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Solving More Complex Quadratic Equations

8.3.3 Solving More Complex Quadratic Equations

8.3.3 • Solving More Complex Quadratic Equations

Activity

Solving an equation of the form a ( x − h ) 2 = k a ( x − h ) 2 = k is a similar process to the equations we have previously solved.

Notice that the quadratic term, x x , in the original form a x 2 = k a x 2 = k is replaced with ( x − h ) ( x − h ) .

On the left, a x squared equals k. An arrow points right to a times (x minus h) squared equals k, with x minus h in red, showing the subsitution of x minus h for the x.

Once the binomial is isolated, the square root property can be used. You can reason the solution by taking the square root and then appropriately isolating the variable using the two different square roots.

Solve the following quadratic equations. Explain or show your reasoning.

1.

144 = ( n + 1 ) 2 144 = ( n + 1 ) 2

Compare your answer:

n = 11 n = 11 and n = − 13 n = − 13

Because 12 and −12 are the two numbers we can square to make 144 and n + 1 n + 1 is the number being squared, we need n n to be 1 less than 12 or −12. So n = 11 n = 11 and n = − 13 n = − 13 .

2.

( u − 6 ) 2 = 64 ( u − 6 ) 2 = 64

Compare your answer:

u = 14 u = 14 and u = − 2 u = − 2

Since 8 and −8 are the two numbers we can square to make 64 and u − 6 u − 6 is the number being squared, we need u u to be 6 more than 8 or −8. So u = 14 u = 14 and u = − 2 u = − 2 .

3.

( n − 5 ) 2 − 30 = 70 ( n − 5 ) 2 − 30 = 70

Compare your answer:

n = 15 n = 15 and n = − 5 n = − 5

To isolate the binomial, add 30 to both sides of the equation. Since 10 and –10 are the two numbers we can square to make 100 and n − 5 n − 5 is the number being squared, we need n n to be 5 more than 10 or –10. So n = 15 n = 15 and n = − 5 n = − 5 .

4.

4 ( x + 6 ) 2 − 5 = 31 4 ( x + 6 ) 2 − 5 = 31

Compare your answer:

x = − 3 x = − 3 and x = − 9 x = − 9

To begin isolating the binomial, add 5 to both sides of the equation. Then divide both sides of the equation by 4. Since 3 and –3 are the two numbers we can square to make 9 and x + 6 x + 6 is the number being squared, we need x x to be 6 less than 3 or –3. So x = − 3 x = − 3 and x = − 9 x = − 9 .

Are you ready for more?

Extending Your Thinking
1.

Solve the following quadratic equation. Be prepared to show your reasoning.

( 4 x − 3 ) 2 + 11 = 47 ( 4 x − 3 ) 2 + 11 = 47

Compare your answer:

Two solutions: x = 9 4 x = 9 4 and x = − 3 4 x = − 3 4 .

To isolate the binomial, subtract 11 from both sides of the equation. This leaves ( 4 x − 3 ) 2 = 36 ( 4 x − 3 ) 2 = 36 . When you take the square root of both sides, you end up with 4 x − 3 = ± 6 4 x − 3 = ± 6 . This leaves two equations to solve:

4 x − 3 = 6 4 x − 3 = − 6 4 x − 3 = 6 4 x − 3 = − 6

By solving each of them, you find the solutions x = 9 4 x = 9 4 and x = − 3 4 x = − 3 4 .

Video: Solving More Complex Quadratic Equations

Watch the following video to learn more about how to solve a quadratic equation of the form a ( x − h ) 2 = k a ( x − h ) 2 = k .

Access multimedia content

Self Check

Solve the following quadratic equation.

( x + 3 ) 2 − 16 = 48

  1. x = 5
  2. x = − 8 and x = 8
  3. x = − 11 and x = 5
  4. x = 11 and x = − 5

Additional Resources

Solving Quadratic Equations of the Form a ( x − h ) 2 = k a ( x − h ) 2 = k

We can use the square root property to solve an equation of the form a ( x − h ) 2 = k a ( x − h ) 2 = k as well.

