118141
The value of \(\sqrt{42+\sqrt{42+\sqrt{42+\ldots . .}}}\) is equal to
1 7
2 -6
3 5
4 4
Explanation:
A \( Let \mathrm{y}=\sqrt{42+\sqrt{42+\sqrt{42+\ldots .}}}\) \(\therefore y=\sqrt{42+y}\) \(y^2=42+y\) \(y^2-y-42=0\) \(y^2-7 y+6 y-42=0\) \(y(y-7)+6(y-7)=0\) \((y+6)(y-7)=0\) \(\therefore y=7\)
AP EAMCET-2004
Complex Numbers and Quadratic Equation
118142
Each of the roots of the equation \(x^3-6 x^2+6 x-5\) \(=0\) are increased by \(h\). So that the new transformed equation does not contain \(x^2\) term, then \(h\) is equal to.
1 1
2 -2
3 \(\frac{1}{2}\)
4 \(\frac{1}{3}\)
Explanation:
B \( Let \alpha, \beta, \gamma \text { are the roots of the equation }\) \(\mathrm{x}^3-6 \mathrm{x}^2+6 \mathrm{x}-5=0\) \(\therefore \alpha+\beta+\gamma=6\) \(\alpha \beta+\beta \gamma+\alpha \gamma=6\) \(\alpha \beta \gamma=5\) \(\text { Now, } \alpha+\mathrm{h}, \beta+\mathrm{h}, \gamma+\mathrm{h} \text { are the roots of the new }\) \(\text { equation. It does not contain the term, } \mathrm{x}^2\) \(\therefore \alpha+\beta+\gamma+3 \mathrm{~h}=0\) \(\text { or } 6+3 \mathrm{~h}=0\) \(\therefore \mathrm{h}=-2\) Now, \(\alpha+\mathrm{h}, \beta+\mathrm{h}, \gamma+\mathrm{h}\) are the roots of the new equation. It does not contain the term, \(\mathrm{x}^2\)
AP EAMCET-2001
Complex Numbers and Quadratic Equation
118143
If \(\alpha, \beta\) are the roots of the equation \(x^2+b x+c\) \(=0\) and \(\alpha+h, \beta+h\) are the roots the equation \(\mathbf{x}^2+\mathbf{q x}+\mathbf{r}=\mathbf{0}\), then \(h\) is equal to.
1 \(b+q\)
2 \(\mathrm{b}-\mathrm{q}\)
3 \(\frac{1}{2}(b+q)\)
4 \(\frac{1}{2}(b-q)\)
Explanation:
D \(: \text { Given } \alpha, \beta \text { are the roots of the equation, }\) \(\mathrm{x}^2+\mathrm{bx}+\mathrm{c}=0\) \(\therefore \alpha+\beta=-\mathrm{b}\) \(\alpha \beta=\mathrm{c}\) \(\text { and also, } \alpha+\mathrm{h}, \beta+\mathrm{h} \text { are the roots of equation, }\) \(\mathrm{x}^2+\mathrm{qx}+\mathrm{r}=0\) \(\text { Sum of the roots, }\) \(\therefore \alpha+\beta+2 \mathrm{~h}=-\mathrm{q} \Rightarrow-\mathrm{b}+2 \mathrm{~h}=-\mathrm{q} \Rightarrow 2 \mathrm{~h}=\mathrm{b}-\mathrm{q}\) \(\therefore \mathrm{h}=\frac{\mathrm{b}-\mathrm{q}}{2}\) Ans: d Exp:D Given \(\alpha, \beta\) are the roots of the equation, \(\mathrm{x}^2+\mathrm{bx}+\mathrm{c}=0\) and also, \(\alpha+\mathrm{h}, \beta+\mathrm{h}\) are the roots of equation, Sum of the roots,
AP EAMCET-2001
Complex Numbers and Quadratic Equation
118144
The cubic equation whose roots are the squares of the roots of \(x^3-2 x^2+10 x-8=0\), is
118141
The value of \(\sqrt{42+\sqrt{42+\sqrt{42+\ldots . .}}}\) is equal to
1 7
2 -6
3 5
4 4
Explanation:
A \( Let \mathrm{y}=\sqrt{42+\sqrt{42+\sqrt{42+\ldots .}}}\) \(\therefore y=\sqrt{42+y}\) \(y^2=42+y\) \(y^2-y-42=0\) \(y^2-7 y+6 y-42=0\) \(y(y-7)+6(y-7)=0\) \((y+6)(y-7)=0\) \(\therefore y=7\)
AP EAMCET-2004
Complex Numbers and Quadratic Equation
118142
Each of the roots of the equation \(x^3-6 x^2+6 x-5\) \(=0\) are increased by \(h\). So that the new transformed equation does not contain \(x^2\) term, then \(h\) is equal to.
