Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

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

Relevance
  • 1 decade ago
    Favorite 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

    • Login to reply the answers
  • 1 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.

    • Login to reply the answers
  • 1 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

    • Login to reply the answers
Still have questions? Get your answers by asking now.