Since, p and q are the roots of the equation x2 + px + q = 0
Then, p + q = - p
and pq = q
Now, pq = q
? p = 1
Putting the value of p in p + q = - p, we get
1 + q = - 1
? q = - 2
Let ? = 2, ? = 1/4
Then, ? + ? = 2 + 1/4 = 9/4
?? = 2.1/4 = 1/2
Equation having the roots ? and ? is
x2 - (? + ?)x + ? ? = 0
? x2 - 9/4x + 1/2 = 0
? 4x2 - 9x + 2 = 0
Let ? and ? be the roots of the quadratic equation 2x2 - 11x + 5 = 0
? ? + ? = -(-11)/2 = 11/2 ...(i)
and ?? = 5/2
Now, (? - ?)2 = (? + ?)2 - 4??
= (11/2)2 - 4(5/2) = 121/4 - 20/2
= 121 - 40/4 = 81/4 = (9/2)2
? Difference of roots = (? - ?) = 9/2 = 4.5
Given equation is 4x2 - 19x + 12 = 0
Let given equation having the roots 1/? and 1/?,
Then required equuation is
12x2 - 19x + 4 = 0
Given that, the roots of the quadrictic equation are 3 and -1.
Let ? = 3 and ? = -1
Sum of roots = ? + ? = 3 - 1 = 2
Products of roots = ? . ? = (3) (-1) = -3
? Required quadric equation is
x2 - (? + ?)x + ? ? = 0
? x2 - (2)x + (-3) = 0
? x2 - 2x - 3 = 0
Given equation is
x2 + 2(k - 4)x + 2k = 0
On comparing with ax2 + bx + c = 0
Here, a = 1, b = 2(k -4), c = 2k
Since, the root are equal, we have D = 0.
b2 - 4ac = 0
? 4(k - 4)2 - 8k = 0
4(k2 + 16 - 8k) - 8k = 0
? 4k2 + 64 - 32k - 8k = 0
? 4k2 - 40k + 64 = 0
? k2 - 10k + 16 = 0
? k2 - 8k - 2k + 16 = 0
? k(k - 8) -2 (k - 8) = 0
? (k - 8) (k - 2) = 0
Hence, the value of k 8 or 2.
We know that, AM ? GM
? ?a + 1/?a ? 2
Here, ?x2 - x + 1 + 1/?x2 - x + 1 ? 2
? 2 - x2 ? 2
? x2 ? 0
? x = 0
Hence, the given equation has only one solution.
Let the roots be ? and ? .
Then, ? + ? = 8 ......(i)
? - ? = 4 ..(ii)
On solving Eqs. (i) and (ii), we get
? = 6, ? = 2
? Required equation is
x2 - (? + ?) x + (? ?) = 0
? x2 - (6 + 2)x + 6 x 2 = 0
? x2 - 8x + 12 = 0
Let the number be x and 1/x,
Then, x + 1/x = 10/3
? (x2 + 1)/x = 10/3
? 3x2 - 10x + 3 = 0
? 3x2 - 9x - x + 3 = 0
? 3x(x - 3) -1 (x - 3) = 0
? (3x - 1) (x - 3) = 0
? x = 1/3, x = 3
? + ? = - p, ? ? = 8
Given, ? - ? = 2
On squaring both sides, we get
(? - ?)2 = 4
? (? + ?)2 - 4?? = 4
? p2 - 32 = 4
? p2 = 36 ? p = ± 6
x2 + x - 20 = 0
[by factorisation method]
? x2 + 5x - 4x - 20 = 0
? x(x + 5) - 4(x + 5) = 0
(x + 5) (x - 4) = 0
? x = -5 or 4
and y2 - y - 30 = 0
? y2 - 6y + 5y - 30 = 0
? y(y - 6) + 5(y - 6) = 0
? (y - 6) (y + 5) = 0
? y = 6 or - 5
Hence, y ? x or x ? y
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