If both the roots of the equation x^2 - 6ax + | vedclass

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

Question

(D) Given equation: $x^2 - 6ax + (9a^2 - 2a + 2) = 0$. Let $f(x) = x^2 - 6ax + 9a^2 - 2a + 2$. For both roots to be greater than $3$,the following con

Vedclass · Accepted Answer

$a > \frac{11}{9}$

Vedclass · Answer

(D) Given equation: $x^2 - 6ax + (9a^2 - 2a + 2) = 0$. Let $f(x) = x^2 - 6ax + 9a^2 - 2a + 2$. For both roots to be greater than $3$,the following conditions must be satisfied: $1)$ Discriminant $D \ge 0$: $D = (-6a)^2 - 4(1)(9a^2 - 2a + 2) = 36a^2 - 36a^2 + 8a - 8 = 8a - 8$. $8a - 8 \ge 0 \Rightarrow a \ge 1$. $2)$ Vertex location: $-\frac{b}{2a} > 3$: $-\frac{-6a}{2(1)} > 3$ $\Rightarrow 3a > 3$ $\Rightarrow a > 1$. $3)$ $f(3) > 0$: $f(3) = 3^2 - 6a(3) + 9a^2 - 2a + 2 > 0$ $9 - 18a + 9a^2 - 2a + 2 > 0$ $9a^2 - 20a + 11 > 0$ $(9a - 11)(a - 1) > 0$. Since $a > 1$,the condition $(9a - 11)(a - 1) > 0$ implies $a > \frac{11}{9}$. Combining all conditions,we get $a > \frac{11}{9}$.

If both the roots of the equation $x^2 - 6ax + 2 - 2a + 9a^2 = 0$ exceed $3$,then

Similar Questions

Let $a, b, c$ be real numbers such that $a+b+c < 0$ and the quadratic equation $a x^{2}+b x+c=0$ has imaginary roots. Then:

If the graph of $y = ax^2 + bx + c$ is as follows,where $\Delta ABC$ is a right-angled isosceles triangle with hypotenuse $AC = 4\sqrt{2} \text{ units}$,then the minimum value of $ax^2 + bx + c$ is:

Number of integral values of $a$ for which both roots of the quadratic equation $x^2 - (2a + 3)x + a^2 + 3a = 0$ lie in the interval $(0, 4)$ is:

For what interval of $m$ do all roots of the quadratic equation $x^2 - 2mx + m^2 - 1 = 0$ lie between $-2$ and $4$?

The real number $k$ for which the equation $2x^2 + 3x + k = 0$ has two distinct real roots in the interval $[0, 1]$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module