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Quadratic Equation and Inequalities question

2014 · Shift 0 · Q40
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  5. /2014 · Shift 0 · Q40

Quadratic Equation and Inequalities question

2014 · Shift 0 · Q40

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
Let α\alphaα and β\betaβ be the roots of equation px2+qx+r=0,pe0.p{x^2} + qx + r = 0,p e 0.px2+qx+r=0,pe0. If p, q, rp,\,q,\,rp,q,r in A.P. and 1α+1β=4,{1 \over \alpha } + {1 \over \beta } = 4,α1​+β1​=4, then the value of ∣α−β∣\left| {\alpha - \beta } \right|∣α−β∣ is :
  1. A
    349{{\sqrt {34} } \over 9}934​​
  2. B
    2139{{2\sqrt 13 } \over 9}921​3​
  3. C
    619{{\sqrt {61} } \over 9}961​​
  4. D
    2179{{2\sqrt 17 } \over 9}921​7​
View written solutionFree

Correct answer: B

  1. Use Vieta’s formulas

For the quadratic equation px2+qx+r=0(p≠0),px^2+qx+r=0 \quad (p\ne 0),px2+qx+r=0(p=0), with roots α,β\alpha,\betaα,β: α+β=−qp,αβ=rp.\alpha+\beta=-\frac{q}{p}, \qquad \alpha\beta=\frac{r}{p}.α+β=−pq​,αβ=pr​.

Also, 1α+1β=α+βαβ=4.\frac1\alpha+\frac1\beta=\frac{\alpha+\beta}{\alpha\beta}=4.α1​+β1​=αβα+β​=4. So, −q/pr/p=−qr=4\frac{-q/p}{r/p}=-\frac{q}{r}=4r/p−q/p​=−rq​=4 which gives q=−4r.q=-4r.q=−4r.

  1. Use the A.P. condition

Since p,q,rp,q,rp,q,r are in A.P., 2q=p+r.2q=p+r.2q=p+r. Substitute q=−4rq=-4rq=−4r: 2(−4r)=p+r2(-4r)=p+r2(−4r)=p+r −8r=p+r-8r=p+r−8r=p+r p=−9r.p=-9r.p=−9r.

  1. Find sum and product of roots

Now, α+β=−qp=−−4r−9r=−49,\alpha+\beta=-\frac{q}{p}=-\frac{-4r}{-9r}=-\frac49,α+β=−pq​=−−9r−4r​=−94​, and αβ=rp=r−9r=−19.\alpha\beta=\frac{r}{p}=\frac{r}{-9r}=-\frac19.αβ=pr​=−9rr​=−91​.

  1. Compute ∣α−β∣|\alpha-\beta|∣α−β∣

Use (α−β)2=(α+β)2−4αβ.(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta.(α−β)2=(α+β)2−4αβ. Thus,

=\frac{16}{81}+\frac{4}{9} =\frac{16}{81}+\frac{36}{81} =\frac{52}{81}.$$ Therefore, $$|\alpha-\beta|=\sqrt{\frac{52}{81}}=\frac{\sqrt{52}}{9}=\frac{2\sqrt{13}}{9}.$$ 5. **Match with options** $$\boxed{\frac{2\sqrt{13}}{9}}$$ which is **Option B**.
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