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30. √(2x+4)+ √(3x+7)=√(12x+9).
31. √(6x + 1) + √2(1 − x) = √(7x+6).
32. 2√(2x+3)- √(5x+1)=√2(x − 1).
33. √(x-1)(x-2)+ √(x-3)(x − 4)=√2.

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34. √(x+1)+

2.

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√(1+x)

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38. √(α-x)+√(b − x)=√(a + b − 2 x).

39. √(ax+b2) + √(bx+ a2) = a - b.

40. √(a + x)+ √(b + x) = √(a+b+2x).

41. √(α − x)+ √(b − x) = √(2 a + 2b).

183. Relations between the roots and the coefficients of a quadratic equation.

We have seen [Art. 122] that a2+p+q can always be expressed in the form (x — «) (x − ß).

We have also seen [Art. 128] that if

x2+px + q = (x − α) (x − ß),

then a and ẞ are the roots of the equation

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Hence in the equation

x2+px + q = 0,

the sum of the roots is

p and the product of the roots is q.

The equation ax2+bx+c=0 becomes, on dividing by a,

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Hence, from the above, if a and ẞ be the roots of

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The above relations between the roots of a quadratic equation and the coefficients of the different powers of the unknown quantity are of great importance.

Analogous relations hold good for equations of the third and of higher degrees. [See Treatise on Algebra, Art. 129.]

These relations may also be deduced as follows:
We have found in Art. 177 that the roots of

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Again, the quotient of a +ẞ by aß is

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184. Special Forms. If one or more of its coefficients vanish, the quadratic equation and the formula for its roots become simplified.

(i.) If c=0, the equation reduces to

x (ax+b)=0,

the roots of which are 0 and - b/a.

ax2

=

(ii.) If c 0 and also b=0, the equation becomes 0, both roots of which are zero.

(iii.) If b= 0, the equation reduces to

ax2 + c = 0,

the roots of which are +√-c/a and −√-c/a. The roots of a quadratic equation are therefore equal and opposite when the coefficient of x is zero.

(iv.) If a, b, and c are all zero, the equation is clearly satisfied for all values of x.

(v.) The case in which either a = =0, or a=0 and b=0 (c not zero), is best studied through the general expressions for the roots of ax2 + bx + c = 0, but in the changed form

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· b − √ (b2 — 4 ac)' −b+√(b2 −4 ac)

That these expressions are respectively equal to

− b + √ (b2 − 4 ac) -b-√(b2 — 4 ac)

2 a

follows from the identity

2 a

{−b+ √(b2 4 ac)} {− b −√(b2-4ac)} = 4 ac. If now a 0, one root becomes -c/b, and the other 2c/0. And if a = = 0 and b = 0, both roots become 2c/0.

A fraction, such as 2c/0, whose numerator is a finite quantity and whose denominator is zero, was called, in Art. 168, an infinite quantity, or infinity, and was denoted by the symbol ∞. Thus 2c/0

=∞.

In accordance with this definition and with the results here obtained, when it becomes necessary to interpret the equations

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we say that the former has one infinite root, the latter two infinite roots. Such equations present themselves in investigations in analytic geometry.

Ex. 1. Solve the equation

√ax + 1+ √x + 4 = 3.

Squaring both members, we have

(a + 1)x + 5 + 2√(ax + 1)(x + 4) = 9 ;

then transposing the rational terms to the right side and squaring again, we obtain

4(ax + 1)(x + 4) = 16 − 8(a + 1)x + (a + 1)2x2,

whence, by the proper reductions and transpositions,

(a− 1)2x2 - 12(2 a + 1)x = 0.

Hence, one root is x = 0, and for this one both square roots must be taken positively. A second root is

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and it requires √x + 4 to be taken negatively.

If a 1, the coefficient of x2 is zero, and the second root is infinite.

Ex. 2. Solve the equation

√(ax)+√(x + 1) = √√b,

and determine for what value of b one root is zero, and for what value of a one root is infinite.

185. Equations with given Roots. Although we cannot in all cases find the roots of a given equation, it is very easy to solve the converse problem, namely, the problem of finding an equation which has given roots.

For example, to find the equation whose roots are 4 and 5.
We are to find an equation which is satisfied when x = 4, or

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