Question

If you don't know the answer don't answer. Find the quadratic equation whose roots are half of the reciprocal of the roots of the equation
ax2 + bx+c=0
(A) 4ax? + 2bx+c=0
(B) 4cx? +2bx + a =0
(C) 2cx? + bx + a = 0
(D) 2cx? +bx+c=0​

Answer 1

Given:-

⚉Quadratic Equation = ax² + bx + c = 0

Find:-

⚉Quadratic Equation whose roots are half of the reciprocal of the roots of the equation.

Solution:-

Let, the roots of quadratic equation ax² + bx + c = 0 are α and β

we, know that

$\large{ \underline{\boxed{ \sf Sum \: of \: zeroes = \alpha + \beta }}}$

$\Longrightarrow \sf \alpha + \beta = \dfrac{ - b}{a}$

Now,

$\large{ \underline{\boxed{ \sf Product \: of \: zeroes = \alpha\beta }}}$

$\Longrightarrow \sf \alpha\beta = \dfrac{c}{a}$

Now, ACCORDING TO QUESTION

The new roots would be

$:\to \sf \alpha' = \dfrac{1}{2 \alpha } \: and \: \beta' =\dfrac{1}{2 \beta }$

we, know that Quadratic Equation =

$\sf :\to {x}^{2} - (sum \: of \: roots)x + (product \: of \: zeroes) = 0$

$\sf :\to {x}^{2} - (\alpha ' + \beta')x + (\alpha'\beta') = 0$

where,

• α' + β' = 1/2α + 1/2β
• α' β' = 1/2α × 1/2β

So,

$\sf :\to {x}^{2} - \bigg ( \dfrac{1}{2 \alpha } + \dfrac{1}{2 \beta } \bigg)x + \bigg( \dfrac{1}{2 \alpha} \times \dfrac{1}{2 \beta } \bigg) = 0$

$\sf :\to {x}^{2} - \bigg ( \dfrac{ \alpha + \beta }{2 \alpha \beta } \bigg)x + \bigg(\dfrac{1}{4 \alpha \beta } \bigg) = 0$

$\sf :\to {x}^{2} - \dfrac{ \alpha + \beta }{2 \alpha \beta } x + \dfrac{1}{4 \alpha \beta } = 0$

$\sf \bigstar Taking \: L.C.M \bigstar$

$\sf :\to \dfrac{4 \alpha \beta {x}^{2} - 2( \alpha + \beta )x + 1}{4 \alpha \beta } = 0$

$\sf :\to 4 \alpha \beta {x}^{2} - 2( \alpha + \beta )x + 1= 0 \times 4 \alpha \beta$

$\sf :\to 4 \alpha \beta {x}^{2} - 2( \alpha + \beta )x + 1= 0$

where,

• $\sf \alpha\beta = \dfrac{c}{a}$
• $\sf \alpha + \beta = \dfrac{ - b}{a}$

So,

$\sf :\to 4(\dfrac{c}{a}){x}^{2} - 2( \dfrac{ - b}{a} )x + 1= 0$

$\sf :\to 4\dfrac{c}{a}{x}^{2} - 2\dfrac{ - b}{a} x + 1= 0$

$\sf :\to 4\dfrac{c}{a}{x}^{2} - 2\dfrac{b}{a} x + 1= 0$

$\sf \bigstar Taking \: L.C.M \: again\bigstar$

$\sf :\to \dfrac{4c {x}^{2} - 2bx + a }{a}= 0$

$\sf :\to 4c {x}^{2} - 2bx + a= 0 \times a$

$\sf :\to 4c {x}^{2} - 2bx + a= 0$

Hence, Quadratic Equation is 4cx² -2bx + a = 0

____________________________

Corrrect Option is:-

Option (B)