118235
If one of the roots of equation \(x^2+a x+3=0\) is 3 and one of the roots of the equation \(x^2+a x+b=0\) is three times the other root, then the value of \(b\) is
1 3
2 4
3 2
4 1
Explanation:
A Given equation is- \(x^2+a x+3=0\) and one root is 3 Let \(p\) and 3 are root of the above equation So, \(p+3=-a\) \(p \times 3=3\) \(p=1\) Then, \(1+3=-a\) \(a=-4\) Now, other equation \(\mathrm{x}^2+\mathrm{ax}+\mathrm{b}=0\) According to the question- roots of above equation are- \(\mathrm{q} \& 3 \mathrm{q}\) Then, \(\quad \mathrm{q}+3 \mathrm{q}=-\mathrm{a}\) \(4 q=-a\) \(4 q=-(-4) \quad\{\because \mathrm{a}=-4\}\) \(\mathrm{q}=1\) and \(\quad \mathrm{q} \times 3 \mathrm{q}=\mathrm{b}\) \(\mathrm{b}=3\)
BCECE-2014
Complex Numbers and Quadratic Equation
118236
If one root of the equation \(x^2-\lambda x+12=0\) is even prime while \(x^2+\lambda x+\mu=0\) has equal roots then \(\mu\) is equal to
1 8
2 16
3 24
4 32
Explanation:
B We know that, only even prime is 2 Then, \(f(2)=0\) \(2^2-\lambda \times 2+12=0\) \(4-2 \lambda+12=0\) \(\lambda=8\) Now, other given equation is- \(x^2+\lambda x+\mu=0 \text { has equal roots }\) Then, \(D=0\) \(\lambda^2-4 \mu=0\) \((8)^2-4 \mu=0\) \(64-4 \mu=0\) \(\mu=16\)
BCECE-2013
Complex Numbers and Quadratic Equation
118237
If \(\alpha\) and \(\beta\) are the roots of \(a x^2+c=b x\), then the equation \((a+c y)^2=b^2 y\) in \(y\) has the roots
1 \(\alpha^{-1}, \beta^{-1}\)
2 \(\alpha^2, \beta^2\)
3 \(\alpha \beta^{-1}, \alpha^{-1} \beta\)
4 \(\alpha^{-2}, \beta^{-2}\)
Explanation:
D Given \(a^2+c=b x \text { and roots are } \alpha, \beta\) \(a x^2-b x+c=0\) \(\alpha+\beta=\frac{b}{a}\) \(\alpha \cdot \beta=c / a\) Other given equation is- \((a+c y)^2=b^2 y\) \(a^2+c^2 y^2+2 a c y=b^2 y\) \(c^2 y^2+\left(2 a c-b^2\right) y+a^2=0\) Divide the above equation by \(\mathrm{a}^2\) \(\frac{c^2}{a^2} y^2+\left(\frac{\left(2 a c-b^2\right)}{a^2}\right) y+1=0\) \(\left(\frac{c}{a}\right)^2 y^2+\left[\frac{2 c}{a}-\left(\frac{b}{a}\right)^2\right] y+1=0\) \((\alpha \beta)^2 y^2+\left[2 \alpha \beta-(\alpha+\beta)^2\right] y+1=0\) \((\alpha \beta)^2 y^2-\left[\alpha^2+\beta^2\right] y+1=0 .\) Now equation (i) is divided by \((\alpha \beta)^2\) \(y^2-\left(\beta^{-2}+\alpha^{-2}\right) y+\alpha^{-2} \beta^{-2}=0\) Hence, from above equation we can see that root of the equation are \(\alpha^{-2} \& \beta^{-2}\)
BCECE-2013
Complex Numbers and Quadratic Equation
118238
If \(\alpha+\beta=-2\) and \(\alpha^3+\beta^3=-56\), then the quadratic equation whose roots are \(\alpha\) and \(\beta\) is
1 \(x^2+2 x-16=0\)
2 \(x^2+2 x+15=0\)
3 \(x^2+2 x-12=0\)
4 \(x^2+2 x-8=0\)
Explanation:
D Given, \(\alpha+\beta=-2\) \(\alpha^3+\beta^3=-56\) Then, \((\alpha+\beta)^3=\alpha^3+\beta^3+3 \alpha \beta(\alpha+\beta)\) \((-2)^3=\alpha^3+\beta^3+3 \alpha \beta(-2)\) \(-8=-56-6 \alpha \beta\) \(48=-6 \alpha \beta\) \(\alpha \beta=-8\) Now, equation whose roots are \(\alpha\) and \(\beta\) \(x^2-(\text { sum of roots }) x+\text { product of roots }=0\) \(x^2-(-2) x+(-8)=0\) \(x^2+2 x-8=0\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Complex Numbers and Quadratic Equation
118235
If one of the roots of equation \(x^2+a x+3=0\) is 3 and one of the roots of the equation \(x^2+a x+b=0\) is three times the other root, then the value of \(b\) is
1 3
2 4
3 2
4 1
Explanation:
A Given equation is- \(x^2+a x+3=0\) and one root is 3 Let \(p\) and 3 are root of the above equation So, \(p+3=-a\) \(p \times 3=3\) \(p=1\) Then, \(1+3=-a\) \(a=-4\) Now, other equation \(\mathrm{x}^2+\mathrm{ax}+\mathrm{b}=0\) According to the question- roots of above equation are- \(\mathrm{q} \& 3 \mathrm{q}\) Then, \(\quad \mathrm{q}+3 \mathrm{q}=-\mathrm{a}\) \(4 q=-a\) \(4 q=-(-4) \quad\{\because \mathrm{a}=-4\}\) \(\mathrm{q}=1\) and \(\quad \mathrm{q} \times 3 \mathrm{q}=\mathrm{b}\) \(\mathrm{b}=3\)
BCECE-2014
Complex Numbers and Quadratic Equation
118236
If one root of the equation \(x^2-\lambda x+12=0\) is even prime while \(x^2+\lambda x+\mu=0\) has equal roots then \(\mu\) is equal to
1 8
2 16
3 24
4 32
Explanation:
B We know that, only even prime is 2 Then, \(f(2)=0\) \(2^2-\lambda \times 2+12=0\) \(4-2 \lambda+12=0\) \(\lambda=8\) Now, other given equation is- \(x^2+\lambda x+\mu=0 \text { has equal roots }\) Then, \(D=0\) \(\lambda^2-4 \mu=0\) \((8)^2-4 \mu=0\) \(64-4 \mu=0\) \(\mu=16\)
BCECE-2013
Complex Numbers and Quadratic Equation
118237
If \(\alpha\) and \(\beta\) are the roots of \(a x^2+c=b x\), then the equation \((a+c y)^2=b^2 y\) in \(y\) has the roots
1 \(\alpha^{-1}, \beta^{-1}\)
2 \(\alpha^2, \beta^2\)
3 \(\alpha \beta^{-1}, \alpha^{-1} \beta\)
4 \(\alpha^{-2}, \beta^{-2}\)
Explanation:
D Given \(a^2+c=b x \text { and roots are } \alpha, \beta\) \(a x^2-b x+c=0\) \(\alpha+\beta=\frac{b}{a}\) \(\alpha \cdot \beta=c / a\) Other given equation is- \((a+c y)^2=b^2 y\) \(a^2+c^2 y^2+2 a c y=b^2 y\) \(c^2 y^2+\left(2 a c-b^2\right) y+a^2=0\) Divide the above equation by \(\mathrm{a}^2\) \(\frac{c^2}{a^2} y^2+\left(\frac{\left(2 a c-b^2\right)}{a^2}\right) y+1=0\) \(\left(\frac{c}{a}\right)^2 y^2+\left[\frac{2 c}{a}-\left(\frac{b}{a}\right)^2\right] y+1=0\) \((\alpha \beta)^2 y^2+\left[2 \alpha \beta-(\alpha+\beta)^2\right] y+1=0\) \((\alpha \beta)^2 y^2-\left[\alpha^2+\beta^2\right] y+1=0 .\) Now equation (i) is divided by \((\alpha \beta)^2\) \(y^2-\left(\beta^{-2}+\alpha^{-2}\right) y+\alpha^{-2} \beta^{-2}=0\) Hence, from above equation we can see that root of the equation are \(\alpha^{-2} \& \beta^{-2}\)
BCECE-2013
Complex Numbers and Quadratic Equation
118238
If \(\alpha+\beta=-2\) and \(\alpha^3+\beta^3=-56\), then the quadratic equation whose roots are \(\alpha\) and \(\beta\) is
1 \(x^2+2 x-16=0\)
2 \(x^2+2 x+15=0\)
3 \(x^2+2 x-12=0\)
4 \(x^2+2 x-8=0\)
Explanation:
D Given, \(\alpha+\beta=-2\) \(\alpha^3+\beta^3=-56\) Then, \((\alpha+\beta)^3=\alpha^3+\beta^3+3 \alpha \beta(\alpha+\beta)\) \((-2)^3=\alpha^3+\beta^3+3 \alpha \beta(-2)\) \(-8=-56-6 \alpha \beta\) \(48=-6 \alpha \beta\) \(\alpha \beta=-8\) Now, equation whose roots are \(\alpha\) and \(\beta\) \(x^2-(\text { sum of roots }) x+\text { product of roots }=0\) \(x^2-(-2) x+(-8)=0\) \(x^2+2 x-8=0\)
118235
If one of the roots of equation \(x^2+a x+3=0\) is 3 and one of the roots of the equation \(x^2+a x+b=0\) is three times the other root, then the value of \(b\) is
1 3
2 4
3 2
4 1
Explanation:
A Given equation is- \(x^2+a x+3=0\) and one root is 3 Let \(p\) and 3 are root of the above equation So, \(p+3=-a\) \(p \times 3=3\) \(p=1\) Then, \(1+3=-a\) \(a=-4\) Now, other equation \(\mathrm{x}^2+\mathrm{ax}+\mathrm{b}=0\) According to the question- roots of above equation are- \(\mathrm{q} \& 3 \mathrm{q}\) Then, \(\quad \mathrm{q}+3 \mathrm{q}=-\mathrm{a}\) \(4 q=-a\) \(4 q=-(-4) \quad\{\because \mathrm{a}=-4\}\) \(\mathrm{q}=1\) and \(\quad \mathrm{q} \times 3 \mathrm{q}=\mathrm{b}\) \(\mathrm{b}=3\)
BCECE-2014
Complex Numbers and Quadratic Equation
118236
If one root of the equation \(x^2-\lambda x+12=0\) is even prime while \(x^2+\lambda x+\mu=0\) has equal roots then \(\mu\) is equal to
1 8
2 16
3 24
4 32
Explanation:
B We know that, only even prime is 2 Then, \(f(2)=0\) \(2^2-\lambda \times 2+12=0\) \(4-2 \lambda+12=0\) \(\lambda=8\) Now, other given equation is- \(x^2+\lambda x+\mu=0 \text { has equal roots }\) Then, \(D=0\) \(\lambda^2-4 \mu=0\) \((8)^2-4 \mu=0\) \(64-4 \mu=0\) \(\mu=16\)
BCECE-2013
Complex Numbers and Quadratic Equation
118237
If \(\alpha\) and \(\beta\) are the roots of \(a x^2+c=b x\), then the equation \((a+c y)^2=b^2 y\) in \(y\) has the roots
1 \(\alpha^{-1}, \beta^{-1}\)
2 \(\alpha^2, \beta^2\)
3 \(\alpha \beta^{-1}, \alpha^{-1} \beta\)
4 \(\alpha^{-2}, \beta^{-2}\)
Explanation:
D Given \(a^2+c=b x \text { and roots are } \alpha, \beta\) \(a x^2-b x+c=0\) \(\alpha+\beta=\frac{b}{a}\) \(\alpha \cdot \beta=c / a\) Other given equation is- \((a+c y)^2=b^2 y\) \(a^2+c^2 y^2+2 a c y=b^2 y\) \(c^2 y^2+\left(2 a c-b^2\right) y+a^2=0\) Divide the above equation by \(\mathrm{a}^2\) \(\frac{c^2}{a^2} y^2+\left(\frac{\left(2 a c-b^2\right)}{a^2}\right) y+1=0\) \(\left(\frac{c}{a}\right)^2 y^2+\left[\frac{2 c}{a}-\left(\frac{b}{a}\right)^2\right] y+1=0\) \((\alpha \beta)^2 y^2+\left[2 \alpha \beta-(\alpha+\beta)^2\right] y+1=0\) \((\alpha \beta)^2 y^2-\left[\alpha^2+\beta^2\right] y+1=0 .\) Now equation (i) is divided by \((\alpha \beta)^2\) \(y^2-\left(\beta^{-2}+\alpha^{-2}\right) y+\alpha^{-2} \beta^{-2}=0\) Hence, from above equation we can see that root of the equation are \(\alpha^{-2} \& \beta^{-2}\)
BCECE-2013
Complex Numbers and Quadratic Equation
118238
If \(\alpha+\beta=-2\) and \(\alpha^3+\beta^3=-56\), then the quadratic equation whose roots are \(\alpha\) and \(\beta\) is
1 \(x^2+2 x-16=0\)
2 \(x^2+2 x+15=0\)
3 \(x^2+2 x-12=0\)
4 \(x^2+2 x-8=0\)
Explanation:
D Given, \(\alpha+\beta=-2\) \(\alpha^3+\beta^3=-56\) Then, \((\alpha+\beta)^3=\alpha^3+\beta^3+3 \alpha \beta(\alpha+\beta)\) \((-2)^3=\alpha^3+\beta^3+3 \alpha \beta(-2)\) \(-8=-56-6 \alpha \beta\) \(48=-6 \alpha \beta\) \(\alpha \beta=-8\) Now, equation whose roots are \(\alpha\) and \(\beta\) \(x^2-(\text { sum of roots }) x+\text { product of roots }=0\) \(x^2-(-2) x+(-8)=0\) \(x^2+2 x-8=0\)
