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Find the roots of the quadratic equation 3x^2 - 4x + 1 = 0

The roots of the quadratic equation

3x24x+1=03x^2 - 4x + 1 = 0

are ( \frac{1}{3} ) and ( 1 ).

To determine the roots of a quadratic equation in the standard form ( ax^2 + bx + c = 0 ), we can utilize the quadratic formula:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

In our equation, we have ( a = 3 ), ( b = -4 ), and ( c = 1 ). Substituting these values into the formula gives us:

x=(4)±(4)24(3)(1)2(3)x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(1)}}{2(3)}

Calculating step-by-step:

  1. Simplify the expression:
x=4±16126x = \frac{4 \pm \sqrt{16 - 12}}{6}
  1. Continue simplifying under the square root:
x=4±46x = \frac{4 \pm \sqrt{4}}{6}
  1. Calculate the square root:
x=4±26x = \frac{4 \pm 2}{6}

This results in two possible values for ( x ):

  1. When we take the positive root:
x=4+26=66=1x = \frac{4 + 2}{6} = \frac{6}{6} = 1
  1. When we take the negative root:
x=426=26=13x = \frac{4 - 2}{6} = \frac{2}{6} = \frac{1}{3}

Thus, the roots of the equation are ( \frac{1}{3} ) and ( 1 ).

To verify these roots, we can substitute them back into the original equation:

  1. Checking ( x = \frac{1}{3} ):
3(13)24(13)+1=03\left(\frac{1}{3}\right)^2 - 4\left(\frac{1}{3}\right) + 1 = 0

Simplifying this gives:

31943+1=03 \cdot \frac{1}{9} - \frac{4}{3} + 1 = 0 1343+33=0\frac{1}{3} - \frac{4}{3} + \frac{3}{3} = 0 14+33=0\frac{1 - 4 + 3}{3} = 0

This simplifies to:

03=0\frac{0}{3} = 0
  1. Checking ( x = 1 ):
3(1)24(1)+1=03(1)^2 - 4(1) + 1 = 0

This simplifies to:

34+1=03 - 4 + 1 = 0

Which is:

0=00 = 0

Both roots satisfy the original quadratic equation, confirming that our calculations are correct.

Answered by: Prof. Peter Brown
IB Maths Tutor
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Our families consistently gain offers from at least one of their target schools, including Eton, Harrow, Wellington and Wycombe Abbey.

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