Question:
If \(\alpha,\beta\) are the roots of
\[
ax^2+bx+c=0,
\]
then the quadratic equation whose roots are \(\sqrt{5}\alpha,\sqrt{5}\beta\) is
If \(\alpha,\beta\) are the roots of
\[
ax^2+bx+c=0,
\]
then the quadratic equation whose roots are \(\sqrt{5}\alpha,\sqrt{5}\beta\) is
Show Hint
If the roots of a quadratic equation are multiplied by a constant \(k\), then the new sum becomes \(k(\alpha+\beta)\) and the new product becomes \(k^2\alpha\beta\).
Updated On: Jun 26, 2026
- \(ax^2+\sqrt{5}bx+5c=0\)
- \(ax^2+\sqrt{5}bx+\sqrt{5}c=0\)
- \(ax^2+5bx+\sqrt{5}c=0\)
- \(ax^2+5bx+5c=0\)
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Use the relation between roots and coefficients.
For the quadratic equation \[ ax^2+bx+c=0, \] if roots are \(\alpha\) and \(\beta\), then \[ \alpha+\beta=-\frac{b}{a} \] and \[ \alpha\beta=\frac{c}{a} \]
Step 2: Find the new roots.
The new roots are \[ \sqrt{5}\alpha \] and \[ \sqrt{5}\beta \] Let the new roots be \(r_1\) and \(r_2\).
So, \[ r_1=\sqrt{5}\alpha,\quad r_2=\sqrt{5}\beta \]
Step 3: Find the sum of the new roots.
\[ r_1+r_2=\sqrt{5}\alpha+\sqrt{5}\beta \] \[ r_1+r_2=\sqrt{5}(\alpha+\beta) \] Using \[ \alpha+\beta=-\frac{b}{a}, \] we get \[ r_1+r_2=-\frac{\sqrt{5}b}{a} \]
Step 4: Find the product of the new roots.
\[ r_1r_2=(\sqrt{5}\alpha)(\sqrt{5}\beta) \] \[ r_1r_2=5\alpha\beta \] Using \[ \alpha\beta=\frac{c}{a}, \] we get \[ r_1r_2=\frac{5c}{a} \]
Step 5: Form the required quadratic equation.
A quadratic equation with roots \(r_1\) and \(r_2\) is \[ x^2-(r_1+r_2)x+r_1r_2=0 \] Substituting the values, \[ x^2-\left(-\frac{\sqrt{5}b}{a}\right)x+\frac{5c}{a}=0 \] \[ x^2+\frac{\sqrt{5}b}{a}x+\frac{5c}{a}=0 \] Multiplying throughout by \(a\), \[ ax^2+\sqrt{5}bx+5c=0 \]
Step 6: Final conclusion.
Therefore, \[ \boxed{ax^2+\sqrt{5}bx+5c=0} \]
For the quadratic equation \[ ax^2+bx+c=0, \] if roots are \(\alpha\) and \(\beta\), then \[ \alpha+\beta=-\frac{b}{a} \] and \[ \alpha\beta=\frac{c}{a} \]
Step 2: Find the new roots.
The new roots are \[ \sqrt{5}\alpha \] and \[ \sqrt{5}\beta \] Let the new roots be \(r_1\) and \(r_2\).
So, \[ r_1=\sqrt{5}\alpha,\quad r_2=\sqrt{5}\beta \]
Step 3: Find the sum of the new roots.
\[ r_1+r_2=\sqrt{5}\alpha+\sqrt{5}\beta \] \[ r_1+r_2=\sqrt{5}(\alpha+\beta) \] Using \[ \alpha+\beta=-\frac{b}{a}, \] we get \[ r_1+r_2=-\frac{\sqrt{5}b}{a} \]
Step 4: Find the product of the new roots.
\[ r_1r_2=(\sqrt{5}\alpha)(\sqrt{5}\beta) \] \[ r_1r_2=5\alpha\beta \] Using \[ \alpha\beta=\frac{c}{a}, \] we get \[ r_1r_2=\frac{5c}{a} \]
Step 5: Form the required quadratic equation.
A quadratic equation with roots \(r_1\) and \(r_2\) is \[ x^2-(r_1+r_2)x+r_1r_2=0 \] Substituting the values, \[ x^2-\left(-\frac{\sqrt{5}b}{a}\right)x+\frac{5c}{a}=0 \] \[ x^2+\frac{\sqrt{5}b}{a}x+\frac{5c}{a}=0 \] Multiplying throughout by \(a\), \[ ax^2+\sqrt{5}bx+5c=0 \]
Step 6: Final conclusion.
Therefore, \[ \boxed{ax^2+\sqrt{5}bx+5c=0} \]
Was this answer helpful?
0
0
Top AP EAPCET Mathematics Questions
- $D = \left\{x\in\mathbb{R}: f\left(x\right) =\sqrt{\frac{x - \left|x\right|}{x - \left[x\right]}} \text{is defined} \right\}$ and $C$ be the range of the real function $g(x) = \frac{2x}{4 + x^{2}}$. Then $D \cap C$
- $f\left(x\right) = \frac{x}{e^{x}-1} + \frac{x}{2} + 2 \cos^{3} \frac{x}{2} $ on $R-\left\{0\right\} $ is
- Consider the following lists.
List-I List-II (A) $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ (I) $[\frac{1}{3} , 1 ]$ (B) $(x)=|[x]|,x \in [R$ (II) Z (C) $h(x) = |x - [x]| , x \in [R$ (III) W (D) $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ (IV) [0, 1) (V) { -1, 1} - If $[x]$ is the greatest integer less than or equal to $x$ and $|x|$ is the modulus of $x$. then the system of three equations $2x + 3 | y | + 5[z] = 0, x + |y| - 2[z] = 4, x + |y| + |z| = 1$ has
- Investigate the values of $\lambda$ and $\mu$ for the system $x+2 y+3 z=6, x+3 y+5 z=9$, $2 x+5 y+\lambda z=\mu$ and match the values in List - I with the items in List - II.
List I List II (A) $\lambda=8, \mu \neq 15$ 1. Infinitely many solutions (B) $\lambda \neq 8, \mu \in R$ 2. No solution (C) $\lambda=8, \mu=15$ 3. Unique solution
View More Questions
Top AP EAPCET Algebra Questions
- If \( \alpha \) and \( \beta \) are the roots of the equation \( 2x^2 - 4x + 3 = 0 \), then \( \frac{2(\alpha^4 + \beta^4) + 3(\alpha^2 + \beta^2)}{\alpha + \beta} = \)
- If \( \alpha \) and \( \beta \) are the roots of the equation \( ax^2 + bx + c = 0 \), then the equation whose roots are \( \alpha + \beta \) and \( \frac{1}{\alpha} + \frac{1}{\beta} \) is
- If the roots of the equation \( 16x^3 - 44x^2 + 36x - 9 = 0 \) are in harmonic progression, then its greatest root is
- If \( \frac{x^4 - 6x^3 + 9x^2 + 5x - 20}{x^2 - x - 2} = f(x) + \frac{a}{x + p} + \frac{b}{x + q} \), then \( 2(a + b) = \)
- The number of solutions of the equations \[ x + y + z = 1,\quad x^2 + y^2 + z^2 = 1,\quad x^3 + y^3 + z^3 = 1 \] is:
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions