Verified Solution • 2025–26 Session moderate

QJEEM03Ma6a28: JEE Main 2003 Mathematics - Class 11 JEEM03Ma6a28 Solution: The value of ' ' for which one root of the quadratic equation…

The value of ' aa ' for which one root of the quadratic equation </mo></mrow><mrowdata−mjx−texclass="INNER"><modata−mjx−texclass="OPEN">(</mo><mrowdata−mjx−texclass="ORD"><mrowdata−mjx−texclass="ORD"><msup><mi>a</mi><mn>2</mn></msup></mrow><mo>−</mo><mn>5</mn><mi>a</mi><mo>+</mo><mn>3</mn></mrow><modata−mjx−texclass="CLOSE">)</mo></mrow><mrowdata−mjx−texclass="ORD"><msup><mi>x</mi><mn>2</mn></msup></mrow><mo>+</mo><mrowdata−mjx−texclass="INNER"><modata−mjx−texclass="OPEN">(</mo><mrowdata−mjx−texclass="ORD"><mn>3</mn><mi>a</mi><mo>−</mo><mn>1</mn></mrow><modata−mjx−texclass="CLOSE">)</mo></mrow><mi>x</mi><mo>+</mo><mn>2</mn><mo>=</mo><mn>0</mn></math></mjx−assistive−mml></mjx−container></mjx−assistive−mml></mjx−container></mjx−assistive−mml></mjx−container></mo></mrow><mrow data-mjx-texclass="INNER"><mo data-mjx-texclass="OPEN">(</mo><mrow data-mjx-texclass="ORD"><mrow data-mjx-texclass="ORD"><msup><mi>a</mi><mn>2</mn></msup></mrow><mo>−</mo><mn>5</mn><mi>a</mi><mo>+</mo><mn>3</mn></mrow><mo data-mjx-texclass="CLOSE">)</mo></mrow><mrow data-mjx-texclass="ORD"><msup><mi>x</mi><mn>2</mn></msup></mrow><mo>+</mo><mrow data-mjx-texclass="INNER"><mo data-mjx-texclass="OPEN">(</mo><mrow data-mjx-texclass="ORD"><mn>3</mn><mi>a</mi><mo>−</mo><mn>1</mn></mrow><mo data-mjx-texclass="CLOSE">)</mo></mrow><mi>x</mi><mo>+</mo><mn>2</mn><mo>=</mo><mn>0</mn></math></mjx-assistive-mml></mjx-container></mjx-assistive-mml></mjx-container></mjx-assistive-mml></mjx-container>
is twice as large as the other is

Multiple Choice Options

  • A −13- {1 \over 3}
  • B 23{2 \over 3}
  • C −23- {2 \over 3}
  • D 13{1 \over 3}

Step-by-Step Verified Solution

Let the roots of given equation be α\alpha and 22 α\alpha then

α+2α=3α=1−3aa2−5a+3\alpha + 2\alpha = 3\alpha = {{1 - 3a} \over {{a^2} - 5a + 3}}

and α.2α=2α2=2a2−5a+3\alpha .2\alpha = 2{\alpha ^2} = {2 \over {{a^2} - 5a + 3}}

⇒α=1−3a3(a2−5a+3)\Rightarrow \alpha = {{1 - 3a} \over {3\left( {{a^2} - 5a + 3} \right)}}

∴\therefore 2[19(1−3a)2(a2−5a+3)2]2\left[ {{1 \over 9}{{{{\left( {1 - 3a} \right)}^2}} \over {{{\left( {{a^2} - 5a + 3} \right)}^2}}}} \right]

=2a2−5a+3= {2 \over {{a^2} - 5a + 3}}

(1−3a)2(a2−5a+3)=9{{{{\left( {1 - 3a} \right)}^2}} \over {\left( {{a^2} - 5a + 3} \right)}} = 9

or 9a2−6a+19{a^2} - 6a + 1

=9a2−45a+27= 9{a^2} - 45a + 27

or 39a=2639a = 26 or a=23a = {2 \over 3}

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