Question

Construct the quadratic equation whose roots are $3$ times the roots $5x^2-3x+3=0$

• A. $5x^2-9x+27=0$
• B. $5x^2+9x+27=0$
• C. $5x^2-9x-27=0$
• D. $5x^2+9x-27=0$

Answer 1

The correct option is A

$5x^2-9x+27=0$

Explanation for correct option:

Step-1: Finding sum and product of the roots of the given data.

we know that the sum and product of the roots of the equation $a{x}^2+bx+c=0$

Sum of roots $=-\frac{b}{a}$

Product of roots $=\frac{c}{a}$

Let, $\alpha ,\beta$ are roots of equation $5x^2-3x+3=0$

$\Rightarrow$$\alpha +\beta =\frac{3}{5}\&\alpha \times \beta =\frac{3}{5}$

Step-2: Finding relation between roots of required quadratic equation.

Let, ${\alpha }_1,{\beta }_1$ are roots of required equation.

$\Rightarrow$${\alpha }_1+{\beta }_1=\frac{3}{5}\&{\alpha }_1\times {\beta }_1=\frac{3}{5}$

Given, $3{\alpha }_1=\alpha \&3{\beta }_1=\beta$

$\Rightarrow$$\frac{\alpha }{3}+\frac{\beta }{3}=\frac{3}{5}\&\frac{\alpha }{3}\times \frac{\beta }{3}=\frac{3}{5}$

$\Rightarrow$$\left(\alpha +\beta \right)=\frac{9}{5}\&\alpha \beta =\frac{27}{5}$

Step-3: Finding required quadratic equation.

we know that quadratic equation is $x^2-\left(\alpha +\beta \right)x+\alpha \beta =0$

$\Rightarrow$$x^2-\left(\frac{9}{5}\right)x+\frac{27}{5}=0$

$\Rightarrow$$5x^2-9x+27=0$

Therefore required quadratic equation is $5x^2-9x+27=0$.

Hence, correct answer is option $\left(A\right)$