Question

Day
Il co
w x² + 2x - 4 = 0; x =3,x = - 2/2 ​

Answer 1

Answer:

Answer: x= -2√2 and x = √2

Solution:

Given, x² + √2x - 4 = 0……………………………….…………………………………(1)

Equation (1) is a quadratic equation in the variable x whose co-efficient is unity. We will solve (1) by the method of completing the square. The process of the method is given below step-wise.

Step 1: Arrange the terms so that the terms in x² and x are on the left-hand side and the constant term on the other. Doing this, we have

x² + √2x = 4

Step 2: Complete the square on the left-hand side by adding to each side of the equation the square of half the co-efficient of x. The co-efficient of x is √2 and square of half of √2 is (√2/2)² = (1/√2)² = 1/2 .

∴ x² + √2x + 1/2 = 4 + 1/2 = (8+1)/2 = 9/2

Or, x² + 2. x. 1/√2 + (1/√2)² = (3/√2)²

Or, (x+ 1/√2 )² = (3/√2 )²

Step 3: Take the square root of each side, and solve the two resulting simple equations. Root extraction gives,

x + 1/√2 = ±3/√2

Step 4:

Hence we have two simple equations

x + 1/√2 = +3/√2 , and x + 1/√2 = -3/√2

∴ x = 3/√2 - 1/√2 , and x = -1/√2 - 3/√2

⇒ x = 2/√2 = 2.√2/√2.√2 = 2.√2/2 = √2

and x = -4/√2 = -4.√2/√2.√2 = -4.√2/2 = -2√2

∴ x = -2√2, or √2

Hence the solution of (1) furnishes two values for x:

-2√2 and √2 (Answer).

Check:

Take x = -2√2 and substitute in the L.H.S. of (1).

L.H.S.=(-2√2)² + √2(-2√2) - 4 = 8 - 4 - 4 = 8–8 = R.H.S.

In like manner, take x = √2 and substitute in the L.H.S. of (1).

L.H.S. = (√2)² + √2.√2 - 4 = 2 + 2 - 4 = 4 - 4 = 0 = R.H.S.

So our results for x are correct.

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Answer 2

Step-by-step explanation:

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