These two last values are impossible.

6. Given 2x-7x=99, to find x. If we let √xy, we shall have

If we take 9 for the value of y, we find 2-81. But if we take

find 121.

4

for the value of y, we

7. Given 2x2+√2x2+1=11, to find the four values of x.

Adding 1 to both members of the equation, we have

The first of these gives

The second gives

x=2, or x = -2.

x=√30, or x=—↓√30.

8. Given x6Sx3-513, to find one of the values of x.

Ans. x=3.

9. Given x+4x2-12, to find two values of x.

Ans. x= √2.

(119.) When, in the general form for quadratics a=1, the rule under Art. 117, is susceptible of considerable modification.

If we substitute 1 for a in formulas (A) and (B), Art. 117, they will become

Therefore, when the coefficient of x is a unit, the value of x can be found from formula (D).

All quadratic equations may be put under the form of (C), by dividing all the terins by the coefficient of x2, so that the expression of (D) for x must be as general as that of (B), Art. 117.

Hence, for the solution of quadratic equations, we have this second

RULE.

Having reduced the equation to the form x2+bx=c, we can find x, by taking half the coefficient of the second term, with its sign changed, plus or minus the square root of the square of the half of the coefficient of the second term, increased by the term independent of x.

EXAMPLES.

1. Given x2-10x=-24, to find x.

In this example, half the coefficient of the second term is 5, which squared and added to -24, the term independent of x is 1. Extracting the square root of 1, we have 1.

This, cleared of fractions, becomes

3x2-5x=7x+420.

4. Given 3x+42x3-3321, to find x.

x2

5. Given x(x—1)—2——(5—x) (1+2)+1, to find x.

7. Given x2-8x=-17, to find the values of x.

Ans. x=4+√=1, or 4—√=1.

(120.) EQUATIONS CONTAINING TWO OR MORE UNKNOWN QUANTITIES, WHICH INVOLVE IN THEIR SOLUTION QUADRATIC EQUATIONS.

Subtracting 4 times the second from the (3), we get

x2-2xy+y2=36.

The square root of (4) is

x—y=±6.

Half the sum of (1) and (5), gives

x=8, or 2.

Half the difference of (1) and (5), gives

y=2, or 8.

(4)

(5)

2. Given

Sx+y=α,
{ x2+ y2=b,

8}}

to find

and y.

Squaring (1), we have

x2+2xy+y2=a2.

Subtracting (2) from (3), we get

2xy=a2—b.

Subtracting (4) from (2), we find

x2-2xy+y=2b-a2.

(5)

Extracting the square root of (5), we get

Taking half the sum of (1) and (6), we get

(6)

If we take the sum of (1), (2), and (3), after expand`ing them, we shall have

2xy+2xx+2yz=a+b+c.

If we subtract twice (3) from (4), we get

2xy=a+b-c.

(4)

(5)