1 1
2 -2
3 \(\frac{1}{2}\)
4 \(\frac{1}{3}\)
Explanation:
B \( Let \alpha, \beta, \gamma \text { are the roots of the equation }\) \(\mathrm{x}^3-6 \mathrm{x}^2+6 \mathrm{x}-5=0\) \(\therefore \alpha+\beta+\gamma=6\) \(\alpha \beta+\beta \gamma+\alpha \gamma=6\) \(\alpha \beta \gamma=5\) \(\text { Now, } \alpha+\mathrm{h}, \beta+\mathrm{h}, \gamma+\mathrm{h} \text { are the roots of the new }\) \(\text { equation. It does not contain the term, } \mathrm{x}^2\) \(\therefore \alpha+\beta+\gamma+3 \mathrm{~h}=0\) \(\text { or } 6+3 \mathrm{~h}=0\) \(\therefore \mathrm{h}=-2\) Now, \(\alpha+\mathrm{h}, \beta+\mathrm{h}, \gamma+\mathrm{h}\) are the roots of the new equation. It does not contain the term, \(\mathrm{x}^2\)
AP EAMCET-2001
Complex Numbers and Quadratic Equation
118143
If \(\alpha, \beta\) are the roots of the equation \(x^2+b x+c\) \(=0\) and \(\alpha+h, \beta+h\) are the roots the equation \(\mathbf{x}^2+\mathbf{q x}+\mathbf{r}=\mathbf{0}\), then \(h\) is equal to.
1 \(b+q\)
2 \(\mathrm{b}-\mathrm{q}\)
3 \(\frac{1}{2}(b+q)\)
4 \(\frac{1}{2}(b-q)\)
Explanation:
D \(: \text { Given } \alpha, \beta \text { are the roots of the equation, }\) \(\mathrm{x}^2+\mathrm{bx}+\mathrm{c}=0\) \(\therefore \alpha+\beta=-\mathrm{b}\) \(\alpha \beta=\mathrm{c}\) \(\text { and also, } \alpha+\mathrm{h}, \beta+\mathrm{h} \text { are the roots of equation, }\) \(\mathrm{x}^2+\mathrm{qx}+\mathrm{r}=0\) \(\text { Sum of the roots, }\) \(\therefore \alpha+\beta+2 \mathrm{~h}=-\mathrm{q} \Rightarrow-\mathrm{b}+2 \mathrm{~h}=-\mathrm{q} \Rightarrow 2 \mathrm{~h}=\mathrm{b}-\mathrm{q}\) \(\therefore \mathrm{h}=\frac{\mathrm{b}-\mathrm{q}}{2}\) Ans: d Exp:D Given \(\alpha, \beta\) are the roots of the equation, \(\mathrm{x}^2+\mathrm{bx}+\mathrm{c}=0\) and also, \(\alpha+\mathrm{h}, \beta+\mathrm{h}\) are the roots of equation, Sum of the roots,
AP EAMCET-2001
Complex Numbers and Quadratic Equation
118144
The cubic equation whose roots are the squares of the roots of \(x^3-2 x^2+10 x-8=0\), is
118141
The value of \(\sqrt{42+\sqrt{42+\sqrt{42+\ldots . .}}}\) is equal to
1 7
2 -6
3 5
4 4
Explanation:
A \( Let \mathrm{y}=\sqrt{42+\sqrt{42+\sqrt{42+\ldots .}}}\) \(\therefore y=\sqrt{42+y}\) \(y^2=42+y\) \(y^2-y-42=0\) \(y^2-7 y+6 y-42=0\) \(y(y-7)+6(y-7)=0\) \((y+6)(y-7)=0\) \(\therefore y=7\)
AP EAMCET-2004
Complex Numbers and Quadratic Equation
118142
Each of the roots of the equation \(x^3-6 x^2+6 x-5\) \(=0\) are increased by \(h\). So that the new transformed equation does not contain \(x^2\) term, then \(h\) is equal to.
