Advertisements
Advertisements
प्रश्न
Discuss the nature of the roots of the following quadratic equations : -2x2 + x + 1 = 0
Advertisements
उत्तर
-2x2 + x + 1 = 0
Here a = -2, b = 1, c = 1
∴ D = b2 - 4ac
= (1)2 - 4 x (-2) x 1
= 1 + 8
= 9
∵ D > 0
∴ Roots are real and distinct.
APPEARS IN
संबंधित प्रश्न
Solve the equation by using the formula method. 3y2 +7y + 4 = 0
Find the values of k for which the quadratic equation (3k + 1) x2 + 2(k + 1) x + 1 = 0 has equal roots. Also, find the roots.
Without solving, examine the nature of roots of the equation 4x2 – 4x + 1 = 0
Find the nature of the roots of the following quadratic equation. If the real roots exist, find them:
2x2 - 6x + 3 = 0
Determine the nature of the roots of the following quadratic equation:
9a2b2x2 - 24abcdx + 16c2d2 = 0
In the following determine the set of values of k for which the given quadratic equation has real roots:
x2 - kx + 9 = 0
The equation `3x^2 – 12x + (n – 5) = 0` has equal roots. Find the value of n.
Determine whether the given quadratic equations have equal roots and if so, find the roots:
`(4)/(3)x^2 - 2x + (3)/(4) = 0`
If b2 – 4ac > 0 and b2 – 4ac < 0, then write the nature of roots of the quadratic equation for each given case.
If –5 is a root of the quadratic equation 2x2 + px – 15 = 0, then:
