118239
The coefficient of \(x\) in the quadratic equation \(a x^2+b x+c=0\) was wrongly taken as 17 in place of 13 and its roots were found to be -2 and -15 , the actual roots of the equation are
1 -2 and 15
2 -3 and -10
3 -4 and -9
4 -5 and -6
Explanation:
B Given, \(a x^2+b x+c=0\) \(\mathrm{~b}=17 \text { and roots are }-2 \&-15\) \((x+2)(x+15)=0\) \(x^2+17 x+30=0\) According to the question- Then, Above equation is compare with equation (i) we get, \(\mathrm{c}=30\) Now quadratic equation can be written with correct value of \(b=13\) and \(c=30\) \(x^2+13 x+30=0\) \((x+10)(x+3)=0\) \(x=-10,-3\) Therefore roots are \(-10 \&-3\)
BCECE-2010
Complex Numbers and Quadratic Equation
118240
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3-6 x^2+11 x+6=0\), then \(\sum \alpha^2 \beta+\sum \alpha \beta^2\) is equal to
118241
The equation \((x-b)(x-c)+(x-a)(x-b)+\) \((x-a)(x-c)=0\) has all its roots
1 positive
2 real
3 imaginary
4 negative
Explanation:
B Given, \((x-b)(x-c)+(x-a)(x-b)\) \(+(x-a)(x-c)=0\) \(x^2-x c-x b+b c+x^2-a x-b x+a b+x^2\) \(-a x-c x+a c=0\) \(3 x^2-2 x(a+b+c)+a b+b c+a c=0\) Now, for nature of root \(b^2-4 a c\) \(=[-2(a+b+c)]^2-4 \times 3(a b+b c+a c)\) \(4\left[a^2+b^2+c^2+2 a b+2 b c+2 c a-3\right.\) \(\left.4\left[a^2+b^2+c^2-a b-b c-a c\right] \quad(a b+b c+a c)\right]\) \(2\left[2 a^2+2 b^2+2 c^2-2 a b-2 b c-2 c a\right]\) \(=2\left[a^2+b^2-2 a b+b^2+c^2-2 b c+c^2+a^2-\right.\) \(=2\left[(a-b)^2+(b-c)^2+(c-a)^2\right]\) In above equation we can see that all values are in square form and sum of all square value always greater than or equal to zero \(\therefore \quad \mathrm{D} \geq 0\) Therefore all roots of given equation are real.
BCECE-2007]
Complex Numbers and Quadratic Equation
118242
The value of a for which the sum of the squares of the roots of the equation \(x^2-(a-2) x-a-1\) \(=0\) assumes the least value is:
1 0
2 1
3 2
4 3
Explanation:
B Let \(\alpha, \beta\) are roots of the given equation \(x^2-(a-2) x-a-1=0\) Then, \(\alpha+\beta=a-2\) \(\alpha \cdot \beta=-(a+1)\) Now, \(\quad \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta\) \(=(a-2)^2+2(a+1)\) \(=a^2-4 a+4+2 a+2\) \(=a^2-2 a+6\) \(\alpha^2+\beta^2=(a-1)^2+5\) \(\alpha^2+\beta^2 \geq 5\)Thus the minimum value of \(\alpha^2+\beta^2=5\) at \(\mathrm{a}=1\)
BCECE-2006
Complex Numbers and Quadratic Equation
118243
The roots of the quadratic equation \(2 x^2+3 x+\) \(\mathbf{1}=\mathbf{0}\) are
1 rational
2 irrational
3 imaginary
4 None of these
Explanation:
Exp: (A): Given, \(2 \mathrm{x}^2+3 \mathrm{x}+1=0\) \(\mathrm{x}=\frac{-3 \pm \sqrt{9-8}}{4}\) \(=\frac{-3 \pm 1}{4}=\frac{-3+1}{4}, \frac{-3-1}{4}=\frac{-1}{2},-1\) \(\therefore \text { Both the roots are rational }\)\(\therefore\) Both the roots are rational
118239
