Let p, q be real numbers. If alpha is a root | vedclass

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

Question

(D) Given that $\alpha$ is a root of $x^{2}+3 p^{2} x+5 q^{2}=0$, we have $\alpha^{2}+3 p^{2} \alpha+5 q^{2}=0$. Since $\beta$ is a root of $x^{2}+9 p

Vedclass · Accepted Answer

$\alpha < \gamma < \beta$

Vedclass · Answer

(D) Given that $\alpha$ is a root of $x^{2}+3 p^{2} x+5 q^{2}=0$, we have $\alpha^{2}+3 p^{2} \alpha+5 q^{2}=0$. Since $\beta$ is a root of $x^{2}+9 p^{2} x+15 q^{2}=0$, we have $\beta^{2}+9 p^{2} \beta+15 q^{2}=0$. Let $f(x)=x^{2}+6 p^{2} x+10 q^{2}$. Evaluating $f(\alpha)$: $f(\alpha)=\alpha^{2}+6 p^{2} \alpha+10 q^{2} = (\alpha^{2}+3 p^{2} \alpha+5 q^{2}) + 3 p^{2} \alpha + 5 q^{2} = 0 + 3 p^{2} \alpha + 5 q^{2}$. Since $\alpha > 0$ and $p, q$ are real numbers (assuming $p, q eq 0$ for non-trivial roots), $f(\alpha) > 0$. Evaluating $f(\beta)$: $f(\beta)=\beta^{2}+6 p^{2} \beta+10 q^{2} = (\beta^{2}+9 p^{2} \beta+15 q^{2}) - (3 p^{2} \beta + 5 q^{2}) = 0 - (3 p^{2} \beta + 5 q^{2})$. Since $\beta > \alpha > 0$, $f(\beta) Since $f(x)$ is a continuous polynomial function and $f(\alpha) > 0$ while $f(\beta) < 0$, by the Intermediate Value Theorem, there must exist a root $\gamma$ such that $\alpha < \gamma < \beta$.

Let $p, q$ be real numbers. If $\alpha$ is a root of $x^{2}+3 p^{2} x+5 q^{2}=0$, $\beta$ is a root of $x^{2}+9 p^{2} x+15 q^{2}=0$ and $0 < \alpha < \beta$, then the equation $x^{2}+6 p^{2} x+10 q^{2}=0$ has a root $\gamma$ that always satisfies:

Similar Questions

The number of solutions of the equation $x^2 + y^2 = a^2 + b^2 + c^2$,where $x, y, a, b, c$ are all prime numbers,is

Suppose $a, b$ denote the distinct real roots of the quadratic polynomial $x^2+20x-2020$ and suppose $c, d$ denote the distinct complex roots of the quadratic polynomial $x^2-20x+2020$. Then the value of $ac(a-c)+ad(a-d)+bc(b-c)+bd(b-d)$ is

If $\alpha$ is a root of the equation $x^2-x+1=0$,then $\left(\alpha+\frac{1}{\alpha}\right)^3+\left(\alpha^2+\frac{1}{\alpha^2}\right)^3+\left(\alpha^3+\frac{1}{\alpha^3}\right)^3+\left(\alpha^4+\frac{1}{\alpha^4}\right)^3=$

For real values of $x$,the range of $\frac{x^2+2x+1}{x^2+2x-1}$ is

If $x$ is real, then the value of $\frac{x^2-3x+4}{x^2+3x+4}$ lies in the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module