118235
If one of the roots of equation \(x^2+a x+3=0\) is 3 and one of the roots of the equation \(x^2+a x+b=0\) is three times the other root, then the value of \(b\) is
1 3
2 4
3 2
4 1
Explanation:
A Given equation is- \(x^2+a x+3=0\) and one root is 3 Let \(p\) and 3 are root of the above equation So, \(p+3=-a\) \(p \times 3=3\) \(p=1\) Then, \(1+3=-a\) \(a=-4\) Now, other equation \(\mathrm{x}^2+\mathrm{ax}+\mathrm{b}=0\) According to the question- roots of above equation are- \(\mathrm{q} \& 3 \mathrm{q}\) Then, \(\quad \mathrm{q}+3 \mathrm{q}=-\mathrm{a}\) \(4 q=-a\) \(4 q=-(-4) \quad\{\because \mathrm{a}=-4\}\) \(\mathrm{q}=1\) and \(\quad \mathrm{q} \times 3 \mathrm{q}=\mathrm{b}\) \(\mathrm{b}=3\)
BCECE-2014
Complex Numbers and Quadratic Equation
118236
If one root of the equation \(x^2-\lambda x+12=0\) is even prime while \(x^2+\lambda x+\mu=0\) has equal roots then \(\mu\) is equal to
1 8
2 16
3 24
4 32
Explanation:
B We know that, only even prime is 2 Then, \(f(2)=0\) \(2^2-\lambda \times 2+12=0\) \(4-2 \lambda+12=0\) \(\lambda=8\) Now, other given equation is- \(x^2+\lambda x+\mu=0 \text { has equal roots }\) Then, \(D=0\) \(\lambda^2-4 \mu=0\) \((8)^2-4 \mu=0\) \(64-4 \mu=0\) \(\mu=16\)
BCECE-2013
Complex Numbers and Quadratic Equation
118237
If \(\alpha\) and \(\beta\) are the roots of \(a x^2+c=b x\), then the equation \((a+c y)^2=b^2 y\) in \(y\) has the roots
1 \(\alpha^{-1}, \beta^{-1}\)
2 \(\alpha^2, \beta^2\)
3 \(\alpha \beta^{-1}, \alpha^{-1} \beta\)
4 \(\alpha^{-2}, \beta^{-2}\)
Explanation:
D Given \(a^2+c=b x \text { and roots are } \alpha, \beta\) \(a x^2-b x+c=0\) \(\alpha+\beta=\frac{b}{a}\) \(\alpha \cdot \beta=c / a\) Other given equation is- \((a+c y)^2=b^2 y\) \(a^2+c^2 y^2+2 a c y=b^2 y\) \(c^2 y^2+\left(2 a c-b^2\right) y+a^2=0\) Divide the above equation by \(\mathrm{a}^2\) \(\frac{c^2}{a^2} y^2+\left(\frac{\left(2 a c-b^2\right)}{a^2}\right) y+1=0\) \(\left(\frac{c}{a}\right)^2 y^2+\left[\frac{2 c}{a}-\left(\frac{b}{a}\right)^2\right] y+1=0\) \((\alpha \beta)^2 y^2+\left[2 \alpha \beta-(\alpha+\beta)^2\right] y+1=0\) \((\alpha \beta)^2 y^2-\left[\alpha^2+\beta^2\right] y+1=0 .\) Now equation (i) is divided by \((\alpha \beta)^2\) \(y^2-\left(\beta^{-2}+\alpha^{-2}\right) y+\alpha^{-2} \beta^{-2}=0\) Hence, from above equation we can see that root of the equation are \(\alpha^{-2} \& \beta^{-2}\)
BCECE-2013
Complex Numbers and Quadratic Equation
118238
If \(\alpha+\beta=-2\) and \(\alpha^3+\beta^3=-56\), then the quadratic equation whose roots are \(\alpha\) and \(\beta\) is
1 \(x^2+2 x-16=0\)
2 \(x^2+2 x+15=0\)
3 \(x^2+2 x-12=0\)
4 \(x^2+2 x-8=0\)
Explanation:
D Given, \(\alpha+\beta=-2\) \(\alpha^3+\beta^3=-56\) Then, \((\alpha+\beta)^3=\alpha^3+\beta^3+3 \alpha \beta(\alpha+\beta)\) \((-2)^3=\alpha^3+\beta^3+3 \alpha \beta(-2)\) \(-8=-56-6 \alpha \beta\) \(48=-6 \alpha \beta\) \(\alpha \beta=-8\) Now, equation whose roots are \(\alpha\) and \(\beta\) \(x^2-(\text { sum of roots }) x+\text { product of roots }=0\) \(x^2-(-2) x+(-8)=0\) \(x^2+2 x-8=0\)