If alpha and beta are the roots of the quadratic equation x-2-12x-35 - 0- then the value of 2begin-pmatrix-dfrac-alpha-beta-dfrac-beta-alpha-end-pmatrix- -dfrac-74-35-dfrac-148-35-dfrac-12-35-dfrac-34-35-

Question

Toppr Admin · Accepted Answer

Let us solve the given quadratic equation $x^{2} - 12 x + 35 = 0$

$⟹$ $x^{2} - 7 x - 5 x + 35 = 0$

$⟹ (x - 7) (x - 5) = 0$

$⟹ x - 7 = 0$ or $x - 5 = 0$

$⟹ x = 7$ or $x = 5$

Thus, $7,$ are roots of the given equation.

Let $α = 7$ and $β = 5$

$∴ 2 (\frac{α}{β} + \frac{β}{α}) = 2 (\frac{7}{5} + \frac{5}{7}) = 2 (\frac{74}{35}) = \frac{148}{35}$
Hence, option B is correct.

If α and β are the roots of the quadratic equation x2−12x+35=0, then find the value of 2(αβ+βα):74351483512353435

Let us solve the given quadratic equation x2−12x+35=0⟹ x2−7x−5x+35=0⟹ (x−7)(x−5)=0⟹ x−7=0 or x−5=0⟹ x=7 or x=5Thus, 7, are roots of the given equation.Let α=7 and β=5∴ 2(αβ+βα)=2(75+57)=2(7435)=14835Hence, option B is correct.

If $α$ and $β$ are the roots of the quadratic equation $x^{2} - 12 x + 35 = 0$ , then find the value of $2 (\begin{matrix} \frac{α}{β} + \frac{β}{α} \end{matrix}) :$
$\frac{74}{35}$
$\frac{148}{35}$
$\frac{12}{35}$
$\frac{34}{35}$