What is the coefficient of X in X 2?

Explanation:

The coefficient of a variable is a constant value that occurs with a term. In the given question, the term is #x/2#.

We can easily make out that #x# is the variable here. The constant value occurring with the variable is #1/2# and hence, this is the coefficient of the given term #-> x/2#

• term #-> x/2#
• variable #-> x#
• the coefficient of the variable #-> 1/2#

Solution:

A coefficient refers to a number or quantity placed with a variable. It is usually an integer that is multiplied by the variable next to it. 

i) 2 + x² + x

Coefficient of x² is 1.

ii) 2 - x² + x³

Coefficient of x² is -1.

iii) (π/2)x² + x

Coefficient of x² is π/2.

iv) √2x - 1

There is no  x² term in the given expression. We can rewrite it as 0x² + √2x - 1 

Therefore, the coefficient of x² = 0, as x² doesn't exist.

☛ Check: Class 9 Maths NCERT Solutions Chapter 2

Video Solution:

Write the coefficients of x² in each of the following: (i) 2 + x² + x (ii) 2 - x² + x³ (iii) (π/2)x² + x (iv) √2x - 1

NCERT Solutions Class 9 Maths Chapter 2 Exercise 2.1 Question 2:

Summary:

The coefficients of x² in each of the following equations 2 + x²+ x, 2 - x² + x³, (π/2) x² + x, and √2x - 1 are 1, -1, π/2, and 0 respectively.

☛ Related Questions:

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• Write the degree of each of the following polynomials:i) 5x3 + 4x2 + 7xii) 4 - y2iii) 5t - √7iv) 3
• Classify the following as linear, quadratic and cubic polynomials:i) x² + x ii) x - x3iii) y + y² + 4iv) 1 + xv) 3tvi) r²vii) 7x3

Reference > Mathematics > Algebra > The Quadratic Formula

This is all well and good, this "completing the square" method (you might say) but what if the coefficient of x2 isn't a perfect square? We can't complete the square in that situation, can we?

Well, not exactly. But there is something we can do. Consider the following:

3x2 - 8x - 3 = 0

3 is not a perfect square, but suppose we took this equation and multiplied both sides by 3...then we would have the following:

9x2 - 24x -9 = 0

What do we need to do to complete the square now? The constant term needs to be 16, and since it's -16, we need to add 25:

9x2 - 24x + 16 = 25
(3x - 4)2 = 52
3x - 4 = 5 or 3x - 4 = -5
3x = 9 or 3x = -1
x = 3 or x = -1/3

Problem #1
Solve for x if 2x2 + x - 15 = 0

Solution #1
If we multiply the entire equation by 2, we get:

4x2 + 2x - 30 = 0

Thus, to complete the square, the constant term must be 1/4. Since it's -30 (or -120/4), we need to add 121/4 to both sides:

4x2 + 2x - 30 + 121/4 = 121/4
4x2 + 2x + 1/4 = 121/4
(2x + 1/2)2 = (11/2)2
2x + 1/2 = 11/2 or 2x + 1/2 = -11/2
2x = 10/2 or 2x = -12/2
x = 5/2 or x = -3

Problem #2
Solve for x if 12x2 - 13x - 4 = 0

Solution #2
You might be tempted to multiply this by 12 to make a perfect square (144), but that's really overkill! Because if we multiply 12 by 3, we get 36, which is also a perfect square. [HINT: in the prime factorization of 12, 3 is the only prime factor which doesn't have an even exponent - that's why I thought to multiply by 3]

So we have 36x2 - 39x - 12 = 0

To have a completed square, the constant term must be (13/4)2 or 169/16. But it's actually -12, or -192/16. Thus we need to add 361/16:

36x2 - 36x -12 + 361/16 = 0 + 361/16
36x2 - 36x + 169/16 = 361/16
(6x - 13/4)2 = (19/4)2
6x - 13/4 = 19/4 or 6x - 13/4 = -19/4
6x = 32/4 or 6x = -6/4
x = 4/3 or x = -1/4

Special Hint
Sometimes you can make the coefficient of x2 a perfect square by dividing instead of multiplying. For example, in 2x2 + 2x - 12 = 0, if you notice that all the coefficients are multiples of 2, you can divide out the two, leaving x2 + x - 6 = 0, which is much simpler.

For each problem below, solve by completing the square.

Questions

8.

(1/2)x2 + (3/2)x + 1 = 0

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