SOLVING QUADRATIC EQUATIONS



Note:




Solve for x in the following equation.

Example 3:text2html_wrap_inline253tex2html_wrap_inline471


The equation is already set to zero.


Almost all precalculus students hate fractions before they renew their love for them. The following steps will transform the equation into an equivalent equation without fractions.


If you have forgotten how to manipulate fractions, click on Fractions for a review.


If you have forgotten what equivalent means, think of a dollar. You can represent the dollar with a dollar bill, 10 dimes, 20 nickels, or 100 pennies. All of these are equivalent because all have the value of a dollar. Got it. If not, click on Equivalence for a review.


Remove all the fractions by writing the equation in an equivalent form without fractional coefficients. In this problem, you can do it by multiplying both sides of the equation by 56. All the denominators 8, 28, and 7 divide into 56 evenly.

eqnarray46


eqnarray56







Method 1:text2html_wrap_inline253Factoring

The equation tex2html_wrap_inline473 can be factored as follows:


eqnarray70


eqnarray74







Method 2:text2html_wrap_inline253Completing the square

Divide both sides of the equation tex2html_wrap_inline475 by 21.


eqnarray84



Add tex2html_wrap_inline477 to both sides of the equation: tex2html_wrap_inline479



Simplify: tex2html_wrap_inline481



Add tex2html_wrap_inline483 to both sides of the equation:


displaymath469



Factor the left side and simplify the right side:

eqnarray135



Take the square root of both sides of the equation:

eqnarray142



Subtract tex2html_wrap_inline485 from both sides of the equation:


eqnarray150








Method 3:text2html_wrap_inline253Quadratic Formula

The quadratic formula is tex2html_wrap_inline487

In the equation tex2html_wrap_inline489 ,a is the coefficient of the tex2html_wrap_inline491 term, b is the coefficient of the x term, and c is the constant. Substitute tex2html_wrap_inline495 for a, tex2html_wrap_inline497 for b, and tex2html_wrap_inline499 for c in the quadratic formula and simplify.



eqnarray200


eqnarray207


eqnarray214







Method 4:text2html_wrap_inline253Graphing


Graph the equation, tex2html_wrap_inline501 (formed by subtracting the right side of the equation from the left side of the equation). Graph tex2html_wrap_inline503 (the x-axis). What you will be looking for is where the graph of tex2html_wrap_inline505 crosses the x-axis. Another way of saying this is that the x-intercepts are the solutions to this equation.


You can see from the graph that there are two x-intercepts, one at tex2html_wrap_inline509 and one at tex2html_wrap_inline511 .


The answers are tex2html_wrap_inline515 and tex2html_wrap_inline517 These answers may or may not be solutions to the original equations. You must verify that these answers are solutions.



Check these answers in the original equation.



Check the solution tex2html_wrap_inline523 by substituting tex2html_wrap_inline515 in the original equation for x. If the left side of the equation

equals the right side of the equation after the substitution, you have found the correct answer.


Since the left side of the original equation is equal to the right side of the original equation after we substitute the value tex2html_wrap_inline515 for x, then tex2html_wrap_inline523 is a solution.

Check the solution tex2html_wrap_inline535 by substituting tex2html_wrap_inline537 in the original equation for x. If the left side of the equation equals the right side of the equation after the substitution, you have found the correct answer.

Since the left side of the original equation is equal to the right side of the original equation after we substitute the value tex2html_wrap_inline537 for x, then tex2html_wrap_inline535 is a solution.







The solutions to the equation tex2html_wrap_inline547 are tex2html_wrap_inline515 and tex2html_wrap_inline551







Comment:text2html_wrap_inline253Recall that when we solved this equation by factoring, we factored the expression tex2html_wrap_inline553 not the original expression tex2html_wrap_inline555 The product of the factors of tex2html_wrap_inline557 does not equal the original expression because the product of the first terms of the factors must equal the first term of the original expression.


We need to add a constant factor.

eqnarray351



Product of factors tex2html_wrap_inline559 equals tex2html_wrap_inline561 What number do I multiply 21 by to get tex2html_wrap_inline565

eqnarray364


Let's us see if tex2html_wrap_inline567 equals the original tex2html_wrap_inline569

eqnarray384


eqnarray391


eqnarray407




The factors of tex2html_wrap_inline571 are tex2html_wrap_inline573



We have illustrated that the solutions to the equation tex2html_wrap_inline575 are tex2html_wrap_inline577 and tex2html_wrap_inline579








Are you secure in your ability to solve this type of equation? If not and if you would like to work another example, click on Example


If you want to verify that you know how to work this type of problem by testing yourself over problems similar to the one above, click on Problem


If you would like to go back to the equation table of contents, click on Contents



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Author: Nancy Marcus

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