The coefficient of \(x\) in the quadratic equation \(a x^2+b x+c=0\) was wrongly taken as 17 in place of 13 and its roots were found to be -2 and -15 , the actual roots of the equation are
1 -2 and 15
2 -3 and -10
3 -4 and -9
4 -5 and -6
Explanation:
B Given, \(a x^2+b x+c=0\) \(\mathrm{~b}=17 \text { and roots are }-2 \&-15\) \((x+2)(x+15)=0\) \(x^2+17 x+30=0\) According to the question- Then, Above equation is compare with equation (i) we get, \(\mathrm{c}=30\) Now quadratic equation can be written with correct value of \(b=13\) and \(c=30\) \(x^2+13 x+30=0\) \((x+10)(x+3)=0\) \(x=-10,-3\) Therefore roots are \(-10 \&-3\)
BCECE-2010
Complex Numbers and Quadratic Equation
118240
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3-6 x^2+11 x+6=0\), then \(\sum \alpha^2 \beta+\sum \alpha \beta^2\) is equal to
118241
The equation \((x-b)(x-c)+(x-a)(x-b)+\) \((x-a)(x-c)=0\) has all its roots
1 positive
2 real
3 imaginary
4 negative
Explanation:
B Given, \((x-b)(x-c)+(x-a)(x-b)\) \(+(x-a)(x-c)=0\) \(x^2-x c-x b+b c+x^2-a x-b x+a b+x^2\) \(-a x-c x+a c=0\) \(3 x^2-2 x(a+b+c)+a b+b c+a c=0\) Now, for nature of root \(b^2-4 a c\) \(=[-2(a+b+c)]^2-4 \times 3(a b+b c+a c)\) \(4\left[a^2+b^2+c^2+2 a b+2 b c+2 c a-3\right.\) \(\left.4\left[a^2+b^2+c^2-a b-b c-a c\right] \quad(a b+b c+a c)\right]\) \(2\left[2 a^2+2 b^2+2 c^2-2 a b-2 b c-2 c a\right]\) \(=2\left[a^2+b^2-2 a b+b^2+c^2-2 b c+c^2+a^2-\right.\) \(=2\left[(a-b)^2+(b-c)^2+(c-a)^2\right]\) In above equation we can see that all values are in square form and sum of all square value always greater than or equal to zero \(\therefore \quad \mathrm{D} \geq 0\) Therefore all roots of given equation are real.
BCECE-2007]
Complex Numbers and Quadratic Equation
118242
The value of a for which the sum of the squares of the roots of the equation \(x^2-(a-2) x-a-1\) \(=0\) assumes the least value is:
1 0
2 1
3 2
4 3
Explanation:
B Let \(\alpha, \beta\) are roots of the given equation \(x^2-(a-2) x-a-1=0\) Then, \(\alpha+\beta=a-2\) \(\alpha \cdot \beta=-(a+1)\) Now, \(\quad \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta\) \(=(a-2)^2+2(a+1)\) \(=a^2-4 a+4+2 a+2\) \(=a^2-2 a+6\) \(\alpha^2+\beta^2=(a-1)^2+5\) \(\alpha^2+\beta^2 \geq 5\)Thus the minimum value of \(\alpha^2+\beta^2=5\) at \(\mathrm{a}=1\)
BCECE-2006
Complex Numbers and Quadratic Equation
118243
The roots of the quadratic equation \(2 x^2+3 x+\) \(\mathbf{1}=\mathbf{0}\) are
1 rational
2 irrational
3 imaginary
4 None of these
Explanation:
Exp: (A): Given, \(2 \mathrm{x}^2+3 \mathrm{x}+1=0\) \(\mathrm{x}=\frac{-3 \pm \sqrt{9-8}}{4}\) \(=\frac{-3 \pm 1}{4}=\frac{-3+1}{4}, \frac{-3-1}{4}=\frac{-1}{2},-1\) \(\therefore \text { Both the roots are rational }\)\(\therefore\) Both the roots are rational
118239
