Number of equations of the form ax^2 + bx + 1 | vedclass

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

Question

(B) For the quadratic equation $ax^2 + bx + 1 = 0$ to have real roots,the discriminant $D$ must be greater than or equal to $0$.$D = b^2 - 4ac \geq 0$

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(B) For the quadratic equation $ax^2 + bx + 1 = 0$ to have real roots,the discriminant $D$ must be greater than or equal to $0$.$D = b^2 - 4ac \geq 0$Given $c = 1$,we have $b^2 - 4a \geq 0$,or $b^2 \geq 4a$.We test values for $b \in \{1, 2, 3, 4\}$:$1$. If $b = 1$,$1^2 \geq 4a \Rightarrow 1 \geq 4a$,which has no solution for $a \in \{1, 2, 3, 4\}$.$2$. If $b = 2$,$2^2 \geq 4a$ $\Rightarrow 4 \geq 4a$ $\Rightarrow a \leq 1$. Thus,$a = 1$ ($1$ solution).$3$. If $b = 3$,$3^2 \geq 4a$ $\Rightarrow 9 \geq 4a$ $\Rightarrow a \leq 2.25$. Thus,$a \in \{1, 2\}$ ($2$ solutions).$4$. If $b = 4$,$4^2 \geq 4a$ $\Rightarrow 16 \geq 4a$ $\Rightarrow a \leq 4$. Thus,$a \in \{1, 2, 3, 4\}$ ($4$ solutions).Total number of equations = $0 + 1 + 2 + 4 = 7$.

Number of equations of the form $ax^2 + bx + 1 = 0$ having real roots,where $a, b \in \{1, 2, 3, 4\}$ is-

Similar Questions

The equation $\sqrt{3x^2 + x + 5} = x - 3$,where $x$ is real,has

The solution to the equation $4^{(x^2 + 2)} - 9 \cdot 2^{(x^2 + 2)} + 8 = 0$ is:

The number of real roots of $|x|^2-5|x|+6=0$ is

When $b=17$,it is found that the roots of the equation $x^2+bx+c=0$ are $-2$ and $-15$. If $\alpha, \beta$ are the roots of the same equation when $b=13$,then $|\alpha-\beta|=$

If the roots of the equation $(m - n)x^2 + (n - l)x + (l - m) = 0$ are equal,then which of the following is true for $l, m,$ and $n$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module