Question

For each of the following equations, show that the given value of x is a solution of the equation by direct substitution.

a) x = 4/3, Equation: 3x^2 + 5x - 12=0

b) x=1 + square root 6 , Equation: x^2 - 2x-5=0

c) x= 1 + 2i, Equation: x^2 - 2x+5=0

Answer 1

a.

The value x = 4/3 is a solution of the equation 3x² + 5x - 12 = 0

Since the equation is 3x² + 5x - 12 = 0, if x = 4/3, is a solution, substituting x = 4/3 into the equation will make the equation will make the left hand side equal zero.

So, 3x² + 5x - 12 = 0

Substituting x = 4/3, we have

3(4/3)² + 5(4/3) - 12

= 3(16/9) + 20/3 - 12

= 16/3 + (20 - 12 × 3)/3

= 16/3 + (20 - 36)/3

= 16/3 - 16/3

= 0

So, x = 4/3 is a solution of the equation 3x² + 5x - 12 = 0

b.

The value x = 1 + √6 is a solution of the equation x² - 2x - 5 = 0

Since the equation is x² - 2x - 5 = 0, if x = 1 + √6, is a solution, substituting x = 1 + √6 into the equation will make the equation will make the left hand side equal zero.

So, x² - 2x - 5 = 0

Substituting x = 4/3, we have

x² - 2x - 5 = (1 + √6)² - 2(1 + √6) - 5

= (1² + 2√6 + (√6)²) - 2 - 2√6) - 5

= 1 + 2√6 + 6 - 2 - 2√6) - 5

= 2√6 + 6 + 1 - 2√6) - 5 - 5

= 2√6 + 7 - 2√6) - 7

=  2√6 - 2√6 + 7 - 7

= 0 + 0

= 0

So, x = 1 + √6 is a solution of the equation x² - 2x - 5 = 0

c.

The value x = 1 + 2i is a solution of the equation x² - 2x + 5 = 0

Since the equation is x² - 2x + 5 = 0, if x = 1 + 2i, is a solution, substituting x = 1 + 2i into the equation will make the equation will make the left hand side equal zero.

So, x² - 2x + 5 = 0

Substituting x = 1 + 2i, we have

x² - 2x + 5 = (1 + 2i)² - 2(1 + 2i) + 5

= (1² + 2(2i) + (2i)²) - 2 - 4i + 5

= 1 + 4i - 4 - 2 - 4i + 5

=  4i + 1 - 4 - 2 - 4i + 5  

= 4i - 5 - 4i + 5  

= 4i - 4i + 5 - 5

= 0 + 0

= 0

So, x = 1 + 2i is a solution of the equation x² - 2x + 5 = 0

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