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QUADRATIC EQUATIONS
A quadratic equation is always written in the form of:
ax^2 + bx + c = 0 where a ≠ 0
The form ax^2 + bx + c = 0 is called the standard form of a quadratic equation.
Examples:
x^2 − 5 x + 6 = 0 This is a quadratic equation written in standard form.
x^2 + 4 x =− 4 This is a quadratic equation that is not written in standard form but can be once we set the equation to: x^2 + 4 x + 4 = 0.
x^2 = x This too can be a quadratic equation once it is set to 0. x^2 − x = 0 (standard form with c=0).
Solving Quadratic Equations by Square Root Property
When x^2 = a , where a is a real number, then your x =± a
Examples: x^2 − 9 = 0 y^2 + 3 = 28 x^2 − 9 = 0 y^2 + 3 − 3 = 28 − 3 x^2 = 9 y^2 = 25 x =± 9 y = ± 25 x =± 3 y = ± 5
Solving Quadratic Equations by Factoring
It can also be solved by factoring the equation. Remember to always check your solutions. You can use direct substitution of the solutions in the equation to see if the solutions satisfy the equation.
Examples: x^2 − 5 x + 6 = 0 ( x − 3 )( x − 2 ) = 0 ← Factoring x
x x
x x
← Set it equal to 0 and solve for x
Now check if, x = 3 and x = 2 are the solutions of x^2 − 5 x + 6 = 0
Check:^3 5 3^6 9 15 6 0
2 x^2 + 7 x − 4 = 0 ( 2 x − 1 )( x + 4 ) = 0 2 1 0 2 1 1 2
x x x
x x
Another method of checking the solutions is by using one of the following statements:
The sum of the solutions = a − b or The product of the solutions = c a where a , b , and c are the coefficients in ax^2 + bx + c = 0.
Now we check if x = 2
(^1) and x = − 4 are the solutions of 2 x (^2) + 7 x − 4 = 0
Check: Using the sum of the solutions = 1 2
Based on the original equation = −^ b = − a
Now by using the product of the solutions = 1 2
Solution Using the Quadratic Formula
Factoring is useful only for those quadratic equations which have whole numbers. When you encounter quadratic equations that can not be easily factored out, use the quadratic formula to find the value of x:
x b^ b^ ac a
= −^ ±^ −
Examples: x^2 − 8 = − 2 x x^2 + 2 x − 8 = 0 ← Rewrite in standard form, where a = 1 , b = 2 and , c = − 8
x = −^2 ±^4 −^ 4 1^ −^8 2 1
← Plug in numbers into the equation
= −^ ±
= −^ ±
= 2 , − 4 ← The two rational solutions
3 x^2 − 13 x + 4 = 0 x = − −(^ )^ ±^ (^ −^ )^ − ( )( ) ( )
2
=^ ±
,^2
, ← The two rational solutions
In some cases you encounter repeated rational solutions. And to prove you have the right values you use the discriminant which gives you information about the nature of the solutions to the equation. Based on the expression b^2 − 4 ac , which is under the radical in the quadratic formula it can be found in the equation ax^2 + bx + c = 0.
I. When the discriminant is equal to 0, the equation has repeated rational solutions.
Example: x^2 − 2 x + 1 = 0
By using the discriminant (^) b^2 − 4 ac = ( − 2 ) 2 − 4 1( )( ) 1 = 0
x = − −(^ )^ ±^ (^ −^ )^ − ( )( ) ( )
2
= 2 ±^0
x = 1 1 , ← Repeated rational solutions
II. When the discriminant is positive and a perfect square, the equation has two distinct rational solutions.
Example: x^2 − 4 x + 3 = 0
By discriminant b^2 − 4 ac = ( − 4 ) 2 − 4 1( )( ) 3 = 4
x = − −(^ )^ ±^ (^ −^ )^ − ( )( ) ( )
2
= 4 ±^4
x = 3 1 , ← Two distinct rational solutions
III. When the discriminant is positive but not a perfect square, the equation has two irrational solutions.
Example: x^2 + 4 x − 6 = 0
The discriminant b^2 − 4 ac = ( ) 4 2 − 4 1( )( − 6 )= 40
x = −^4 ±^4 −^ 4 1^ −^6 2 1
( )^2 ( )( )
= −^4 ±^40
x = − 2 ± 10 ← Two irrational solutions IV. When the discriminant is negative, the equation has two complex number solutions.
