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Quadratic Equation and Inequalities question

2022 · 28 Jul · Shift 2 · Q22
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  5. /2022 · 28 Jul · Shift 2 · Q22

Quadratic Equation and Inequalities question

2022 · 28 Jul · Shift 2 · Q22

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
Let α\alphaα, β\betaβ be the roots of the equation x2−2x+6=0x^{2}-\sqrt{2} x+\sqrt{6}=0x2−2​x+6​=0 and 1α2+1,1β2+1\frac{1}{\alpha^{2}}+1, \frac{1}{\beta^{2}}+1α21​+1,β21​+1 be the roots of the equation x2+ax+b=0x^{2}+a x+b=0x2+ax+b=0. Then the roots of the equation x2−(a+b−2)x+(a+b+2)=0x^{2}-(a+b-2) x+(a+b+2)=0x2−(a+b−2)x+(a+b+2)=0 are :
  1. A
    non-real complex numbers
  2. B
    real and both negative
  3. C
    real and both positive
  4. D
    real and exactly one of them is positive
View written solutionFree

Correct answer: B

  1. Given quadratic and its roots

The roots α,β\alpha,\betaα,β satisfy x2−2x+6=0.x^2-\sqrt{2}x+\sqrt{6}=0.x2−2​x+6​=0. So by Vieta's formulas, α+β=2,αβ=6.\alpha+\beta=\sqrt{2}, \qquad \alpha\beta=\sqrt{6}.α+β=2​,αβ=6​.

  1. Form the new roots

We are told that 1α2+1and1β2+1\frac{1}{\alpha^2}+1 \quad \text{and} \quad \frac{1}{\beta^2}+1α21​+1andβ21​+1 are the roots of x2+ax+b=0.x^2+ax+b=0.x2+ax+b=0. Let r1=1+1α2,r2=1+1β2.r_1=1+\frac{1}{\alpha^2}, \qquad r_2=1+\frac{1}{\beta^2}.r1​=1+α21​,r2​=1+β21​.

Then r1+r2=−a,r1r2=b.r_1+r_2=-a, \qquad r_1r_2=b.r1​+r2​=−a,r1​r2​=b.

We compute these.

  1. Find 1α2+1β2\dfrac{1}{\alpha^2}+\dfrac{1}{\beta^2}α21​+β21​

First, α2+β2=(α+β)2−2αβ=2−26.\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=2-2\sqrt{6}.α2+β2=(α+β)2−2αβ=2−26​. Also, α2β2=(αβ)2=6.\alpha^2\beta^2=(\alpha\beta)^2=6.α2β2=(αβ)2=6. Hence,

=\frac{2-2\sqrt{6}}{6}=\frac{1-\sqrt{6}}{3}.$$ Therefore, $$r_1+r_2=2+\frac{1-\sqrt{6}}{3}=\frac{7-\sqrt{6}}{3}.$$ So, $$-a=\frac{7-\sqrt{6}}{3} \quad\Rightarrow\quad a=-\frac{7-\sqrt{6}}{3}=\frac{\sqrt{6}-7}{3}.$$ --- 4. **Find $b$** $$b=r_1r_2=\left(1+\frac{1}{\alpha^2}\right)\left(1+\frac{1}{\beta^2}\right).$$ Expand: $$b=1+\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\alpha^2\beta^2}.$$ Now, $$\frac{1}{\alpha^2\beta^2}=\frac{1}{(\alpha\beta)^2}=\frac{1}{6}.$$ So, $$b=1+\frac{1-\sqrt{6}}{3}+\frac{1}{6}.$$ Taking LCM $6$, $$b=\frac{6}{6}+\frac{2-2\sqrt{6}}{6}+\frac{1}{6} =\frac{9-2\sqrt{6}}{6}.$$ --- 5. **Compute $a+b$** $$a+b=\frac{\sqrt{6}-7}{3}+\frac{9-2\sqrt{6}}{6}.$$ Taking denominator $6$, $$a+b=\frac{2\sqrt{6}-14+9-2\sqrt{6}}{6}=\frac{-5}{6}.$$ Thus, $$a+b=-\frac{5}{6}.$$ --- 6. **Form the required quadratic** We need the roots of $$x^2-(a+b-2)x+(a+b+2)=0.$$ Since $a+b=-\frac{5}{6}$, $$a+b-2=-\frac{5}{6}-2=-\frac{17}{6},$$ $$a+b+2=-\frac{5}{6}+2=\frac{7}{6}.$$ Therefore the equation becomes $$x^2-\left(-\frac{17}{6}\right)x+\frac{7}{6}=0,$$ that is, $$x^2+\frac{17}{6}x+\frac{7}{6}=0.$$ Multiply by $6$: $$6x^2+17x+7=0.$$ --- 7. **Nature and sign of roots** Discriminant: $$\Delta=17^2-4\cdot 6\cdot 7=289-168=121>0.$$ So the roots are real. Now, $$\text{sum of roots}=-\frac{17}{6}<0, \qquad \text{product of roots}=\frac{7}{6}>0.$$ If two real roots have positive product and negative sum, both roots are negative. In fact, $$6x^2+17x+7=(3x+7)(2x+1),$$ so the roots are $$x=-\frac{7}{3}, \qquad x=-\frac{1}{2},$$ both negative. --- 8. **Evaluate options** - **A:** non-real complex numbers $\rightarrow$ False - **B:** real and both negative $\rightarrow$ True - **C:** real and both positive $\rightarrow$ False - **D:** real and exactly one positive $\rightarrow$ False Therefore, the correct option is $$\boxed{\text{B}}.$$
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