Notice that the quadratic term, x x , in the original form a x 2 = k a x 2 = k is replaced with ( x − h ) ( x − h ) .

a x 2 = k a ( x − h ) 2 = k a x 2 = k a ( x − h ) 2 = k

Like before, the first step is to isolate the term that has the variable squared. In this case, a binomial is being squared.

Once the binomial is isolated, by dividing each side by the coefficient of a a , the square root property can be used on ( x − h ) 2 ( x − h ) 2 .

Solve 4 ( y − 7 ) 2 = 48 4 ( y − 7 ) 2 = 48 .

Step 1 - Isolate the quadratic term and make its coefficient 1.

So, divide both sides of the equation by 4.

4 ( y − 7 ) 2 = 48 4 ( y − 7 ) 2 = 48

( y − 7 ) 2 = 12 ( y − 7 ) 2 = 12

Step 2 - Use the square root property on the binomial.

y − 7 = ± 12 y − 7 = ± 12

Step 3 - Simplify the radical by pulling out the factors that are perfect squares.

Solve for y y . Rewrite the two solutions.

y − 7 = ± 12 y − 7 = ± 12

y − 7 = ± ( 4 · 3 ) y − 7 = ± ( 4 · 3 )

y − 7 = ± 2 3 y − 7 = ± 2 3

y = 7 ± 2 3 y = 7 ± 2 3

y = 7 + 2 3 y = 7 + 2 3 , y = 7 − 2 3 y = 7 − 2 3

Step 4 Check the solutions.

4 ( y − 7 ) 2 = 48 4 ( y − 7 ) 2 = 48 4 ( 7 + 2 3 − 7 ) 2 = ? 48 4 ( 7 − 2 3 − 7 ) 2 = ? 48 4 ( 2 3 ) 2 = ? 48 4 ( − 2 3 ) 2 = ? 48 4 ( 12 ) 2 = ? 48 4 ( 12 ) 2 = ? 48 48 = 48 ✓ 48 = 48 ✓ 4 ( y − 7 ) 2 = 48 4 ( y − 7 ) 2 = 48 4 ( 7 + 2 3 − 7 ) 2 = ? 48 4 ( 7 − 2 3 − 7 ) 2 = ? 48 4 ( 2 3 ) 2 = ? 48 4 ( − 2 3 ) 2 = ? 48 4 ( 12 ) 2 = ? 48 4 ( 12 ) 2 = ? 48 48 = 48 ✓ 48 = 48 ✓

Try it

Solving Quadratic Equations of the Form a ( x − h ) 2 = k a ( x − h ) 2 = k

Solve the following quadratic equations.

1. 3 ( a − 3 ) 2 = 363 3 ( a − 3 ) 2 = 363

2. 1 2 ( b + 2 ) 2 = 200 1 2 ( b + 2 ) 2 = 200

Here is how to solve these quadratic equations of the form a ( x − h ) 2 = k a ( x − h ) 2 = k :

  1. 3 ( a − 3 ) 2 = 363 3 ( a − 3 ) 2 = 363
    ( a − 3 ) 2 = 121 ( a − 3 ) 2 = 121
    a − 3 = ± 11 a − 3 = ± 11
    a − 3 = 11 a − 3 = 11
    a − 3 = − 11 a − 3 = − 11
    a = 14 a = 14 and a = − 8 a = − 8
  1. 1 2 ( b + 2 ) 2 = 200 1 2 ( b + 2 ) 2 = 200
    ( b + 2 ) 2 = 400 ( b + 2 ) 2 = 400
    b + 2 = ± 20 b + 2 = ± 20
    b + 2 = 20 b + 2 = 20
    b + 2 = − 20 b + 2 = − 20
    b = 18 b = 18 and b = − 22 b = − 22