The coefficient of \(x\) in the quadratic equation \(a x^2+b x+c=0\) was wrongly taken as 17 in place of 13 and its roots were found to be -2 and -15 , the actual roots of the equation are
1 -2 and 15
2 -3 and -10
3 -4 and -9
4 -5 and -6
Explanation:
B Given, \(a x^2+b x+c=0\) \(\mathrm{~b}=17 \text { and roots are }-2 \&-15\) \((x+2)(x+15)=0\) \(x^2+17 x+30=0\) According to the question- Then, Above equation is compare with equation (i) we get, \(\mathrm{c}=30\) Now quadratic equation can be written with correct value of \(b=13\) and \(c=30\) \(x^2+13 x+30=0\) \((x+10)(x+3)=0\) \(x=-10,-3\) Therefore roots are \(-10 \&-3\)
BCECE-2010
Complex Numbers and Quadratic Equation
118240
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3-6 x^2+11 x+6=0\), then \(\sum \alpha^2 \beta+\sum \alpha \beta^2\) is equal to
118241
The equation \((x-b)(x-c)+(x-a)(x-b)+\) \((x-a)(x-c)=0\) has all its roots
1 positive
2 real
3 imaginary
4 negative
Explanation:
B Given, \((x-b)(x-c)+(x-a)(x-b)\) \(+(x-a)(x-c)=0\) \(x^2-x c-x b+b c+x^2-a x-b x+a b+x^2\) \(-a x-c x+a c=0\) \(3 x^2-2 x(a+b+c)+a b+b c+a c=0\) Now, for nature of root \(b^2-4 a c\) \(=[-2(a+b+c)]^2-4 \times 3(a b+b c+a c)\) \(4\left[a^2+b^2+c^2+2 a b+2 b c+2 c a-3\right.\) \(\left.4\left[a^2+b^2+c^2-a b-b c-a c\right] \quad(a b+b c+a c)\right]\) \(2\left[2 a^2+2 b^2+2 c^2-2 a b-2 b c-2 c a\right]\) \(=2\left[a^2+b^2-2 a b+b^2+c^2-2 b c+c^2+a^2-\right.\) \(=2\left[(a-b)^2+(b-c)^2+(c-a)^2\right]\) In above equation we can see that all values are in square form and sum of all square value always greater than or equal to zero \(\therefore \quad \mathrm{D} \geq 0\) Therefore all roots of given equation are real.
BCECE-2007]
Complex Numbers and Quadratic Equation
118242
The value of a for which the sum of the squares of the roots of the equation \(x^2-(a-2) x-a-1\) \(=0\) assumes the least value is:
1 0
2 1
3 2
4 3
Explanation:
B Let \(\alpha, \beta\) are roots of the given equation \(x^2-(a-2) x-a-1=0\) Then, \(\alpha+\beta=a-2\) \(\alpha \cdot \beta=-(a+1)\) Now, \(\quad \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta\) \(=(a-2)^2+2(a+1)\) \(=a^2-4 a+4+2 a+2\) \(=a^2-2 a+6\) \(\alpha^2+\beta^2=(a-1)^2+5\) \(\alpha^2+\beta^2 \geq 5\)Thus the minimum value of \(\alpha^2+\beta^2=5\) at \(\mathrm{a}=1\)
BCECE-2006
Complex Numbers and Quadratic Equation
118243
The roots of the quadratic equation \(2 x^2+3 x+\) \(\mathbf{1}=\mathbf{0}\) are
1 rational
2 irrational
3 imaginary
4 None of these
Explanation:
Exp: (A): Given, \(2 \mathrm{x}^2+3 \mathrm{x}+1=0\) \(\mathrm{x}=\frac{-3 \pm \sqrt{9-8}}{4}\) \(=\frac{-3 \pm 1}{4}=\frac{-3+1}{4}, \frac{-3-1}{4}=\frac{-1}{2},-1\) \(\therefore \text { Both the roots are rational }\)\(\therefore\) Both the roots are rational
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Complex Numbers and Quadratic Equation
