Verification. 17(8)—2(8)2=32—3(8), or + 8=+ 8.

17(2)—2(2)2—32—3(2), or +26=+26.

2. Given 3x2-2x=65, to find x.

Dividing by 3, . . . x2—3x=65;

Completing the square, x2-3x+(})2=65+(3)2=196.
Extracting the root, x =±

x=45, or -4.

Both of which values verify the equation.

3. Given 4a2-2x2+2ax=18ab-1862, to find x.

4. Given x+√(5x+10)=8, to find x.

Completing the square, x2-21x+(21)2=441-54-225;

Extracting the root,

Whence,

x=2±15=36=18, or §=3.

These two values of x are the roots of the equation, x2-21x=-54, but they will not both verify the original equation.

For, the proposed equation might have been x±√(5x+10)=8; and the operations which have been employed would result in the same equation, x2—21x——54, whether the sign of the radical part be + or

Hence, in the equation x+√(5x+10)=8, the value of x is 3; but in the equation —√(5x+10)=8, the value is 18.

28. (a—b)x2-(a+b)x+2b=0. Ans. x=1, or

a

26

Ρ

m

232. Hindoo Method of Solving Quadratics.-When an equation is brought to the form ax2+bx=c, it may be reduced to a simple equation, without dividing by the coefficient of x2, thus avoiding fractions.

If we multiply every term of the equation ax2+bx=c, by four times the coefficient of the first term, and add to both sides the square of the coëfficient of the second term, we shall have

4a2x2+4abx+b2=4ac+b2.

Now, the first member of this equation is a perfect square, and by extracting the square root of both sides, we have

Jax+b=±√/4ac+62, which is a simple equation. Hence, the

Hindoo Rule for the Solution of Quadratic Equations.-1st. Reduce the equation to the form ax2+bx=c. 2d. Multiply both sides by four times the coefficient of x2. 3d. Add the square of the coefficient of x to each side, extract the square root, and finish the solution.

1. Given 2x2-5x-3, to find x.

Multiplying both sides by 8, four times the coëfficient of x2,
We have
16x2-40x 24.

Adding to each side 25, which is the square of the coëfficient of x, 16x2-40x+25=49;

We have

.

Extracting the root,

4x-5=7; whence, x=3, or

Find the value of the unknown quantity in each of the following examples by the Hindoo Rule:

By an inspection of the forms given in Art. 231, it will be seen that the value of the unknown quantity may be found without the formality of completing the square, by the following

Rule. Reduce the equation to the form x2+2px=q. The unknown quantity will then be equal to one half the coefficient of its first power taken with a contrary sign, plus or minus the square root of the square of the number last written together with the known quantity in the second member of the equation taken with its proper sign.

Thus, let x2+16x=-60.

Then, x=-8±√64—60=-8±√/4=-8±2.

x=-6, or -10.

After some exercise in completing the square, it is best to employ this last method.

PROBLEMS PRODUCING AFFECTED EQUATIONS.

233.-1. A person bought a certain number of sheep for $40, and if he had bought 2 more for the same sum they would have cost $1 apiece less. Required the number of sheep, and the price of each.

Solving this eq., x=—1±9—8, or -10, number of sheep.

The negative value, -10, to fulfill the conditions of the question in an arithmetical sense, must be modified, on the principles explained in Art. 164, thus:

A person sells a certain number of sheep for $40. If he had. sold 2 fewer for the same sum he would have received $1 apiece more for them. Required the number sold..

2. Find a number such, that if 17 times the number be diminished by its square, the remainder shall be 70.

Let x the number.
Then, 17x-x2-70.
Or; x2-17x=-70.

Whence, x=7, or 10.

In this case, both values of x satisfy the question in its arithmetical sense.

Thus, 17X7-72-119- 49-70.

Or,

17X10-102-170-100-70.

3. Of a number of bees, after

and the square root

of of them, had flown away, there were two remaining; what was the number at first?