Solve this quadratic equation by completing the square: − 2 t 2 + 3 t + 2 = 0

Question

Solve this quadratic equation by completing the square:  −  2      t          2        +  3  t      +        2    =  0

kuthiwenihca · Accepted Answer

Move the constant to the right side of the equation and change its sign:

- 2 t^{2} + 3 t = - 2

We want to isolate the unknown variable on one side, so divide both sides of the equation by

- 2

:

t^{2} - \frac{3}{2} t = 1

To complete the square while preserving the relation between the sides of the equation, we need to add the same value to both sides (there’s the addition and subtraction property of equality again):

t^{2} - \frac{3}{2} t + ? = 1 + ?

Write the expression as a product with the factors

2

and

t

so that our expression has the same structure as the formula we want to use:

t^{2} - 2 \times t \times \frac{3}{4} + ? = 1 + ?

Since

\frac{3}{4}

is part of the middle term, add

{(\frac{3}{4})}^{2}

to both sides of the equation:

t^{2} - 2 \times t \times \frac{3}{4} + {(\frac{3}{4})}^{2} = 1 + {(\frac{3}{4})}^{2}

The square of the sum formula is

a^{2} + 2 a b + b^{2} = (a + b)^{2}

so let’s use it to factor the expression on the left-hand side:

{(t - \frac{3}{4})}^{2} = 1 + {(\frac{3}{4})}^{2}

Evaluate the power:

{(t - \frac{3}{4})}^{2} = 1 + \frac{9}{16}

Calculate the sum:

{(t - \frac{3}{4})}^{2} = \frac{25}{16}

Take the square root of both sides of the equation, remembering to use both positive and negative roots. We can do this because of the rule stating if two expressions are equal, then their square roots are also equal:

t - \frac{3}{4} = \pm \frac{5}{4}

Separate the equation into

2

possible cases (one with the minus root and one with the plus root):

t - \frac{3}{4} = - \frac{5}{4}

t - \frac{3}{4} = \frac{5}{4}

Move the constants to the right-hand side of the equation and change their signs:

t = - \frac{5}{4} + \frac{3}{4}

t = \frac{5}{4} + \frac{3}{4}

Add the fractions:

t = - \frac{1}{2}

t = 2

There we go! The equation has

2

solutions:

t_{1} = - \frac{1}{2}, t_{2} = 2

Solve this quadratic equation by completing the square: -2t^(2) + 3t +2 = 0

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