118239
The coefficient of \(x\) in the quadratic equation \(a x^2+b x+c=0\) was wrongly taken as 17 in place of 13 and its roots were found to be -2 and -15 , the actual roots of the equation are
1 -2 and 15
2 -3 and -10
3 -4 and -9
4 -5 and -6
Explanation:
B Given, \(a x^2+b x+c=0\) \(\mathrm{~b}=17 \text { and roots are }-2 \&-15\) \((x+2)(x+15)=0\) \(x^2+17 x+30=0\) According to the question- Then, Above equation is compare with equation (i) we get, \(\mathrm{c}=30\) Now quadratic equation can be written with correct value of \(b=13\) and \(c=30\) \(x^2+13 x+30=0\) \((x+10)(x+3)=0\) \(x=-10,-3\) Therefore roots are \(-10 \&-3\)
BCECE-2010
Complex Numbers and Quadratic Equation
118240
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3-6 x^2+11 x+6=0\), then \(\sum \alpha^2 \beta+\sum \alpha \beta^2\) is equal to
118241
The equation \((x-b)(x-c)+(x-a)(x-b)+\) \((x-a)(x-c)=0\) has all its roots
1 positive
2 real
3 imaginary
4 negative
Explanation:
B Given, \((x-b)(x-c)+(x-a)(x-b)\) \(+(x-a)(x-c)=0\) \(x^2-x c-x b+b c+x^2-a x-b x+a b+x^2\) \(-a x-c x+a c=0\) \(3 x^2-2 x(a+b+c)+a b+b c+a c=0\) Now, for nature of root \(b^2-4 a c\) \(=[-2(a+b+c)]^2-4 \times 3(a b+b c+a c)\) \(4\left[a^2+b^2+c^2+2 a b+2 b c+2 c a-3\right.\) \(\left.4\left[a^2+b^2+c^2-a b-b c-a c\right] \quad(a b+b c+a c)\right]\) \(2\left[2 a^2+2 b^2+2 c^2-2 a b-2 b c-2 c a\right]\) \(=2\left[a^2+b^2-2 a b+b^2+c^2-2 b c+c^2+a^2-\right.\) \(=2\left[(a-b)^2+(b-c)^2+(c-a)^2\right]\) In above equation we can see that all values are in square form and sum of all square value always greater than or equal to zero \(\therefore \quad \mathrm{D} \geq 0\) Therefore all roots of given equation are real.
BCECE-2007]
Complex Numbers and Quadratic Equation
118242
The value of a for which the sum of the squares of the roots of the equation \(x^2-(a-2) x-a-1\) \(=0\) assumes the least value is:
1 0
2 1
3 2
4 3
Explanation:
B Let \(\alpha, \beta\) are roots of the given equation \(x^2-(a-2) x-a-1=0\) Then, \(\alpha+\beta=a-2\) \(\alpha \cdot \beta=-(a+1)\) Now, \(\quad \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta\) \(=(a-2)^2+2(a+1)\) \(=a^2-4 a+4+2 a+2\) \(=a^2-2 a+6\) \(\alpha^2+\beta^2=(a-1)^2+5\) \(\alpha^2+\beta^2 \geq 5\)Thus the minimum value of \(\alpha^2+\beta^2=5\) at \(\mathrm{a}=1\)
BCECE-2006
Complex Numbers and Quadratic Equation
118243
The roots of the quadratic equation \(2 x^2+3 x+\) \(\mathbf{1}=\mathbf{0}\) are
1 rational
2 irrational
3 imaginary
4 None of these
Explanation:
Exp: (A): Given, \(2 \mathrm{x}^2+3 \mathrm{x}+1=0\) \(\mathrm{x}=\frac{-3 \pm \sqrt{9-8}}{4}\) \(=\frac{-3 \pm 1}{4}=\frac{-3+1}{4}, \frac{-3-1}{4}=\frac{-1}{2},-1\) \(\therefore \text { Both the roots are rational }\)\(\therefore\) Both the roots are rational