Example: x^2 + 4 x + 6 = 0
The discriminant b^2 − 4 ac = ( ) 4 2 − 4 1( )( ) 6 = − 8
x = −^4 ±^4 −4 1^6 2 1
x = −^ ±^ − ← Two complex number solutions
- Check your solution.
2 2
x x x
← Both solutions are true: ( ) ( )
2 2
x x x
Let’s keep practicing with one more.
Example:
4 x^2 − 2 x − 5 = 0 ← One way to make the leading coefficient 1 is
( 2 ) (^ )
x − x − = by multiplying both sides of the equation by 1 4 (^2 1 5 ) 2 4
x − x − = ← Move the constant to the right side of the equation 2 1 5 2 4
x − x = 2 2 2 1 1 5 1 2 4 4 4
x − x + ^ ^ = +^
← Divide the middle term coefficient by 2, square it, and add it to both sides of the equation:
÷ = × = → ^
2 2 1 1 5 1 2 4 4 16
x − x + ^ = + 2 2 1 1 21 2 4 16 x − x + ^ =
(^) x − =
← Write the squared number in binomial form.
(^) x − = ±
← Find the square root of both sides and don’t forget the ± sign.
1 21 (^4 )
x − = ± ← Send the other number to the right side of the
equation. 1 21 4 4
x − = ± ← Try to solve it using the square root. If not possible
leave it in radical form.
21 1 4 4 x = ± + ← Solve for x.
1 21 4
x = ± ← Final answer.
QUADRATIC EQUATIONS – ANSWER TO EXERCISES
- x^2 − 2 x + 1 = 0 2. x^2 + 9 x + 20 = 0 ( x − 1 )( x − 1 ) = 0 ( x + 5 )( x + 4 ) = 0
x x
x x
x x
x x
- 3 x^2 − 5 x − 12 = 0 4. 6 x^2 + 9 x − 6 = 0 ( 3 x + 4 )( x − 3 ) = 0 3 2( x^2 + 3 x − 2 )= 0 3 2( x − 1 )( x + 2 )= 0 3 4 0 4 3
x x
x x
x x
x x
- x^2 + 3 x − 28 = 0 6. 3 x^2 − 2 x = 2 x + 7 ( x + 7 )( x − 4 )= 0 3 x^2 − 2 x − 2 x − 7 = 0 3 x^2 − 4 x − 7 = 0 x x
x x
( 3 x − 7 )( x + 1 ) = 0
3 7 0 7 3
x x
x x
- 4 x^2 − 12 x = 16 8. x^2 + 3 x = 0 4 x^2 − 12 x − 16 = 0 x x ( + 3 ) = 0 4 ( x^2 − 3 x − 4 )= 0 4 ( x − 4 )( x + 1 ) = 0 x = 0 x + 3 = 0 x = − 3
x x
x x
9. 3 + 1 =^102
x x
- 3 y^2 − y − 4 = 0 3 x^2 + x = 10 ( 3 y − 4 )( y + 1 ) = 0 3 x^2 + x − 10 = 0 ( 3 x − 5 )( x + 2 ) = 0
y y
y y
x x
x x
- y^2 + 2 y + 1 = 0 12. x^2 − 2 x − 8 = 0 ( y + 1 ) 2 = 0 ( x − 4 )( x + 2 )= 0 y y
x x
x x
- x^2 + 4 = 0 14. x^2 + x = − 1
x = −^0 ±^0 −4 1^4 2
( )( ) (^) x (^2) + x + 1 = 0
16 2
x = ± − ( )( ) ( )
x
x =− ±^ −
- 9 y^2 + 6 y − 8 = 0 16. y^2 − 25 = 0
x = −^6 ±^36 −^ 4 9^ −^8 2 9
( y − 5 )( y + 5 ) = 0
6 324 18
= −^ ± y y
y y
=^ −^ ±
x
x
=^ +
x
x
=^ −
- 6 y^2 − 13 y + 6 = 0 18. x x
( 3 y − 2 )( 2 y − 3 ) = 0 3 x^2 − 4 = − x 3 x^2 + x − 4 = 0 3 2 0 2 3
y y
y y
( 3 x + 4 )( x − 1 ) = 0
3 4 0 4 3
x x
x x
