class 11 Mathematics: JEE based questions:

Q1 If one real root of the quadratic equation $49x^2+kx+64=0$ is the square of the other root, then find $k$.

Solution:

• Let roots be $\alpha$ and ${\alpha }^2$.
• Sum of roots = $-k\mathrm{/}49$. $\alpha +{\alpha }^2=-k\mathrm{/}49$.
• Product of roots = $64\mathrm{/}49$. $\alpha \cdot {\alpha }^2={\alpha }^3=64\mathrm{/}49$.
• So, ${\alpha }^3=64\mathrm{/}49⟹\alpha =\frac{4}{\sqrt[3]{49}}$.
• Substitute into sum: $\alpha +{\alpha }^2=\frac{4}{\sqrt[3]{49}}+\frac{16}{\sqrt[3]{{49}^2}}$.
• Hence, $k=-49(\alpha +{\alpha }^2)$.
• Ans: $k=-49(\frac{4}{\sqrt[3]{49}}+\frac{16}{\sqrt[3]{{49}^2}})$.

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Q2 If $3+2i$ is one root of $x^3-7x^2+kx-15=0$, find the real root.

Solution:

• Complex roots occur in conjugate pairs → other root is $3-2i$.
• Product of roots = constant term / coefficient of $x^3$ = $(-15)\mathrm{/}1=-15$.
• So, $(3+2i)(3-2i)(\text{real root})=-15$.
• $(3+2i)(3-2i)=9+4=13$.
• $13\cdot (\text{real root})=-15$.
• Real root = $-15\mathrm{/}13$.
• Ans: Real root = $-\frac{15}{13}$.

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Q3 If $x^2+3x+2=0$ and $a{x}^2+bx+c=0$ have a common root, find $a:b:c$.

Solution:

• Roots of first equation: $-1,-2$.
• Suppose common root = $-1$.
• Substitution: $a(-1{)}^2+b(-1)+c=0⟹a-b+c=0$.
• Ratio condition: choose $a=1,b=1,c=0$.
• So $a:b:c=1:1:0$.
• Ans: $a:b:c=1:1:0$.

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Q4 If ${\alpha }^2=4\alpha -2$ and ${\beta }^2=4\beta -2$, find equation whose roots are $\alpha \mathrm{/}\beta$ and $\beta \mathrm{/}\alpha$.

Solution:

• Equation: $x^2-(\alpha \mathrm{/}\beta +\beta \mathrm{/}\alpha )x+1=0$.
• From given: ${\alpha }^2-4\alpha +2=0$.
• Roots: $\alpha =2\pm \sqrt{2}$.
• Similarly, $\beta =2\pm \sqrt{2}$.
• Take distinct roots: $\alpha =2+\sqrt{2},\beta =2-\sqrt{2}$.
• Compute: $\alpha \mathrm{/}\beta +\beta \mathrm{/}\alpha =\frac{(2+\sqrt{2}{)}^2+(2-\sqrt{2}{)}^2}{(2+\sqrt{2})(2-\sqrt{2})}$.
• Numerator = $(6+4\sqrt{2})+(6-4\sqrt{2})=12$.
• Denominator = $(4-2)=2$.
• So sum = $12\mathrm{/}2=6$.
• Equation: $x^2-6x+1=0$.
• Ans: $x^2-6x+1=0$.

Q5 If $p$ and $q$ are roots of $x^2+2x+3=0$, then find possible values of $p,q$.

Solution:

• Equation: $x^2+2x+3=0$.
• Roots: $\frac{-2\pm \sqrt{4-12}}{2}=\frac{-2\pm \sqrt{-8}}{2}$.
• = $\frac{-2\pm 2\sqrt{2}i}{2}=-1\pm \sqrt{2}i$
• Ans: $p=-1+\sqrt{2}i,q=-1-\sqrt{2}i$.

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Q.6 Equation $49x^2+kx+64=0$ Condition: One root is the square of the other.

Step 1: Let roots be $\alpha$ and ${\alpha }^2$.

Step 2: Relations

• Sum of roots = $-k\mathrm{/}49$. $\alpha +{\alpha }^2=-k\mathrm{/}49$.
• Product of roots = $64\mathrm{/}49$. $\alpha \cdot {\alpha }^2={\alpha }^3=64\mathrm{/}49$.

Step 3: Solve for $\alpha$ ${\alpha }^3=\frac{64}{49}⟹\alpha =\frac{4}{\sqrt[3]{49}}$.

Step 4: Substitute in sum $\alpha +{\alpha }^2=\frac{4}{\sqrt[3]{49}}+\frac{16}{\sqrt[3]{{49}^2}}$.

Step 5: Find $k$ $k=-49(\alpha +{\alpha }^2)$.

Ans: $k=-49(\frac{4}{\sqrt[3]{49}}+\frac{16}{\sqrt[3]{{49}^2}})$.

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Q.7 Equation $x^3-7x^2+kx-15=0$. Given root: $3+2i$.

Step 1: Conjugate root Other root = $3-2i$.

Step 2: Product of roots Product of all roots = constant term / coefficient of $x^3$. $(-15)\mathrm{/}1=-15$.

Step 3: Multiply complex roots $(3+2i)(3-2i)=9+4=13$.

Step 4: Real root $13\cdot (\text{real root})=-15⟹\text{real root}=-\frac{15}{13}$.

Ans: Real root = $-\frac{15}{13}$.

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Q.8 Given: ${\alpha }^2=4\alpha -2$, ${\beta }^2=4\beta -2$. Find equation with roots $\alpha \mathrm{/}\beta$ and $\beta \mathrm{/}\alpha$.

Step 1: Roots of quadratic ${\alpha }^2-4\alpha +2=0$. $\alpha =2\pm \sqrt{2}$. Similarly, $\beta =2\pm \sqrt{2}$.

Step 2: Take distinct roots $\alpha =2+\sqrt{2},\beta =2-\sqrt{2}$.

Step 3: Compute sum $\alpha \mathrm{/}\beta +\beta \mathrm{/}\alpha =\frac{(2+\sqrt{2}{)}^2+(2-\sqrt{2}{)}^2}{(2+\sqrt{2})(2-\sqrt{2})}$. Numerator = $6+4\sqrt{2+}6-4\sqrt{2}=12$. Denominator = $4-2=2$. So sum = $12\mathrm{/}2=6$.

Step 4: Equation Equation = $x^2-6x+1=0$.

Ans: $x^2-6x+1=0$.

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Q.9 Equation: $x^2+2x+3=0$. Find roots.

Step 1: Quadratic formula $x=\frac{-2\pm \sqrt{4-12}}{2}$. $=\frac{-2\pm \sqrt{-8}}{2}$. $=\frac{-2\pm 2\sqrt{2}i}{2}$. $=-1\pm \sqrt{2}i$.

Ans: Roots are $p=-1+\sqrt{2}i,q=-1-\sqrt{2}i$.

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