RESOLUTION OF QUADRATICS INTO FACTORS.

262. The principles already set forth in Article (259) serve for the resolution of quadratic expressions of the form ax + bee into factors, each of which shall contain the first degree of only.

The equation ax2 + bx + c = 0 may be written

Letting r and r2 represent the roots, we have

Factor, by applying the foregoing principles, the expression 2x2+7x-15=0. Placing the expression equal to zero, we

Factor the quadratic expressions by the principles set forth:

CHAPTER XXVII.

SIMULTANEOUS EQUATIONS INVOLVING
QUADRATICS.

263. Various methods are employed for solving simultaneous equations where one or more may be of a degree higher than the first. The student must learn the proper application of the methods by experience. In no other section of Algebra can a student's ingenuity be so well exercised as in the solution of simultaneous quadratics.

The expressions

may be read

or,

and the expressions may be read

or,

x=5, y = ± 6,

x = +5, y =+ 6,
x=-5, y=-6;

x = ± 5, y = 6,

x=5, y=6,

x=-5, y=+6.
− 5, y = + 6.

We shall now explain by means of examples the different cases of solution of simultaneous quadratics.

264. CASE I.. SUBSTITUTION.

That is, the value of one

unknown quantity is substituted in terms of the other.

265. CASE II.. - ADDITION AND SUBTRACTION. That is, one unknown quantity is eliminated by addition or subtraction, either in the first instance, or after some simple operation has been performed.

266. CASE III.

SUBSTITUTING nx FOR y.

When the equa

tions are of the second degree and homogeneous.

267. CASE IV.

AND Y.

and y.

REMOVING THE HIGHEST POWERS OF X When the equations are symmetrical with respect to x

268. CASE V. SUBSTITUTING a+b FOR X AND α b FOR Y. When the equations are symmetrical with respect to x and y.

269. CASE VI. - DIVISION AND FACTORING. When the equations admit of factoring or the exact division of one into the other.

Substitute this value of 2 in (2), and we have,

Simplify,

2(6432y+4y2) + y2 = 17.

9y64y-111.

Completing the square, we have,

9 y2 — 64 y + (3,2)2 = − 111 +(32)2.

Taking the square root of each member, we have,

Reducing,

3 y 32 = ± 5.

y =

37 or 3.

Substitute these values of y in (1), and we have,

(1)

(2)

(3)

As the equations in the above examples are of the second degree and homogeneous, a solution may also be had by Case III.

Substituting these values of x in (2), we have,

Substituting in (4), we obtain other values of x, and substituting these values in (1) or (2), we obtain other values

of y.

Case IV. Solve the equations,

x + y = 3, 24+ y2 = 17.

(1)

(2)

Raising (1) to the fourth power, we have,