JEE AdvancedMathematicsQuadratic Equation and InequalitiesMCQ+3 / −1
Suppose a, b denote the distinct real roots of the quadratic polynomial x2 + 20x 2020 and suppose c, d denote the distinct complex roots of the quadratic polynomial x2 20x + 2020. Then the value of ac(a c) + ad(a d) + bc(b c) + bd(b d) is
- A0
- B8000
- C8080
- D16000
View written solutionFree
Correct answer: D
- Let and its distinct real roots be .
Using Vieta's formulas:
\qquad ab=-2020.$$ 2. Let $$Q(x)=x^2-20x+2020$$ and its distinct complex roots be $c,d$. Again by Vieta's formulas: $$c+d=20, \qquad cd=2020.$$ 3. We need to evaluate $$S=ac(a-c)+ad(a-d)+bc(b-c)+bd(b-d).$$ Group terms with $a$ and $b$: $$S=a\big(c(a-c)+d(a-d)\big)+b\big(c(b-c)+d(b-d)\big).$$ Now simplify inside: $$c(a-c)+d(a-d)=a(c+d)-(c^2+d^2),$$ $$c(b-c)+d(b-d)=b(c+d)-(c^2+d^2).$$ So, $$S=a\big(a(c+d)-(c^2+d^2)\big)+b\big(b(c+d)-(c^2+d^2)\big).$$ Hence, $$S=(a^2+b^2)(c+d)-(a+b)(c^2+d^2).$$ 4. Compute $a^2+b^2$ and $c^2+d^2$. For $a,b$: $$a^2+b^2=(a+b)^2-2ab=(-20)^2-2(-2020)=400+4040=4440.$$ For $c,d$: $$c^2+d^2=(c+d)^2-2cd=20^2-2(2020)=400-4040=-3640.$$ 5. Substitute: $$S=4440(20)-(-20)(-3640).$$ Now, $$4440\cdot 20=88800,$$ $$(-20)(-3640)=72800.$$ Therefore, $$S=88800-72800=16000.$$ 6. So the correct option is $$\boxed{16000}$$ which is option $\boxed{D}$. 7. Comparison with stored answer: Stored correct answer is $D$, which matches our result.More from Quadratic Equation and Inequalities
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