Steps in solving quadratic equation by extracting the square root


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Were the steps of solving quadratic equations by extracting square roots helpful to you? Why?

Answered by admin @



Answer:

Recall that a quadratic equation is in standard form1 if it is equal to 0:

ax2+bx+c=0

where a,b, and c are real numbers and a≠0. A solution to such an equation is a root of the quadratic function defined by f(x)=ax2+bx+c. Quadratic equations can have two real solutions, one real solution, or no real solution—in which case there will be two complex solutions. If the quadratic expression factors, then we can solve the equation by factoring. For example, we can solve 4x2−9=0 by factoring as follows:

4x2−9 =0 (2x+3)(2x−3) =0

2x+3=0 or 2x−3=0 2x=−32x=3 x=−

3

2

x=

3

2

The two solutions are ±

3

2

. Here we use ± to write the two solutions in a more compact form. The goal in this section is to develop an alternative method that can be used to easily solve equations where b=0, giving the form

ax2+c=0

The equation 4x2−9=0 is in this form and can be solved by first isolating x2.

4x2−9 =0 4x2 =9 x2 =

9

4

If we take the square root of both sides of this equation, we obtain the following:

x2

=

9

4

|x| =

3

2

Here we see that x=±

3

2

are solutions to the resulting equation. In general, this describes the square root property2; for any real number k,

Ifx2=k, then x=±

k

Applying the square root property as a means of solving a quadratic equation is called extracting the root3. This method allows us to solve equations that do not factor.

Example 6.1.1:

Solve: 9x2−8=0.

Solution

Notice that the quadratic expression on the left does not factor. However, it is in the form ax2+c=0 and so we can solve it by extracting the roots. Begin by isolating x2.

9x2−8 =0 9x2 =8 x2 =

8

9

Next, apply the square root property. Remember to include the ± and simplify.

x =±

8

9

2

2

3

For completeness, check that these two real solutions solve the original quadratic equation.

Check x=−

2

2

3

Check x=

2

2

3

9x2−8 =0 9(−

2

2

3

)2−8 =0 9(

4.2

9

)−8 =0 8−8 =0 0 =0✓

9x2−8 =0 9(

2

2

3

)2−8 =0 9(

4.2

9

)−8 =0 8−8 =0 0 =0✓

Table 6.1.1

Answer:

Two real solutions, ±

2

2

3

Sometimes quadratic equations have no real solution. In this case, the solutions will be complex numbers.

Example 6.1.2:

Solve: x2+25=0.

Solution

Begin by isolating x2 and then apply the square root property.

x2+25 =0 x2 =−25 x =±

−25

After applying the square root property, we are left with the square root of a negative number. Therefore, there is no real solution to this equation; the solutions are complex. We can write these solutions in terms of the imaginary unit i=

−1

.

x =±

−25

−1⋅25

=±i⋅5 =±5i

Check x=−5i

Check x=5i

x2+25 =0 (−5i)2+25 =0 25i2+25 =0 25(−1)+25 =0 −25+25 =0 0 =0✓

x2+25 =0 (−5i)2+25 =0 25i2+25 =0 25(−1)+25 =0 −25+25 =0 0 =0✓

Table 6.1.2

Answer:

Two complex solutions, ±5i.

Exercise 6.1.1

Solve: 2x2−3=0.

Answer

Consider solving the following equation:

(x+5)2=9

To solve this equation by factoring, first square x+5 and then put the equation in standard form, equal to zero, by subtracting 9 from both sides.

(x+5)2 =9 x2+10x+25 =9 x2+10x+16 =0

Factor and then apply the zero-product property.

x2+10x+16 =0 (x+8)(x+2) =0 x+8 =0 or x+2=0 x =−8x=−2

The two solutions are −8 and −2. When an equation is in this form, we can obtain the solutions in fewer steps by


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