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# MATHS POLYNOMIALS HELP?

MATHS - POLYNOMIALS. HELP PLEASE?

Let p(x)=x^2 - 6x - 3 and q(x)= x^2 - 2x + 4

The polynomial p(x) + Aq(x), where A is a constant, is a perfect square.

- calculate the two possible values of A.

### 3 Answers

- Scrander berryLv 71 decade agoFavorite Answer
p(x) + Aq(x) = (ax + b)²

p(x) + Aq(x)

= x² - 6x - 3 + A(x² - 2x + 4)

= x² - 6x - 3 + Ax² - 2Ax + 4A

= (1 + A)x² - (6 + 2A)x + (4A - 3)

Complete the square

= x² - (6 + 2A)x/(1 + A) + (4A - 3)/(1 + A)

To be a perfect square:

[(6 + 2A)/(2 + 2A)]² = (4A - 3)/(1 + A)

(6 + 2A)²/(2 + 2A)² = (4A - 3)/(1 + A)

(6 + 2A)² = (4A - 3)(2 + 2A)²/(1 + A)

36 + 24A + 4A² = 4(4A - 3)(1 + A)²/(1 + A)

36 + 24A + 4A² = 4(4A - 3)(1 + A)

36 + 24A + 4A² = 4A - 12 + 16A²

12A² - 20A - 48 = 0

3A² - 5A - 12 = 0

(3A + 4)(A - 3) = 0

A = 3 or A = -4/3

Since you asked twice I answered twice

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- BhaskarLv 41 decade ago
p(x) = x^2 - 6x - 3

q(x) = x^2 - 2x + 4.

p(x) + a*q(x) = x^2(a+1) - x (6 + 2a) + 4a - 3

= (a+1)x^2 - 2x(3 + a) + 4a - 3.

since this polynomial has equal roots, its discriminant must be 0 that is:

[2*(3+a)]^2 - 4(a+1)(4a-3) = 0

(a+3)^2 - (a+1)(4a-3) = 0

a^2 + 6a + 9 - [4a^2 + a - 3] = 0

-3a^2 + 5a + 12 = 0

3a^2 - 5a - 12 = 0

solving this equation,

3a^2 + 4a - 9a - 12 = 0

a(3a + 4) - 3 (3a + 4) = 0

(3a+4)(a-3) = 0

=> a = -4/3 or a = 3.

hence, the two possible values of a are -4/3 and 3.

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- Jerome JLv 71 decade ago
p(x) + Aq(x)

= x^2 - 6x - 3 + A(x^2 - 2x + 4) = 0

x^2 - 6x - 3 + Ax^2 - 2Ax + 4A = 0

Ax^2 + x^2 - 2Ax - 6x+ 4A - 3 = 0

(A + 1)x^2 - (2a +6)x + (4A - 3) = 0

In a quadratic expression, b^2 - 4ac = 0 to be a perfect square

a = (A+1), b = (2a+6), c = (4A - 3)

(2A+6)^2 - 4(A+1)(4A-3) = 0

4A^2 + 24A + 36 - 4(4A^2 + A - 3) = 0

4A^2 + 24A + 36 - 16A^2 - 4A + 12 = 0

-12A^2 + 20A + 48 = 0

3A^2 - 5A - 12 = 0

(A - 3)(3A + 4) = 0

A = 3 or A = -4/3

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