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Algebra 1

8.3.3 Solving More Complex Quadratic Equations

Algebra 18.3.3 Solving More Complex Quadratic Equations

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Activity

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

Notice that the quadratic term, xx, in the original form ax2=kax2=k is replaced with (xh)(xh).

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

2.

( u 6 ) 2 = 64 ( u 6 ) 2 = 64

3.

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

4.

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

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

Video: Solving More Complex Quadratic Equations

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

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(xh)2=ka(xh)2=k

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

Notice that the quadratic term, xx, in the original form ax2=kax2=k is replaced with (xh)(xh).

ax2=ka(xh)2=kax2=ka(xh)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 aa, the square root property can be used on (xh)2(xh)2.

Solve 4(y7)2=484(y7)2=48.

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

So, divide both sides of the equation by 4.

4(y7)2=484(y7)2=48

(y7)2=12(y7)2=12

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

y7=±12y7=±12

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

Solve for yy. Rewrite the two solutions.

y7=±12y7=±12

y7=±(4·3)y7=±(4·3)

y7=±23y7=±23

y=7±23y=7±23

y=7+23y=7+23, y=723y=723

Step 4 Check the solutions.

4(y7)2=484(y7)2=484(7+237)2 = ? 484(7237)2 = ? 484(23)2 = ? 484(23)2 = ? 484(12)2 = ? 484(12)2 = ? 4848=4848=484(y7)2=484(y7)2=484(7+237)2 = ? 484(7237)2 = ? 484(23)2 = ? 484(23)2 = ? 484(12)2 = ? 484(12)2 = ? 4848=4848=48

Try it

Try It: Solving Quadratic Equations of the Form a(xh)2=ka(xh)2=k

Solve the following quadratic equations.

1. 3(a3)2=3633(a3)2=363

2. 12(b+2)2=20012(b+2)2=200

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