Linear and Quadratic Equations
Mathematics
FIRST DEGREE EQUATIONS AND SECOND DEGREE EQUATIONS 1
FIRST DEGREE EQUATIONS
A first degree equation (or linear equation) is an expression with two polynomials of first degree separated by an equal sign (=). The equality is true only for particular values of the variables. For example, the linear equation: x + 6 = 2x + 1 is true only when x = 5, because: 5+ 6=2·5+1 11 = 11 FIRST EXPRESSION
SECOND EXPRESSION
x + 6 = 2x + 1 TERMS
On each side of the equaI sign there is an expression; the addends are called terms; x is the variable or unknown. The value for which the equation is true is called solution (in the previous example the solution is x = 5).
2
HOW TO SOLVE LINEAR EQUATIONS To “solve” the equation is to calculate the value of “x” for which the equality is true.
To solve a first degree equation we find a simpler equivalent equation (with the same solution). To obtain an equivalent equation: •
If we add (or subtract) a variable or a constant from both expressions in an equation, the result is another equivalent equation.
•
If we multiply (or divide) both expressions in an equation by the same number, the result is another equivalent equation.
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Linear and Quadratic Equations
Mathematics
EXAMPLE Solve:
4x – 6 = x 4x – 6 + 6 = x + 6 4x = x + 6 4x – x = x – x + 6 3x = 6 3x /3 = 6 /3 x=2
These rules must be followed, usually in this order, to solve easy linear equations: 1 If there are brackets we must remove them (expand the expression). 2 Any x-term or number that is adding (positive sign) moves to the other side subtracting (negative sign) and vice versa. 3 If there are x-terms on both sides, collect them on one side and do the same with the numbers. 4 A number that is multiplying (the whole expression) on one side moves to the other side dividing and vice versa. EXAMPLE Solve:
6x – 3 = 3(x + 1) 6x – 3 = 3x + 3 6x – 3x = 3 + 3 3x = 6 x =6/3=2
3
LINEAR EQUATIONS WITH DENOMINATORS Solving a complex equation requires organising the calculations in these steps: 1 If there are brackets remove them.
2 Remove the denominators multiplying both sides by the LCM (least common multiple) of the denominators.
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Linear and Quadratic Equations
Mathematics
3 Transpose the like terms (move to one side the x-terms and the numbers to the other side). 4 Combine like terms. 5 Isolate the unknown (move the coefficient of the x to the other side). EXAMPLE x+ 4 x−1 x+ 1 − = 3 2 6
Solve:
l. c. m (3,2,6) = 2·3 = 6
1 Remove denominators: 6·
( ) ( ) ( ) x+ 4 x−1 x+ 1 −6 · =6· 3 2 6
2 · ( x+ 4) −3 · ( x−1 ) =x+ 1 2 Remove brackets: 2 x+ 8−3 x+ 3=x+ 1
3 Tranpose and combine like terms:
8+ 3−1=3x−2x+ x 10=2x 5 Isolate the unknown: x=
3/8
10 =5 2
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Linear and Quadratic Equations
4
Mathematics
SECOND DEGREE EQUATIONS Second degree equations ( or quadratic equations) with one unknown verify: a. They have got an only unknown. b. They always have a term (no terms of higher degree); x2 in general they have also a x- term and a number. Generally they have two different solutions. General expression: 2
a x + bx + c = 0 a≠ 0
To solve quadratic equations we will distinguish four cases: CASE 1. If b = 0:
a x2 + c = 0 ax 2=−c x2 =
−c a
And we can obtain: •
If x2 = −c > 0 , there are two different solutions. a x1 =−
•
√
−c a
x2 =+
√
−c a
If x2 = −c < 0 , there is not any solution because the square of all a number gives a positive number.
EXAMPLE Solve the quadratic equation: 5x 2 − 20=0 → 5x 2 = 20 → x 2 = 4/8
5x 2−20=0 20 = 4 → x =±√ 4 5 Maths Department, IES Al-qázeres
Linear and Quadratic Equations
Mathematics
So there are two different solutions:
x1 =+ 2 x2 =−2
2
CASE 2. If c = 0:
a x + bx = 0
We factorise:
x ·( ax+ b)=0
The product of two numbers can only be zero if one or both are zero, so: x=0 and
ax + b = 0 → x =
−b a
x1 =0
There are two different solutions:
x2 =
−b a
EXAMPLE 2
5x −2x=0
Solve the quadratic equation:
5x 2 − 2x=0 → x (5x −2)=0 →
{
x =0 5x −2= 0 → 5x =2 → x =
2 5
}
So there are two different solutions: x1 =0 x2 =
CASE 3. If b, c = 0:
2
a x =0
2 5
a≠ 0
The only solution is x = 0, because all number multiplied by 0 is zero. 5/8
Maths Department, IES Al-qázeres
Linear and Quadratic Equations
Mathematics
EXAMPLE Solve the quadratic equation: 2
2
3x =0
2
3x =0 → x =0 → x=0
a x 2 + bx + c = 0
CASE 4. General expression:
We calculate the solutions with this formula: x=
−b ± √ b2 − 4ac 2a
•
If
2 (b − 4ac) > 0 we obtain 2 solutions.
•
If
(b2 − 4ac) = 0 we obtain an only solution.
•
If
(b2 − 4ac) < 0 there is not any solution, because we can´t calculate the square root of a negative number.
EXAMPLE Solve the quadratic equation:
2
x + 2x − 3 = 0
x2 + 2x − 3 = 0 → a = 1, b = 2, c = −3 −b ± √ b 2 − 4ac −2 ± √ 2 − 4 · 1 · (−3) −2 ± √ 16 = = 2a 2 · 1 2 2
x=
So our two solutions are:
{ 6/8
−2+ 4 2 = =1 2 2 −2 − 4 −6 x 2= = =−3 2 2 x1 =
}
Maths Department, IES Al-qázeres
Linear and Quadratic Equations
5
Mathematics
SOLVING PROBLEMS WITH LINEAR EQUATIONS AND QUADRATIC EQUATIONS
You must follow these 4 steps: I. Read the problem carefully and identify the unknown “x”. II. Write the problem in the form of an equation. III. Solve the equation. IV. Check your solution using the problem and give the answer in words.
EXAMPLE 1. Anne bought two pens and a marker for a total of five euros. What was the price of each of the items if a marker costs fifty cents more than a pen? PRICE (€) Pen
x
Marker
x + 0,50
Two pens and one marker
5
The price of two pens is 2x, so the equation is:
2x + x + 0,50 = 5
Now we solve the equation: 3x + 0,50 = 5 → 3x = 5 − 0,5 → 3x = 4,5 → x =
4,5 = 1,5 3
SOLUTION: a pen costs 1.5 € and a marker costs 2 €. ( We check the solution: two pens cost 2·1,5=3 € and the marker 2 €, so 3 €+2 € = 5 €). 2. Two natural numbers which differ 3, have a product of 88. Find them. The numbers are x and x+3. Equation:
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x · ( x + 3) = 88
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Linear and Quadratic Equations
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We solve the equation:
x · ( x + 3) = 88 → x 2 + 3x = 88 → x 2 + 3x − 88 = 0 −3 ± √ 32 − 4 · 1 · (−88) −3 ± √ 9 + 352 −3 ± √ 361 −3 ± 19 x= = = = 2· 1 2 2 2
{
−3 + 19 16 = =8 2 2 −3 −19 −22 x 2= = =−11 2 2 x1 =
}
We have to reject x = -11 because it is not a natural number so our numbers are x = 8 and x + 3 = 11. (We check: 8 · 11 = 88)
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