1 1
2 -2
3 \(\frac{1}{2}\)
4 \(\frac{1}{3}\)
Explanation:
B \( Let \alpha, \beta, \gamma \text { are the roots of the equation }\) \(\mathrm{x}^3-6 \mathrm{x}^2+6 \mathrm{x}-5=0\) \(\therefore \alpha+\beta+\gamma=6\) \(\alpha \beta+\beta \gamma+\alpha \gamma=6\) \(\alpha \beta \gamma=5\) \(\text { Now, } \alpha+\mathrm{h}, \beta+\mathrm{h}, \gamma+\mathrm{h} \text { are the roots of the new }\) \(\text { equation. It does not contain the term, } \mathrm{x}^2\) \(\therefore \alpha+\beta+\gamma+3 \mathrm{~h}=0\) \(\text { or } 6+3 \mathrm{~h}=0\) \(\therefore \mathrm{h}=-2\) Now, \(\alpha+\mathrm{h}, \beta+\mathrm{h}, \gamma+\mathrm{h}\) are the roots of the new equation. It does not contain the term, \(\mathrm{x}^2\)
AP EAMCET-2001
Complex Numbers and Quadratic Equation
118143
If \(\alpha, \beta\) are the roots of the equation \(x^2+b x+c\) \(=0\) and \(\alpha+h, \beta+h\) are the roots the equation \(\mathbf{x}^2+\mathbf{q x}+\mathbf{r}=\mathbf{0}\), then \(h\) is equal to.
1 \(b+q\)
2 \(\mathrm{b}-\mathrm{q}\)
3 \(\frac{1}{2}(b+q)\)
4 \(\frac{1}{2}(b-q)\)
Explanation:
D \(: \text { Given } \alpha, \beta \text { are the roots of the equation, }\) \(\mathrm{x}^2+\mathrm{bx}+\mathrm{c}=0\) \(\therefore \alpha+\beta=-\mathrm{b}\) \(\alpha \beta=\mathrm{c}\) \(\text { and also, } \alpha+\mathrm{h}, \beta+\mathrm{h} \text { are the roots of equation, }\) \(\mathrm{x}^2+\mathrm{qx}+\mathrm{r}=0\) \(\text { Sum of the roots, }\) \(\therefore \alpha+\beta+2 \mathrm{~h}=-\mathrm{q} \Rightarrow-\mathrm{b}+2 \mathrm{~h}=-\mathrm{q} \Rightarrow 2 \mathrm{~h}=\mathrm{b}-\mathrm{q}\) \(\therefore \mathrm{h}=\frac{\mathrm{b}-\mathrm{q}}{2}\) Ans: d Exp:D Given \(\alpha, \beta\) are the roots of the equation, \(\mathrm{x}^2+\mathrm{bx}+\mathrm{c}=0\) and also, \(\alpha+\mathrm{h}, \beta+\mathrm{h}\) are the roots of equation, Sum of the roots,
AP EAMCET-2001
Complex Numbers and Quadratic Equation
118144
The cubic equation whose roots are the squares of the roots of \(x^3-2 x^2+10 x-8=0\), is
118141
The value of \(\sqrt{42+\sqrt{42+\sqrt{42+\ldots . .}}}\) is equal to
1 7
2 -6
3 5
4 4
Explanation:
A \( Let \mathrm{y}=\sqrt{42+\sqrt{42+\sqrt{42+\ldots .}}}\) \(\therefore y=\sqrt{42+y}\) \(y^2=42+y\) \(y^2-y-42=0\) \(y^2-7 y+6 y-42=0\) \(y(y-7)+6(y-7)=0\) \((y+6)(y-7)=0\) \(\therefore y=7\)
AP EAMCET-2004
Complex Numbers and Quadratic Equation
118142
Each of the roots of the equation \(x^3-6 x^2+6 x-5\) \(=0\) are increased by \(h\). So that the new transformed equation does not contain \(x^2\) term, then \(h\) is equal to.
