Algebra Honors ( Period 6 & 7)

Quadratic Equations with Perfect Squares 12-1

In Chapter 5 you learned how to solve certain quadratic equations by factoring and in Chapter 11 you learned how to solve quadratics in the for
x² = k
as in x² = 49
x = ±7

This lesson extend to any quadratic equation involving a perfect square

if we have x² = k
if:
k > 0 then x² = k has two real-numbered roots with x = ±√k
if k = 0 then x² = k has one real numbered root x = 0
if k<0 nbsp="" sup="" then="" x="">2
  = k has no real numbered roots
m² = 49
√(m²) = ±√49
m = ±7
{-7, 7}


5r²= 45

5r² /5= 45/5
r²= 9
√r²= ±√9
r = ±3
{-3,3}

(x + 6)² = 64
√(x+6)² = ±√64
x+6 = ±8

be careful here
you now have
x = -6±8 which means

x = -6-8 = -14 AND x = -6+8 = 2
{-14, 2}

9r² = 121

9r²/9 = 121/9
r² = 121/9
√r²= ±√121/9
r = ±11/3

Make sure to check your solutions to insure that they both work

(x-3)² = 100
√ (x-3)² = ±√100
(x-3) = ±10
x = 3 ±10
x = 3 +10 = 13 and x = 3-10 = -7
{-7, 13}

5(x-4)² = 40
5(x-4)²/5 = 40/5
(x-4)² = 8
√ (x-4)² = ±√8
Now you need to simplify the Radical

√ (x-4)² = ±2√2
(x-4) = ±2√2
x = 4 ±2√2

{4 - 2√2, 4 + 2√2}



7(x - 8)² = -28
7(x-8)²/7 = -28/7
(x-8)² = -4
WAIT!!! that can't happen no real numbered solutions...

An equation that has a negative on one side and a perfect square as the other has NO REAL NUMBER Solutions

y² + 6y + 9 = 49
(y+3)² = 49
√(y+3)² = ±√49
(y+3) = ±7
y = -3 ±7
y = -3-7 = -10 and y = -3 +7 = 4
{-10. 4}

Perfect Squares like (x -4) ² and ( y + 3) ²

The square of any real number is always a non negative real number.
2(3x-5)² + 15 = 7 becomes
2(3x -5)² = -8
or (3x -5)² = -4... NO REAL number solution!!