JEE Main 2023 — Mathematics Algebra

Given $\alpha$α and $\beta$β are roots of the below equation

$$14{x}^{2}-31x+3\lambda =0$$14x2−31x+3λ=0

So, $\alpha +\beta =\frac{31}{14}\ldots (1)$α+β=1431​…(1)

And, $\alpha \beta =\frac{3\lambda }{14}\ldots (2)$αβ=143λ​…(2)

Given $\alpha$α and $\gamma$γ are roots of the below equation

$$35{x}^{2}-53x+4\lambda =0$$35x2−53x+4λ=0

So, $\alpha +\gamma =\frac{53}{35}\ldots (3)$α+γ=3553​…(3)

And, $\alpha \gamma =\frac{4\lambda }{35}\ldots (4)$αγ=354λ​…(4)

On dividing equation$(2)$(2) and equation $(4)$(4),

$$\frac{\beta }{\gamma }=\frac{3\times 35}{4\times 14}=\frac{15}{8}\Rightarrow \beta =\frac{15}{8}\gamma$$γβ​=4×143×35​=815​⇒β=815​γ

On subtracting equation $(1)$(1) and equation $(3)$(3),

$$\Rightarrow \beta -\gamma =\frac{31}{14}-\frac{53}{35}=\frac{155-106}{70}=\frac{7}{10}$$⇒β−γ=1431​−3553​=70155−106​=107​

$$\Rightarrow \frac{15}{8}\gamma -\gamma =\frac{7}{10}\Rightarrow \gamma =\frac{4}{5}$$⇒815​γ−γ=107​⇒γ=54​

$$\Rightarrow \beta =\frac{15}{8}\times \frac{4}{5}=\frac{3}{2}$$⇒β=815​×54​=23​

So, $\alpha =\frac{31}{14}-\beta =\frac{31}{14}-\frac{3}{2}=\frac{5}{7}$α=1431​−β=1431​−23​=75​

And $\lambda =\frac{14}{3}\alpha \beta =\frac{14}{3}\times \frac{5}{7}\times \frac{3}{2}=5$λ=314​αβ=314​×75​×23​=5

So, the sum of the roots of the required equation,

$$\frac{3\alpha }{\beta }+\frac{4\alpha }{\gamma }=(\frac{3\alpha \gamma +4\alpha \beta }{\beta \gamma })$$β3α​+γ4α​=(βγ3αγ+4αβ​)

$$=\frac{(3\times \frac{4\lambda }{35}+4\times \frac{3\lambda }{14})}{\beta \gamma }=\frac{12\lambda (14+35)}{14\times 35\beta \gamma }$$=βγ(3×354λ​+4×143λ​)​=14×35βγ12λ(14+35)​

$$=\frac{49\times 12\times 5}{490\times \frac{3}{2}\times \frac{4}{5}}=5$$=490×23​×54​49×12×5​=5

And the product of the roots of the required equation,

$$=\frac{3\alpha }{\beta }\times \frac{4\alpha }{\gamma }=\frac{12{\alpha }^{2}}{\beta \gamma }=\frac{12\times \frac{25}{49}}{\frac{3}{2}\times \frac{4}{5}}=\frac{250}{49}$$=β3α​×γ4α​=βγ12α2​=23​×54​12×4925​​=49250​

So, the required equation is

$${x}^{2}-(\text{sum of roots})x+\text{product of roots}=0$$x2−(sum of roots)x+product of roots=0

$$\Rightarrow {x}^{2}-5x+\frac{250}{49}=0$$⇒x2−5x+49250​=0

$$\Rightarrow 49{x}^{2}-245x+250=0$$⇒49x2−245x+250=0