1 1
2 -2
3 \(\frac{1}{2}\)
4 \(\frac{1}{3}\)
Explanation:
B \( Let \alpha, \beta, \gamma \text { are the roots of the equation }\) \(\mathrm{x}^3-6 \mathrm{x}^2+6 \mathrm{x}-5=0\) \(\therefore \alpha+\beta+\gamma=6\) \(\alpha \beta+\beta \gamma+\alpha \gamma=6\) \(\alpha \beta \gamma=5\) \(\text { Now, } \alpha+\mathrm{h}, \beta+\mathrm{h}, \gamma+\mathrm{h} \text { are the roots of the new }\) \(\text { equation. It does not contain the term, } \mathrm{x}^2\) \(\therefore \alpha+\beta+\gamma+3 \mathrm{~h}=0\) \(\text { or } 6+3 \mathrm{~h}=0\) \(\therefore \mathrm{h}=-2\) Now, \(\alpha+\mathrm{h}, \beta+\mathrm{h}, \gamma+\mathrm{h}\) are the roots of the new equation. It does not contain the term, \(\mathrm{x}^2\)
AP EAMCET-2001
Complex Numbers and Quadratic Equation
118143
If \(\alpha, \beta\) are the roots of the equation \(x^2+b x+c\) \(=0\) and \(\alpha+h, \beta+h\) are the roots the equation \(\mathbf{x}^2+\mathbf{q x}+\mathbf{r}=\mathbf{0}\), then \(h\) is equal to.
1 \(b+q\)
2 \(\mathrm{b}-\mathrm{q}\)
3 \(\frac{1}{2}(b+q)\)
4 \(\frac{1}{2}(b-q)\)
Explanation:
D \(: \text { Given } \alpha, \beta \text { are the roots of the equation, }\) \(\mathrm{x}^2+\mathrm{bx}+\mathrm{c}=0\) \(\therefore \alpha+\beta=-\mathrm{b}\) \(\alpha \beta=\mathrm{c}\) \(\text { and also, } \alpha+\mathrm{h}, \beta+\mathrm{h} \text { are the roots of equation, }\) \(\mathrm{x}^2+\mathrm{qx}+\mathrm{r}=0\) \(\text { Sum of the roots, }\) \(\therefore \alpha+\beta+2 \mathrm{~h}=-\mathrm{q} \Rightarrow-\mathrm{b}+2 \mathrm{~h}=-\mathrm{q} \Rightarrow 2 \mathrm{~h}=\mathrm{b}-\mathrm{q}\) \(\therefore \mathrm{h}=\frac{\mathrm{b}-\mathrm{q}}{2}\) Ans: d Exp:D Given \(\alpha, \beta\) are the roots of the equation, \(\mathrm{x}^2+\mathrm{bx}+\mathrm{c}=0\) and also, \(\alpha+\mathrm{h}, \beta+\mathrm{h}\) are the roots of equation, Sum of the roots,
AP EAMCET-2001
Complex Numbers and Quadratic Equation
118144
The cubic equation whose roots are the squares of the roots of \(x^3-2 x^2+10 x-8=0\), is
