Aberystwyth Mathematics Club Aberystwyth University
IMAPS
Quadratic equations. Preliminary notes.
1. Square roots.We consider a simple problem: if the square of some number is equal to 16, find this number. Solution of this exercise reduces to finding x which satisfies the equation x2 = 16 or
x2− 16 = 0. Writing the left hand side of the equation in (linear) factors, we obtain
(1) (x − 4)(x + 4) = 0.
Then there are two solutions of the problem: 4 and −4. This follows from the fact that when two numbers are multiplied, the result (their product) can only be zero if one of the numbers is zero.
The Arithmetic quadratic root of a number a is defined to be a non-negative number whose square is equal to a. The Arithmetic quadratic root is defined only for a ≥ 0 and it is denoted by√a.
From the definition it follows that √a2 = |a|. For example, √25 = 5, √0.01 = 0.1,
q
(1 −√5)2=√5 − 1.
Observe that √16 = 4 but equation x2 = 16 has two solutions: one of which coincides
with the arithmetic quadratic root of 16, i.e. equal to 4, while the second root differs from the first one by sign and is equal to −√16, i.e. −4.
Remember that the following statements are true:
(i) If a ≥ 0 and b ≥ 0, then √ab =√a√b;
(ii) If a ≥ 0 and b > 0, then r a b = √ a √ b; (iii) If a > b ≥ 0, then√a >√b.
Example 1. Find, for which values of x the expression√x − 2 +√6 − x is defined.
Solution. Arithmetic quadratic root √x − 2 is defined for x − 2 ≥ 0, i.e. for x ≥ 2; the root√6 − x is defined for 6 − x ≥ 0, i.e. for x ≤ 6.
Answer: 2 ≤ x ≤ 6.
Example 2. Pick out the multiplier of the root √−2a3b3, if a < 0 and b > 0.
Solution. As√a2= |a|, then √−2a3b3 = |ab|√−2ab = −ab√−2ab.
Example 3. Pick in multiplier under the root in the expression (√5 − 3) q 1 +√2. Solution. (√5 − 3)p1 +√2 = −(3 −√5)p1 +√2 = − q (3 −√5)2(1 +√2).
Example 4. Simplify the expression p(x − 2)2+ 8x −p(x + 1)2− 4x.
Solution. Since
(x − 2)2+ 8x = x2− 4x + 4 + 8x = x2+ 4x + 4 = (x + 2)2,
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we can conclude that p(x − 2)2+ 8x = |x + 2|. On the other hand,
(x + 1)2− 4x = x2+ 2x + 1 − 4x = x2− 2x + 1 = (x − 1)2, and, as a result,p(x + 1)2− 4x = |x − 1|.
Let us consider three cases: if x ≥ 1, then p (x − 2)2+ 8x −p(x + 1)2− 4x = (x + 2) − (x − 1) = 3; if −2 ≤ x < 1, then p (x − 2)2+ 8x −p(x + 1)2− 4x = (x + 2) + (x − 1) = 2x + 1;
and, finally, if x < −2, then p
(x − 2)2+ 8x −p(x + 1)2− 4x = −(x + 2) + (x − 1) = −3.
Answer: The result obtained can be written in an equivalent form: p (x − 2)2+ 8x −p(x + 1)2− 4x = −(x + 2) + (x − 1) = 3 x ≥ 1 2x + 1 −2 ≤ x < 1. −3 x < 2
Example 5. Simplify the fraction √ 1
2 +√5 in such a way that no irrational numbers are present in the denominator.
Solution. Multiply both the numerator and the denominator with algebraic “conjugate to √
2 +√5” which is √2 −√5 (coming from the rule of conjugates: (a + b)(a − b) = a2− b2)
and get √ 2 −√5 (√2)2− (√5)2 = √ 2 −√5 −3 = √ 5 −√2 3 .
2. Equations and rules for their transformations
We call an equation with variable (or unknown) x an equality of two expressions
(2) f (x) = g(x).
For example,
x2+ 1 = x − 3, x2− 1 = 0, |x| − 3 = 0, 2x − 1
x + 3 = x + 1.
