Question

If roots of the quadratic equation $\bf 2x^2-6x+3=0$ are $\bf \alpha \:,\:\beta$
Then prove that

$\Large{\sf \dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}+5\left(\dfrac{1}{\alpha}+\dfrac{1}{\beta}\right)+2\alpha\beta=17}$​

Answer 1

Answer:-

Given:-

α & β are the roots of 2x² - 6x + 3.

On comparing with the standard form of a quadratic equation i.e., ax² + bx + c = 0 we get,

• a = 2
• b = - 6
• c = 3

We know that,

Sum of the roots = - b/a

⟹ α + β = - ( - 6/2)

⟹ α + β = 3 -- equation (1)

And,

Product of the roots = c/a

⟹ αβ = 3/2 -- equation (2)

Now,

We have to prove:

$\implies \sf \dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}+5\bigg(\dfrac{1}{\alpha}+\dfrac{1}{\beta}\bigg)+2\alpha\beta=17 \\ \\ \\ \implies \sf \: \frac{ { \alpha }^{2} + { \beta }^{2} }{ \alpha \beta } + 5 \bigg( \frac{ \beta + \alpha }{ \alpha \beta } \bigg) + 2 \alpha \beta = 17$

using a² + b² = (a + b)² - 2ab in LHS we get,

$\implies \sf \: \dfrac{( { \alpha + \beta )}^{2} - 2 \alpha \beta }{ \frac{3}{2} } + 5 \bigg( \frac{3}{ \frac{3}{2} } \bigg) + 2 \times \frac{3}{2} = 17 \: \: \: \: \: \: \: \: \: \: \: ( from \: (1) \& \: (2)) \\ \\ \\ \implies \sf \: \bigg( {(3)}^{2} - 2 \times \frac{3}{2} \bigg) \times \frac{2}{3} + 5 \times 3 \times \frac{2}{3} + 3 = 17 \\ \\ \\ \implies \sf \: (9 - 3) \times \frac{2}{3} + 10 + 3 = 17 \\ \\ \\ \implies \sf \:6 \times \frac{2}{3} + 13 = 17 \\ \\ \\ \implies \sf \: 4 + 13 = 17 \\ \\ \\ \implies \sf \:17 = 17$

Hence, Proved.

Answer 2

Given,

☛ Quadratic equation is,

$\bullet\:\tt{2x^2-6x+3=0}$

☛ Roots of the quadratic equation are α & β.

As we know that,

✯ Sum of roots = $\tt\red{\dfrac{-b}{a}}$

➵ α + β = $\tt{\dfrac{-(-6)}{2}}$

➵ α + β = $\tt{\dfrac{6}{2}}$

➵ α + β = $\bf{3}$

✯ Product of roots = $\tt\red{\dfrac{c}{a}}$

➵ α.β = $\tt{\dfrac{3}{2}}$

➵ α.β = $\bf{1.5}$

Now,

☛ Let us assume L.H.S given as,

$\checkmark\:\tt {\dfrac{\alpha}{\beta}\:+\:\dfrac{\beta}{\alpha}\:+\:5\bigg(\dfrac{1}{\alpha}\:+\:\dfrac{1}{\beta}\bigg)\:+\:2\alpha\beta\:}$

$:\implies\:\tt {\dfrac{\alpha^2\:+\:\beta^2}{\alpha\:\beta}\:+\:5\bigg(\dfrac{\beta\:+\:\alpha}{\alpha\:\beta}\bigg)\:+\:2\alpha\beta\:}$

$:\implies\:\tt {\dfrac{\alpha^2\:+\:\beta^2\:+\:2\alpha\beta\:-\:2\alpha\beta}{\alpha\:\beta}\:+\:5\bigg(\dfrac{\alpha\:+\:\beta}{\alpha\:\beta}\bigg)\:+\:2\alpha\beta\:}$

$:\implies\:\tt {\dfrac{(\alpha\:+\:\beta)^2\:-\:2\alpha\beta}{\alpha\:\beta}\:+\:5\bigg(\dfrac{\alpha\:+\:\beta}{\alpha\:\beta}\bigg)\:+\:2\alpha\beta\:}$

$:\implies\:\tt {\dfrac{3^2\:-\:(2\times{1.5})}{1.5}\:+\:5\bigg(\dfrac{3}{1.5}\bigg)\:+\:2\times{1.5}\:}$

$:\implies\:\tt {\dfrac{9\:-\:3}{1.5}\:+\:5\times{2}\:+\:3\:}$

$:\implies\:\tt {\dfrac{6}{1.5}\:+\:10\:+\:3\:}$

$:\implies\:\tt {4\:+\:10\:+\:3\:}$

$:\implies\:\bf\pink{17\:} ~~~[R.H.S]~$

∴ L.H.S = R.H.S ⠀⠀⠀[Hence proved]