Question

a,b,c,d are real n distinct.

a and b are roots of x square -2cx-5d =0

c and d are roots of x square - 2ax -5b =0

Find numerical value of a+b+c+d​

Answer 1

Answer:

$\bold\red{a+b+c+d=30}$

Step-by-step explanation:

Given,

Quadratic equations,

$\green{{x}^{2}-2cx-5d=0}$

'a' and 'b' are roots of this equation.

Therefore,

a + b = 2c .....(i)

ab = -5d .......(ii)

Also,

'c' and 'd' are roots of Quadratic equation

$\green{{x}^{2}-2ax-5b=0}$

Therefore,

c + d = 2a .......(iii)

cd = -5b .........(iv)

From (i) and (iii),

we get,

a + b + c + d = 2(a+c)

=> a + c = b + d

From (ii) and (iv),

we get,

abcd = 25bd

=> ac = 25

Now,

Multiply a with (i) and c with (iii) and adding,

we get,

${a}^{2} + ab + {c}^{2} + cd = 2ac + 2ca = 4ac \\ \\ = > {a}^{2} + {c}^{2} - 5b - 5d = 100 \\ \\ = > {a}^{2} + {c}^{2} - 5(b + d) - 100 = 0 \\ \\ = > {a}^{2} + {c}^{2} - 5(a + c) - 100 = 0 \\ \\ = > {(a + c)}^{2} - 5(a + c) - 150 = 0$

Now,

it is a Quadratic equation of (a+c) as variable

Therefore,

we get,

$= > (a + c) = \frac{5± \sqrt{25 + 600} }{2} \\ \\ = > (a + c) = \frac{5± \sqrt{625} }{2} \\ \\ = > (a + c) = \frac{5± 25 }{2} \\ \\ = >(a + c) = \frac{5 + 25 }{2} \: \: or \: \: \frac{5 - 25}{2} \\ \\ = >(a + c) = 15 \: \: or \: \: - 10$

But, it is given that,

'a' and 'c' are distinct,

therefore,

-10 is not possible.

Therefore,

(a+c ) = (b +d) = 15

Hence,

a + b + c + d = 30