Solveeit Logo
Login

Question

Question: If a quadratic equation \({{x}^{2}}-10ax-11b=0\) has roots c and d, or, an another quadratic equatio...

If a quadratic equation x2−10ax−11b=0{{x}^{2}}-10ax-11b=0 has roots c and d, or, an another quadratic equation x2−10cx−11d=0{{x}^{2}}-10cx-11d=0 has roots a and b, then a+b+c+d=.
(a). 1220
(b). 1110
(c). 1210
(d). 1310

Explanation

Solution

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 ax2+bx+c=0a{{x}^{2}}+bx+c=0 , the roots are given by:
x=−b±b2−4ac2ax=\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 = −(coefficient of x)coefficient of x2=−ba\dfrac{-\left( \text{coefficient of x} \right)}{\text{coefficient of }{{\text{x}}^{2}}}=\dfrac{-b}{a} .
Product of the roots of quadratic equation = constant termcoefficient of x2=ca\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 x2−10ax−11b=0{{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.
c2−10ac−11b=0.........(i){{c}^{2}}-10ac-11b=0.........(i)
Using the relation of the sum of roots for the quadratic equation, we get
c+d=10a.......(ii)c+d=10a.......(ii)
Now, let us find some results for the quadratic equation x2−10cx−11d=0{{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.
a2−10ac−11d=0.........(iii){{a}^{2}}-10ac-11d=0.........(iii)
Using the relation of sum of roots for the quadratic equation, we get
a+b=10c.......(iv)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
a2−c2−11d+11b=0{{a}^{2}}-{{c}^{2}}-11d+11b=0
Substituting the value of (b-d) from equation (v), we get
a2−c2+11×11(c−a)=0{{a}^{2}}-{{c}^{2}}+11\times 11\left( c-a \right)=0
⇒(a+c)(a−c)=11×11(a−c)\Rightarrow \left( a+c \right)\left( a-c \right)=11\times 11\left( a-c \right)
⇒(a+c)=121\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)= 10×121=121010\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.