A number a is called zero (or solution) of a given equation (2) with variable x, if after the substitution of a in both sides of the equation (2) we get a true equality, i.e, if for x = a the sides of the equation are defined and have equal values. We say x = a simplifies the equation. Examples:
a) The equation 3x2 = 0 has the unique solution x = 0, while the equation x2+ 2 = 0 does
not have any, as x2+ 2 ≥ 2 for all real values of the variable x.
b) The equation (x + 1)(x − 2) = 0 has only two solutions x = −1 and x = 2. For any x different from 1 and −2, the left hand side differs from zero, consequently, the equation does not have other solutions but −1 and 2 (see (1).
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c) The solutions of the equation x − 2
x − 2 = 1 are all numbers except x = 2. The number 2 is not a solution of the equation as for x = 2 the left hand side of the equation is undefined. This equation has (infinitely many) an infinite number of solutions.
d) The equation 5x = 5x is satisfied by all the real numbers, while the solutions of the equation |x| = x are all the nonnegative numbers.
To solve a given equation means to find all its zeros (solutions) or to prove that there are none.
Two equations with the same unknowns are called equivalent, if they they have the same set of solutions, i.e. if any solution of the first equation is also solution to the second one, and vice versa; or both equations do not have any solutions.
For example, equations 2x − 1 = 3 and (x − 2)(x2+ x + 1) = 0 are equivalent as each of
them has the real unique solution x = 2. The equations (x − 3)(x − 4) = 0 and (x − 3)2= 0 are not equivalent, as the number 4 is solution to the first equation but not to the second equation.
Let us formulate some rules which apply when transforming the equations, commonly used to solve the equations.
Rule Nr 1. If the function h(x) is defined for all x for which functions f (x) and g(x) are defined, then equations
f (x) = g(x) and f (x) + h(x) = g(x) + h(x) are equivalent. In particularly, the equations
f (x) = g(x) and f (x) − g(x) = 0 are equivalent.
By this rule, any summand can be carried over from one side of the equation to the other with the opposite sign, and the resulting transformed equation is equivalent to the original one.
Rule Nr 2. If the function h(x) is defined for all x, for which expressions f (x) and g(x) are defined, then any solution to the equation f (x) = g(x) is also solution to the equation
f (x) · h(x) = g(x) · h(x). If, moreover, h(x) 6= 0, for all x, then equations
f (x) = g(x) and f (x) · h(x) = g(x) · h(x) are equivalent.
Rule Nr 3. Each solution to the equation f (x) = g(x) is also a solution to the equation (f (x))n= (g(x))n for any natural number n.
3. Linear equations An equation of the form
(3) ax + b = 0,
where a and b are some given real numbers, is called linear equation.
• If a 6= 0, then equation (3) has the unique solution x = −b/a.
• If a = 0, but b 6= 0, then the equation (3) does not have any solutions;
• if a = 0 and b = 0 then any real number is a solution to (3).
Example 6. Solve
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Solutions.
a) Carry over the summand 4x to the left hand side of the equation, and the summand 7 over to the right hand side, changing their signs.: 3x − 4x = 3 − 7. This equation has the unique solution x = 4, consequently, the original equation has the unique solution x = 4.
b) Open parentheses and carry over the summand containing x from the right hand side of equation to the left hand side, the summand 6 over to the right hand side. Doing this we remember to change the signs of these summands. The result we get is 3x − 3x = 3 − 6. This equation is equivalent to the equation 0 · x = −3, which does not have any solutions, as consequence, also original equation does not have any solutions.
c) We transform this equation: 3x + 8 = 3x + 8. Any real number satisfies the obtained equation; consequently, the given equation is satisfied by any real number.
Example 7. Find, which pair of the equations given below are equivalent:
a) 2x + 1 = 7x − 9, and 3(2x + 1) = 7x + 1 b) 3x + 5 = x + 7, and 2x − (x + 3) = x − 1. Solutions.
a) The first equation is equivalent to the equation 2x − 7x = −1 − 9, with the solution x = 2. The second equation is equivalent to the equation 6x − 7x = −3 + 1, with unique solution x = 2. Thus, the given equations are equivalent.
b) The solution to the first equation is the number x = 1. The second equation is equivalent to the equation 0·x = 2, which does not have any solutions. Consequently, the given equations are not equivalent.
Advice. In attempt to solve an equation, try if possible to perform only equivalent transfor-mations. This usually helps and can sometimes shorten time. However, if you are not sure whether the transformations are equivalent or it is difficult to perform such transformation -do not worry -do your best in any other way. However, ensure that all possible solutions of the original equation are also zeros of the new one. Obtaining a solution for the last one, you will need only to check whether it is simultaneously a solution for the original equation.
