Question

The value of a for which one root of the quadratic equations (a²-5a+3)x² + (3a-2)x+2 = 0 is twice as large other is

Answer 1

EXPLANATION.

one roots of the quadratic equation

=> ( a² - 5a + 3 )x² + ( 3a - 2 )x + 2 = 0

=> twice as larger other.

$\sf : \implies \: let \: we \: assume \: that \: \alpha and \: 2 \alpha \: be \: the \: roots \: of \: equation. \\ \\ \sf : \implies \: sum \: of \: roots \: of \: quadratic \: equation \: = \alpha + \beta = \frac{ - b}{a} \\ \\ \sf : \implies \: \alpha + 2 \alpha = 3 \alpha = \frac{ 1 - 3 a }{ {a}^{2} - 5a + 3} \: \: \: ......(1)$

$\sf : \implies \: product \: of \: quadratic \: equation \: = \alpha \beta = \dfrac{c}{a} \\ \\ \sf : \implies \: \alpha (2 \alpha ) = 2 { \alpha }^{2} = \frac{2}{ {a}^{2} - 5a + 3} \: \: \: \: ......(2)$

$\sf : \implies \: on \: squaring \: equation \: (1) \: \: we \: \: get \\ \\ \sf : \implies \: 9 { \alpha }^{2} = \frac{( 1- 3a) {}^{2} }{( {a}^{2} - 5a + 3) {}^{2} } \\ \\ \sf : \implies \: dividing \: equation \: (1) \: \: and \: \: (2) \: \: we \: \: get$

$\sf : \implies \: \dfrac{9 { \alpha }^{2} }{2 { \alpha }^{2} } = \dfrac{(1 - 3a) {}^{2} }{( {a}^{2} - 5a + 3) {}^{2} } \times \dfrac{( {a}^{2} - 5a + 3)}{2} \\ \\ \sf : \implies \: \frac{9}{2} = \frac{(1 - 3a) {}^{2} }{2( {a}^{2} - 5a + 3) } \\ \\ \sf : \implies \: 9( {a}^{2} - 5a + 3) = (1 - 3a) {}^{2}$

$\sf : \implies \: 9 {a}^{2} - 45a + 27 = 1 + 9 {a}^{2} - 6a \\ \\ \sf : \implies \: 26 = 39a \\ \\ \sf : \implies \: a \: = \frac{2}{3}$

Answer 2

Answer:

✳️ Given ✳️

✒️ The value of 'a' for which one root of the quadratic equation (a² - 5a + 3)x² + (3a - 1)x + 2 = 0

is twice as large .

✳️ To Find ✳️

✒️ What is the value of a.

✳️ Solution ✳️

✍️ Let the roots be $\sf\alpha{_1}$$\beta$, given $\beta$ = 2a

$\implies$ 3$\alpha$ = $\dfrac{-(3a - 1)}{a² - 5a + 3}$ ...... Equation no 1 ( Sum of roots)

$\implies$ 2$\alpha$² = $\dfrac{2}{a² - 5a + 3}$ ..... Equation no 2 (Product of roots)

$\implies$ $\dfrac{(3a - 1)²}{(a² - 5a + 3)²}$ = $\dfrac{9}{a² - 5a + 3}$

$\implies$ $\dfrac{(3a - 1)² - 9(a² - 5a + 3)}{(a² - 5a + 3)²}$ = 0

$\implies$ 9a² - 6a + 1 - 9a² + 45a - 27 = 0

$\implies$ 39a - 26 = 0

$\implies$ 39a = 26

$\implies$ a = $\sf\dfrac{\cancel{26}}{\cancel{39}}$

$\dashrightarrow$ a = $\dfrac{2}{3}$

$\therefore$ The value of a = $\dfrac{2}{3}$

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