Question

If a quadratic equation ${{x}^{2}}-10ax-11b=0$ has roots c and d, or, an another quadratic equation ${{x}^{2}}-10cx-11d=0$ has roots a and b, then a+b+c+d=.
(a). 1220
(b). 1110
(c). 1210
(d). 1310

Answer 1

Hint: Use the relation between the zeroes and the coefficients of a polynomial. Also, use the property that a root of an equation always satisfies the equation.

Complete step-by-step solution -
 We know, for standard quadratic equation $a{{x}^{2}}+bx+c=0$ , the roots are given by:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Also, the relation between the coefficients and the roots of a general quadratic equation comes out to be:
Sum of the roots of quadratic equation = $\dfrac{-\left( \text{coefficient of x} \right)}{\text{coefficient of }{{\text{x}}^{2}}}=\dfrac{-b}{a}$ .
Product of the roots of quadratic equation = \[\dfrac{\text{constant term}}{\text{coefficient of }{{\text{x}}^{2}}}\text{=}\dfrac{\text{c}}{\text{a}}\] .
First, let us find some results for the quadratic equation ${{x}^{2}}-10ax-11b=0$ whose roots are c and d. So, we put x=c in the quadratic equation, and as it is the root of the equation, it must satisfy the given polynomial.
${{c}^{2}}-10ac-11b=0.........(i)$
Using the relation of the sum of roots for the quadratic equation, we get
$c+d=10a.......(ii)$
Now, let us find some results for the quadratic equation ${{x}^{2}}-10cx-11d=0$ whose roots are a and b. So, we put x=a in the quadratic equation, and as it is the root of the equation, it must satisfy the given polynomial.
${{a}^{2}}-10ac-11d=0.........(iii)$
Using the relation of sum of roots for the quadratic equation, we get
$a+b=10c.......(iv)$
If we subtract the equation (ii) from the equation (iv), we get
a-c+b-d=10c-10a
b-d=11(c-a)………..(v)
Now let us subtract equation (i) from equation (iii). On doing so, we get
${{a}^{2}}-{{c}^{2}}-11d+11b=0$
Substituting the value of (b-d) from equation (v), we get
${{a}^{2}}-{{c}^{2}}+11\times 11\left( c-a \right)=0$
$\Rightarrow \left( a+c \right)\left( a-c \right)=11\times 11\left( a-c \right)$
$\Rightarrow \left( a+c \right)=121$
Now we will add the equation (ii) and equation (iv). On doing so, we get
a+b+c+d=10(a+c)= $10\times 121=1210$
Therefore, the answer to the above question is option (c).

Note: Be careful about the calculations and signs. Also, remember all the relations related to the roots and coefficients of the polynomial as they are used in almost every question related to a polynomial, as we have seen in the above question.