Question

pls answer fast .
I will mark as brainliest​

Answer 1

QuEsTion :-

Given x₁ > 0 and x₂ = 4 x₁ are the solutions of the quadratic equation ax₂+bx+c = 0. If 3 a = 2 ( c - b ) then the value of x₁ is equal to :

Options :-

$\sf{(A)\;\; \frac{1}{4}\:\:\:\:\:\:\:\:\:\:(B)\;\;\frac{3}{2}}$

$\sf{(C)\;\;\frac{2}{3}\:\:\:\:\:\:\:\:\:(D)\;\;\frac{1}{9}}$

Knowledge required :-

▶ Sum of the zeroes of quadratic polynomial is given by

$\sf{\frac{-\;coefficient\:of\:x}{coefficient\:of\:x^2}}$

▶ Product of zeroes of quadratic polynomial is given by

$\sf{\frac{constant\:term}{coefficient\:of\:x^2}}$

▶ We should know how to find the roots any quadratic equation

SoluTion :-

Given the two zeroes of ax²+bx+c are

• x₁ and x₂ = 4 x₁

Also given that

• 3 a = 2 ( c - b )

⇨ Using Sum of the zeroes of ax²+bx+c

$\sf{\implies x_1+x_2=\frac{-b}{a}}$

$\implies\sf{x_1+4 x_1=\frac{-b}{a}}$

$\implies\sf{5x_1=\frac{-b}{a}}$

$\implies\underline{\underline{\pink{\sf{-b=a\;5x_1}}}}$

⇨ Using product of the zeroes of ax²+bx+c

$\implies\sf{x_1\;x_2=\frac{c}{a}}$

$\implies\sf{x_1\;4x_1=\frac{c}{a}}$

$\implies\sf{4\;{x_1}^2=\frac{c}{a}}$

$\implies\underline{\underline{\sf{\pink{c=4a\;{x_1}^2}}}}$

⇨ Substituting the value of (-b) and c

in 3a = (c - b)

$\longmapsto \sf{3a=2(c-b)}$

$\longmapsto \sf{3a=2(4a\;{x_1}^2+a\;5x_1)}$

$\longmapsto\sf{3a=2a(4{x_1}^2+5x_1)}$

$\longmapsto\sf{3\cancel{a}=2\cancel{a}(4{x_1}^2+5x_1)}$

$\longmapsto\underline{\underline{\pink{\sf{8{x_1}^2+10x_1-3=0}}}}$

⇨ Finding the roots of found quadratic eqn

$\rightarrowtail \sf{8{x_1}^2+10x_1-3=0}$

$\rightarrowtail \sf{8{x_1}^2+12x_1-2x_1-3=0}$

$\rightarrowtail \sf{4x_1(2x_1+3)-1(2x_1+3)=0}$

$\rightarrowtail \sf{(4x_1-1)(2x_1+3)=0}$

From here we will get

$\rightarrowtail\underline{\underline{\pink{\sf{x_1=\frac{1}{4}\;\;and\;\;x_1=\frac{-3}{2}}}}}$

But, $\sf{x_1=\frac{-3}{2} \;is\:\:less\:than\:zero}$

and given that x₁ > 0

so,

The correct answer is

$\boxed{\sf{\red{option\:(A)\:\frac{1}{4}}}}$

Answer 2

answer is a....means1/४