Real roots maths question?

Given that the equation

kx²-4x+(k-3)=0

has real roots, show that

k²-3k-4≤0

Find the range of values of k satisfying this inequality.

3 Answers

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  • Puggy
    Lv 7
    1 decade ago
    Best Answer

    Given kx² - 4x + (k - 3)=0 has real roots, it follows that its discriminant is greater or equal to 0. The discriminant is (b² - 4ac).

    b² - 4ac = (-4)^2 - 4(k)(k - 3)

    = 16 - 4k(k - 3)

    = 16 - 4k^2 + 12k

    = -4k^2 + 12k + 16

    And, as stated above, this must be greater than or equal to 0. That is,

    -4k^2 + 12k + 16 ≥ 0

    Factoring out a -4 both sides,

    -4(k^2 - 3k - 4) ≥ 0

    Divide (-4) both sides, which will flip the inequality.

    k^2 - 3k - 4 ≤ 0

    To find the range of values satisfying this inequality, factor the left hand side.

    (k - 4)(k + 1) ≤ 0

    To solve this inequality, make a number line consisting of the solution to the corresponding equality, (k - 4)(k + 1) = 0.

    k = {-1, 4}.

    Now, make your number line.

    . . . . . . .(-1) . . . . . . . . (4) . . . . . . . . . .

    And test (k - 4)(k + 1) for each region around the critical values.

    Test (-2): (-2 - 4)(-2 + 1) = (-6)(-1) = 6, which is positive. Mark the region as positive.

    . . .{+}. . .(-1) . . . . . . . . (4) . . . . . . . . . .

    Test 0: (0 - 4)(0 + 1) = (-4)(1) = -4, which is negative. Mark that region as negative.

    . . .{+}. . .(-1) . . . .{-}. . . (4) . . . . . . . . . .

    Test 5: (5 - 4)(5 + 1) = 1(6) = 6, which is positive.

    . . .{+}. . .(-1) . . . .{-}. . . (4) . . . .{+}. . . . .

    Since our inequality is less than _or equal to_ 0, we look for any negative regions. The "or equal to" means we *include* our critical values -1 and 4. Therefore, the range of values satisfying k are:

    -1 ≤ k ≤ 4

  • Tim
    Lv 4
    1 decade ago

    You can solve this using the quadratic formula. Since the quadratic formula is: \frac{-b +/- \sqrt{b^2 - 4*a*c}}{2*a}, in order to be a real solution, the square root cannot be negative. This defines: b^2 - 4*a*c>=0

    From your equation given:

    a = k

    b = -4

    c = k - 3

    Plugging into the equation above:

    (-4)^2 - 4*k*(k-3)>=0

    16 - 4*k^2 + 12*k >= 0

    4*(4 - k^2 + 3*k) >= 0

    -k^2 + 3*k + 4 >= 0

    Note: When you multiply an inequality by a negative number, the inequality switches direction.

    k^2 - 3*k - 4 <= 0

  • Anonymous
    1 decade ago

    Figure out the discriminant (b²-4ac) and you'll arrive at that inequality.

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