CHAPTER VII

EQUATIONS WHICH ARE REDUCIBLE TO THE SECOND DEGREE

BIQUADRATIC EQUATIONS. ax1 + bx2 + c = 0

425. Special Case of the Biquadratic Equation

ax1 +4 bx3 + 6 cx2 + 4 dx + e =

0.

The solutions of many equations, not really quadratics, may be reduced to the solutions of a quadratic equation. For example, suppose that

-25 x2 + 144 = 0.

Examples of this kind are equations of the general biquadratic

If both values of z are real and positive, then the values of x, x1, x, x, x1, are all real; if both values of z are real, one positive and one negative, then two values of x are real and the other two imaginary; in case both values of z are negative or imaginary, then all the values of x, x,, x2, x, x, are imaginary.

IRRATIONAL EQUATIONS

426. An equation is said to be irrational when the unknown quantity appears under the radical.

where P is a polynomial involving x to the first power and Qa polynomial of the second degree.

and both members are squared, an equation free from radicals will result

which in general will be an equation of the second degree in x, because is of the second degree, P of the first, and P2 of the second. Every root of equation (1) is a root of equation (2), but the converse is not true. For, if the second term of equation (1) had the - sign, on transposing the result would be

Hence the roots of equation (2) may satisfy either equation P+ √ Q = 0, or P-VQ= 0, or both equations. This point will be illustrated immediately.

As examples illustrating a doubtful solution, consider the following equations:

On substituting these values in the equation, it is seen that x = = 3 satisfies the equation, but 18 does not. However, 18 satisfies the equation

√5x+10= 8

x.

This equation is included in the discussion above, for, on squaring, the same result is obtained as on squaring 15x + 10 = 8 - X. Hence it is not possible to be sure that the values of x which are finally found will satisfy the given equation; they may satisfy the equation formed by changing the sign of one or more radicals.

2. Solve the equation,

√x + 2 + √2 x + 2 = x.

√2x+2=x — - Vic + 2,

2x+2x-2x √ x+2+x+2;

Transpose, then

square,

x=3±19+7=3+4=7 or 1.

The value 7 satisfies the equation, but 1 will not; for

which is impossible. However, x 1 will satisfy the equation,

x= —

− 1 x + 2 + 1 2x + 2 = x.

It follows from these two examples that, in case an equation has been reduced to a rational form by squaring it, it will be necessary to observe whether the values found for x will satisfy the given equation in its original form.

This caution applies, for example, to equations like (9), (10), (12), 412.

Solution of Equations which can be Reduced to the Form

ax2+2bx+211 ax2 + 2 bx + c = p

427. Solve 2 x2 3x+81/2x2 3x-4=13.

Put the equation in the quadratic form:

(2x2 - 3x-4)+ 8(2x2 -3x-4)= 9,

and, solving, get

(2x2 - 3x-4)= −4±1/16 +9=-4+5=1 or -9.

In general, on adding c to both members of the general equation above, it follows that

ax2+2bx + c + 211 ax2 + 2 bx + c = p + c,

and on solving

Vax2 + 2 bx + c = − l± √ F + (p + c).

Hence there are two equations to solve:

ax2+2bx+c=(-l+V P2 + (p+ c))2=202+p+c−2lv/l2+p+c, and ax2+2bx+c=(−l—V ́l2+(p+c))*=2l2+p+c+2lv/F+p+c. One must be careful to select the roots which satisfy the given equation (426).

Solution of the Equation ax + bx" + c = 0

428. Here n may be an integer, or a fraction which is positive or negative. Put x” = z, x2” = 22; then the equation ax2+ bx”+ c=0

If n is an even integer, every positive value of z=

-b± √ b2-4ac 2 a

furnishes two pairs of real and equal values of r with contrary signs; in case one value of the z is positive and the other negative, two of the values of x are real, equal, and of contrary signs, and the other two are imaginary; if both values of z are negative or imaginary, all the values of x are imaginary. In case n is odd, every real value of z gives a real value of x of the same sign, and only one.

Put

P = 1 9

XP

1. Suppose, for example, that x + 4√x-21= 0. x=z; whence x = z2 and the equation becomes

[126]

-2+5=3 or

- 7,

x-3,

=

Put z = x2, z=x-6; then the equation becomes

429. The solution of a very important class of equations of the fourth degree (biquadratic equations) can be reduced to the solution of a quadratic equation. Consider the equations,

in which the coefficients equally distant from the ends are numer