Advertisements
Advertisements
प्रश्न
Find the value(s) of k so that the quadratic equation x2 − 4kx + k = 0 has equal roots.
उत्तर
Given equation is x2 − 4kx + k = 0
For the given equation to have equal roots, D = 0.
`rArr b^2-4ac=0 `
`rArr (-4k)^2-4(1)(k)=0`
`rArr 16k^2-4k=0`
`rArr 4k(4k-1)=0`
`rArr4k=0`or `k=1/4`
Hence, for k = 0 or `1/4`, the given equation will have equal roots.
APPEARS IN
संबंधित प्रश्न
The vertices of a ΔABC are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that `(AD)/(AB) = (AE)/(AC) = 1/4`Calculate the area of the ΔADE and compare it with the area of ΔABC. (Recall Converse of basic proportionality theorem and Theorem 6.6 related to ratio of areas of two similar triangles)
Find the missing value:
Base | Height | Area of triangle |
15 cm | ______ | 87 cm2 |
Four points A (6, 3), B (−3, 5), C(4, −2) and D (x, 3x) are given in such a way that `(ΔDBG) /(ΔABG)=1/2,` find x
Show that the points O(0,0), A`( 3,sqrt(3)) and B (3,-sqrt(3))` are the vertices of an equilateral triangle. Find the area of this triangle.
The coordinates of the point P dividing the line segment joining the points A (1, 3) and B (4, 6) in the ratio 2 : 1 are:
Using integration, find the area of the triangle whose vertices are (2, 3), (3, 5) and (4, 4).
The points (0, 5), (0, –9) and (3, 6) are collinear.
Find the coordinates of the point Q on the x-axis which lies on the perpendicular bisector of the line segment joining the points A(–5, –2) and B(4, –2). Name the type of triangle formed by the points Q, A and B.
Let a vector `αhati + βhatj` be obtained by rotating the vector `sqrt(3)hati + hatj` by an angle 45° about the origin in counter-clockwise direction in the first quadrant. Then the area of triangle having vertices (α, β), (0, β) and (0, 0) is equal to ______.
Using determinants, find the area of ΔPQR with vertices P(3, 1), Q(9, 3) and R(5, 7). Also, find the equation of line PQ using determinants.