Class 11Mathematics4-2.Quadratic Equations and InequationsSolution of quadratic equations and Nature of roots
EasyIf the roots of the equation $x^2 - bx + c = 0$ are two consecutive integers,then $b^2 - 4c = ......$
- A$1$
- B$2$
- C$3$
- D$4$
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Similar Questions
Solve the equation $\sqrt{2} x^{2} + x + \sqrt{2} = 0$.
Medium
View SolutionLet $\alpha$ and $\beta$ be the roots of the quadratic equation $a x^2+b x+c=0$. Observe the lists given below:
List-$I$ List-$II$ $(i)$ $\alpha = \beta$ $(A)$ $(ac^2)^{1/3} + (a^2c)^{1/3} + b = 0$ $(ii)$ $\alpha = 2\beta$ $(B)$ $2b^2 = 9ac$ $(iii)$ $\alpha = 3\beta$ $(C)$ $b^2 = 6ac$ $(iv)$ $\alpha = \beta^2$ $(D)$ $3b^2 = 16ac$ $(E)$ $b^2 = 4ac$ $(F)$ $(ac^2)^{1/3} + (a^2c)^{1/3} = b$
The correct match of List-$I$ from List-$II$ is:
| List-$I$ | List-$II$ |
| $(i)$ $\alpha = \beta$ | $(A)$ $(ac^2)^{1/3} + (a^2c)^{1/3} + b = 0$ |
| $(ii)$ $\alpha = 2\beta$ | $(B)$ $2b^2 = 9ac$ |
| $(iii)$ $\alpha = 3\beta$ | $(C)$ $b^2 = 6ac$ |
| $(iv)$ $\alpha = \beta^2$ | $(D)$ $3b^2 = 16ac$ |
| $(E)$ $b^2 = 4ac$ | |
| $(F)$ $(ac^2)^{1/3} + (a^2c)^{1/3} = b$ |
The correct match of List-$I$ from List-$II$ is:
If $x^2 + ax + b$ is an integer for every integer $x$,then which of the following is true?
Difficult
View SolutionThe equation $4^{(x^2 + 2)} - 9 \cdot 2^{(x^2 + 2)} + 8 = 0$ has the solution:
Difficult
View SolutionIf $-1$ is a twice repeated root of the equation $a(x^3+x^2)+bx+c=0$,then $a:b:c=$
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