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Quadratic Equation and Inequalities question

2002 · Shift 0 · Q100

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Mathematics
Quadratic Equation and Inequalities
2002 · Shift 0 · Q100

Quadratic Equation and Inequalities question

2002 · Shift 0 · Q100

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1

If

\alpha e \beta

but

{\alpha ^2} = 5\alpha - 3

and

{\beta ^2} = 5\beta - 3

then the equation having

\alpha /\beta

and

\beta /\alpha \,\,

as its roots is

A
$3{x^2} - 19x + 3 = 0$
B
$3{x^2} + 19x - 3 = 0$
C
$3{x^2} - 19x - 3 = 0$
D
${x^2} - 5x + 3 = 0$

View written solutionFree

Correct answer: A

Since both $\alpha$ and $\beta$ satisfy $x^2=5x-3,$ we rewrite it as $x^2-5x+3=0.$
Hence $\alpha,\beta$ are the roots of the quadratic $x^2-5x+3=0.$ So by Vieta's formulas, $\alpha+\beta=5, \qquad \alpha\beta=3.$
We need the equation whose roots are $\frac{\alpha}{\beta} \quad \text{and} \quad \frac{\beta}{\alpha}.$ Let these roots be $r_1$ and $r_2$ .

Then

=\frac{\alpha^2+\beta^2}{\alpha\beta}.$$ Now, $$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=5^2-2\cdot 3=25-6=19.$$ Thus, $$r_1+r_2=\frac{19}{3}.$$ 4. Their product is $$r_1r_2=\frac{\alpha}{\beta}\cdot\frac{\beta}{\alpha}=1.$$ 5. Therefore the required quadratic equation is $$x^2-(r_1+r_2)x+r_1r_2=0,$$ so $$x^2-\frac{19}{3}x+1=0.$$ Multiplying by $3$, $$3x^2-19x+3=0.$$ 6. Comparing with the options, this is **Option A**.

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