Question

How do you solve $ 4{c^2} + 10c = - 7 $ by completing the square?

Answer 1

Hint: First of all factor out the coefficient of then work with the quadratic expression which has the coefficient of which is equal to $ 1 $ and then check the coefficient of “c” in the new quadratic expression and take half of it and this is the number which goes into the complete square bracket and then balance the constant term by subtracting the square of the number and then place in the constant from the quadratic expression and simplify for the resultant value for “c”.

Complete step by step solution:
Take the given expression: $ 4{c^2} + 10c = - 7 $
Move the term from the right hand side of the equation on the left
Take out the coefficient of common from the above equation.
 $ 4\left( {{c^2} + \dfrac{{10}}{4}c + \dfrac{7}{4}} \right) = 0 $
Now to create the trinomial square on the left hand side of the equation find the value which is equal to the square of half of the “middle term”, “b”
 $ {\left( {\dfrac{b}{2}} \right)^2} = {\left( {\dfrac{{10}}{{4 \times 2}}} \right)^2} $
Note, $ 4\left[ {{{\left( {c + \dfrac{{10}}{{4.2}}} \right)}^2} - {{\left( {\dfrac{5}{4}} \right)}^2} + \dfrac{7}{4}} \right] = 0 $
Common factors from the numerator and the denominator cancel each other. Also, simplify the square of the terms.
 $ 4\left[ {{{\left( {c + \dfrac{5}{4}} \right)}^2} - \left( {\dfrac{{25}}{{16}}} \right) + \dfrac{7}{4}} \right] = 0 $
Simplify among the like terms.
 $ 4\left[ {{{\left( {c + \dfrac{5}{4}} \right)}^2}\underline { - \dfrac{{25}}{{16}} + \dfrac{{28}}{{16}}} } \right] = 0 $
When denominators are the same, numerators are directly simplified.
 $ 4\left[ {{{\left( {{c^2} + \dfrac{5}{4}} \right)}^2} + \dfrac{3}{{16}}} \right] = 0 $
Now, multiply the term outside the bracket with both the terms inside the bracket.
 $ \left[ {4{{\left( {c + \dfrac{5}{4}} \right)}^2} + 4.\dfrac{3}{{16}}} \right] = 0 $
Now, move the constant on the right hand side of the equation.
 $ 4{\left( {c + \dfrac{5}{4}} \right)^2} = - \dfrac{{4.3}}{{16}} $
Common multiples from both the sides of the equation cancel each other.
 $ {\left( {c + \dfrac{5}{4}} \right)^2} = - \dfrac{3}{{16}} $
Take the square root on both sides of the equation.
 $ \sqrt {{{\left( {c + \dfrac{5}{4}} \right)}^2}} = \sqrt {\dfrac{{ - 3}}{{16}}} $
Square and square root cancel each other on the left hand side of the equation.
 $ \left( {c + \dfrac{5}{4}} \right) = \pm \dfrac{{\sqrt { - 3} }}{4} $
Make the required unknown the subject “c”
 $ c = - \dfrac{5}{4} \pm \dfrac{{\sqrt { - 3} }}{4} $
This is the required solution.
So, the correct answer is “ $ c = - \dfrac{5}{4} \pm \dfrac{{\sqrt { - 3} }}{4} $ ”.

Note: Always remember that when the square of negative number is taken, it always gives the positive term. Square is the number multiplied by itself twice. Square and square root always cancel each other. Always remember when you move any term from one side to the another, then the sign of the term also changes. Positive term is changed to negative and the negative term becomes positive.