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Hint: We will use the direct condition to find the common roots of the given equation. For part (a) we can directly solve the equation by eliminating ‘a’ from both the equations. For part (b) we can use the formula, as we know that the common root of the equations ${a_1}{x^2} + {b_1}x + {c_1} = 0$ and ${a_2}{x^2} + {b_2}x + {c_2} = 0$ is given by $\dfrac{{{\alpha ^2}}}{{{b_1}{c_2} - {b_2}{c_1}}} = \dfrac{\alpha }{{{c_1}{a_2} - {a_1}{c_2}}} = \dfrac{1}{{{a_1}{b_2} - {a_2}{b_1}}}$. Using this formula will get the result. Part (c) can be solved by taking conjugate of z.
Now from the question we have,
a) we have equations $(2a - 5){x^2} - 4x - 15 = 0$ or,
$2a{x^2} - 5{x^2} - 4x - 15 = 0$ ….(1)
$(3a - 8){x^2} - 5x - 21 = 0$ or,
$3a{x^2} - 8{x^2} - 5x - 21 = 0$ ….(2)
Multiplying equation (1) by 3 and equation (2) by 2 to eliminate the term a, we get
${x^2} - 2x - 3 = 0$, solving this equation we get
x = 3 and x = -1
putting x = 3 and x = -1 in the any equation (1) and (2), we get
a = 4 and a = 8, which is the required answer.
b) the given equations are ${x^2} - x - p = 0$ and ${x^2} - 2xp - 12 = 0$
let the root be $\alpha $, then putting x = $\alpha $ in the above two equation we get,
${\alpha ^2} - \alpha - p = 0$, ${\alpha ^2} + 2p\alpha - 12 = 0$
Now solving the above equations we get,
$\alpha $ = 2, which is the required answer.
c) we have given the equation
$\overline z = z$ as z is real.
Taking conjugate in the whole equation we have
${z^2} + \overline \alpha z + \overline \beta = 0$
$\therefore \dfrac{{{z^2}}}{{\alpha \overline \beta }} = \dfrac{z}{{\beta - \overline \beta }} = \dfrac{1}{{\overline \alpha - \alpha }}$
$\therefore (\beta - \overline \beta ) = (\overline \alpha - \alpha )(\alpha \overline \beta - \overline \alpha \beta )$
Which is the required condition.
NOTE Quadratic equation is any equation that can be arranged in the form of $a{x^2} + bx + c = 0$
Where , x represents an unknown, and a, b, and c represent known numbers, where a $ \ne $ 0. If a = 0, then the equation will become linear not quadratic, as there is no $a{x^2}$ term. The numbers a, b, c are the coefficients of the equation and may be distinguished by calling them, respectively, the quadratic coefficient, the linear coefficient and the constant term.
Now from the question we have,
a) we have equations $(2a - 5){x^2} - 4x - 15 = 0$ or,
$2a{x^2} - 5{x^2} - 4x - 15 = 0$ ….(1)
$(3a - 8){x^2} - 5x - 21 = 0$ or,
$3a{x^2} - 8{x^2} - 5x - 21 = 0$ ….(2)
Multiplying equation (1) by 3 and equation (2) by 2 to eliminate the term a, we get
${x^2} - 2x - 3 = 0$, solving this equation we get
x = 3 and x = -1
putting x = 3 and x = -1 in the any equation (1) and (2), we get
a = 4 and a = 8, which is the required answer.
b) the given equations are ${x^2} - x - p = 0$ and ${x^2} - 2xp - 12 = 0$
let the root be $\alpha $, then putting x = $\alpha $ in the above two equation we get,
${\alpha ^2} - \alpha - p = 0$, ${\alpha ^2} + 2p\alpha - 12 = 0$
Now solving the above equations we get,
$\alpha $ = 2, which is the required answer.
c) we have given the equation
$\overline z = z$ as z is real.
Taking conjugate in the whole equation we have
${z^2} + \overline \alpha z + \overline \beta = 0$
$\therefore \dfrac{{{z^2}}}{{\alpha \overline \beta }} = \dfrac{z}{{\beta - \overline \beta }} = \dfrac{1}{{\overline \alpha - \alpha }}$
$\therefore (\beta - \overline \beta ) = (\overline \alpha - \alpha )(\alpha \overline \beta - \overline \alpha \beta )$
Which is the required condition.
NOTE Quadratic equation is any equation that can be arranged in the form of $a{x^2} + bx + c = 0$
Where , x represents an unknown, and a, b, and c represent known numbers, where a $ \ne $ 0. If a = 0, then the equation will become linear not quadratic, as there is no $a{x^2}$ term. The numbers a, b, c are the coefficients of the equation and may be distinguished by calling them, respectively, the quadratic coefficient, the linear coefficient and the constant term.
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