118239
The coefficient of \(x\) in the quadratic equation \(a x^2+b x+c=0\) was wrongly taken as 17 in place of 13 and its roots were found to be -2 and -15 , the actual roots of the equation are
1 -2 and 15
2 -3 and -10
3 -4 and -9
4 -5 and -6
Explanation:
B Given, \(a x^2+b x+c=0\) \(\mathrm{~b}=17 \text { and roots are }-2 \&-15\) \((x+2)(x+15)=0\) \(x^2+17 x+30=0\) According to the question- Then, Above equation is compare with equation (i) we get, \(\mathrm{c}=30\) Now quadratic equation can be written with correct value of \(b=13\) and \(c=30\) \(x^2+13 x+30=0\) \((x+10)(x+3)=0\) \(x=-10,-3\) Therefore roots are \(-10 \&-3\)
BCECE-2010
Complex Numbers and Quadratic Equation
118240
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3-6 x^2+11 x+6=0\), then \(\sum \alpha^2 \beta+\sum \alpha \beta^2\) is equal to
118241
The equation \((x-b)(x-c)+(x-a)(x-b)+\) \((x-a)(x-c)=0\) has all its roots
1 positive
2 real
3 imaginary
4 negative
Explanation:
B Given, \((x-b)(x-c)+(x-a)(x-b)\) \(+(x-a)(x-c)=0\) \(x^2-x c-x b+b c+x^2-a x-b x+a b+x^2\) \(-a x-c x+a c=0\) \(3 x^2-2 x(a+b+c)+a b+b c+a c=0\) Now, for nature of root \(b^2-4 a c\) \(=[-2(a+b+c)]^2-4 \times 3(a b+b c+a c)\) \(4\left[a^2+b^2+c^2+2 a b+2 b c+2 c a-3\right.\) \(\left.4\left[a^2+b^2+c^2-a b-b c-a c\right] \quad(a b+b c+a c)\right]\) \(2\left[2 a^2+2 b^2+2 c^2-2 a b-2 b c-2 c a\right]\) \(=2\left[a^2+b^2-2 a b+b^2+c^2-2 b c+c^2+a^2-\right.\) \(=2\left[(a-b)^2+(b-c)^2+(c-a)^2\right]\) In above equation we can see that all values are in square form and sum of all square value always greater than or equal to zero \(\therefore \quad \mathrm{D} \geq 0\) Therefore all roots of given equation are real.
BCECE-2007]
Complex Numbers and Quadratic Equation
118242
The value of a for which the sum of the squares of the roots of the equation \(x^2-(a-2) x-a-1\) \(=0\) assumes the least value is:
1 0
2 1
3 2
4 3
Explanation:
B Let \(\alpha, \beta\) are roots of the given equation \(x^2-(a-2) x-a-1=0\) Then, \(\alpha+\beta=a-2\) \(\alpha \cdot \beta=-(a+1)\) Now, \(\quad \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta\) \(=(a-2)^2+2(a+1)\) \(=a^2-4 a+4+2 a+2\) \(=a^2-2 a+6\) \(\alpha^2+\beta^2=(a-1)^2+5\) \(\alpha^2+\beta^2 \geq 5\)Thus the minimum value of \(\alpha^2+\beta^2=5\) at \(\mathrm{a}=1\)
BCECE-2006
Complex Numbers and Quadratic Equation
118243
The roots of the quadratic equation \(2 x^2+3 x+\) \(\mathbf{1}=\mathbf{0}\) are
1 rational
2 irrational
3 imaginary
4 None of these
Explanation:
Exp: (A): Given, \(2 \mathrm{x}^2+3 \mathrm{x}+1=0\) \(\mathrm{x}=\frac{-3 \pm \sqrt{9-8}}{4}\) \(=\frac{-3 \pm 1}{4}=\frac{-3+1}{4}, \frac{-3-1}{4}=\frac{-1}{2},-1\) \(\therefore \text { Both the roots are rational }\)\(\therefore\) Both the roots are rational