If a, b in {1, 2, 3} and the equation ax^{2} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The given quadratic equation is $ax^{2} + bx + 1 = 0$.For the equation to have real roots,the discriminant $D$ must be greater than or equal to $0

Vedclass · Accepted Answer

number of possible ordered pairs $(a, b)$ is $3$

Vedclass · Answer

(C) The given quadratic equation is $ax^{2} + bx + 1 = 0$.For the equation to have real roots,the discriminant $D$ must be greater than or equal to $0$.$D = b^{2} - 4ac \geq 0$.Substituting $c = 1$,we get $b^{2} - 4a \geq 0$,which implies $b^{2} \geq 4a$.Given $a, b \in \{1, 2, 3\}$,we test the possible values:If $a = 1$,$b^{2} \geq 4 \implies b \in \{2, 3\}$. Pairs: $(1, 2), (1, 3)$.If $a = 2$,$b^{2} \geq 8 \implies b = 3$. Pair: $(2, 3)$.If $a = 3$,$b^{2} \geq 12 \implies$ no value of $b$ satisfies this.The possible ordered pairs $(a, b)$ are $(1, 2), (1, 3), (2, 3)$.Thus,the number of possible ordered pairs is $3$.

If $a, b \in \{1, 2, 3\}$ and the equation $ax^{2} + bx + 1 = 0$ has real roots,then

Similar Questions

Solve the equation $x^{2}+x+\frac{1}{\sqrt{2}}=0$.

Let $f(x) = x^2 + ax + b$,where $a, b \in R$. If $f(x) = 0$ has all its roots imaginary,then the roots of $f(x) + f'(x) + f''(x) = 0$ are

Let $\alpha$ and $\beta$ be the roots of the quadratic equation $x^2 \sin \theta - x(\sin \theta \cos \theta + 1) + \cos \theta = 0$ where $0 < \theta < 45^\circ$,and $\alpha < \beta$. Then $\sum_{n=0}^\infty (\alpha^n + \frac{(-1)^n}{\beta^n})$ is equal to

The number of real roots of the equation $\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=\frac{13}{6}$ is

If a root of the equation $x^2 + px + 12 = 0$ is $4$,while the roots of the equation $x^2 + px + q = 0$ are equal,then the value of $q$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module