/6

    =>

    x=[4+/-sqr(100)]/6

    =>

    x=[4+/-10]/6

    =>

    x=14/6=7/3

    or

    x=-6/6=-1

    same as before.

  • David
    Lv 7
    1 month ago

    Yes, by using the quadratic equation formula 3x^2 -4x -7 = 0 the value of x = -1 or 7/3

  • ted s
    Lv 7
    1 month ago

    COMMENT  : you really have 3 x² + (-4) x + ( - 7 ) = 0...I may already told you that subtraction and division are not operations on the real numbers...subtraction of W is the addition of the opposite , + ( - W) , and division by K is multiplication by a ' multiplicative inverse ' , A / K = A B where KB = 1

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  • rotchm
    Lv 7
    1 month ago

    Yes. Ax² + Bx + C = 0 can be solved with the quadratic eqs, no matter the A,B & C  (as long A <> 0). 

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