118102
The quadratic equation whose roots are \(\frac{1}{3+\sqrt{2}}\) and \(\frac{1}{3-\sqrt{2}}\), willbe
1 \(7 \mathrm{x}^2+6 \mathrm{x}+1=0\)
2 \(6 x^2-7 x+1=0\)
3 \(\mathrm{x}^2-6 \mathrm{x}+7=0\)
4 \(x^2-7 x+6=0\)
Explanation:
A Given, root are \(\frac{1}{3+\sqrt{2}}\) and \(\frac{1}{3-\sqrt{2}}\) \(\text { i.e. } \frac{1}{(3+\sqrt{2})} \times \frac{(3-\sqrt{2})}{(3-\sqrt{2})} \text { and } \frac{1}{(3-\sqrt{2})} \times \frac{(3+\sqrt{2})}{(3+\sqrt{2})}\) \(\Rightarrow \quad \frac{3-\sqrt{2}}{7} \text { and } \frac{3+\sqrt{2}}{7}\) So, the required quadratic equation is \(\mathrm{x}^2-\) (Sum of the roots) \(\mathrm{x}+\) Product of the roots \(=0\) \(\Rightarrow x^2-\left[\frac{3-\sqrt{2}}{7}+\frac{3+\sqrt{2}}{7}\right] x+\left(\frac{3-\sqrt{2}}{7}\right)\left(\frac{3+\sqrt{2}}{7}\right)=0\) \(\Rightarrow x^2+\left(\frac{6}{7}\right) x+\frac{9-2}{49}=0\) \(\Rightarrow x^2+\frac{6}{7} x+\frac{7}{49}=0\) \(\Rightarrow 7 x^2+6 x+1=0\)
CG PET- 2015
Complex Numbers and Quadratic Equation
118104
Let \(\alpha\) and \(\beta\) be the roots of the equation \(x^2+(2 \mathrm{i}\) - 1) \(=0\). Then, the value of \(\left|\alpha^8+\beta^8\right|\) is equal to :
1 50
2 250
3 1250
4 1500
Explanation:
A Given, \(x^2+(2 i-1)=0\) \(x^2=1-2 i\) \(\alpha\) and \(\beta\) are the roots. Therefore, \(\alpha\) and \(\beta\) will satisfy the equation, \(\therefore \quad \alpha^2=(1-2 \mathrm{i})\) \(\alpha^4= (1-2 \mathrm{i})^2\) \(=1+4 \mathrm{i}^2-4 \mathrm{i}\) \(=-3-4 \mathrm{i}\) \(\Rightarrow \quad \alpha^8=9+16 \mathrm{i}^2+24 \mathrm{i}\) \(=-7+24 \mathrm{i}\) Similarly \(\beta^2=(1-2 i)\) \(\therefore \beta^8=-7+24 i\) \(\therefore \alpha^8+\beta^8=-14+48 \mathrm{i}\) \(\therefore \left|\alpha^8+\beta^8\right|=|-14+48 \mathrm{i}|=50\)
JEE Main-29.06.2022
Complex Numbers and Quadratic Equation
118105
If \(f(x)\) is a second order polynomial (or quadratic expression in \(x\) ) and \(\int_a^b f(x) d x=(a-b)\left(a^2+b^2+a b+2\right)\), then \(f(x)\) with be of the form
1 \(3 x^2+x+2\)
2 \(2 x^2-x\)
3 \(-3 x^2-2\)
4 \(3 x^2+2\)
Explanation:
C Given, \(\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=(\mathrm{a}-\mathrm{b})\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{ab}+2\right)\) Let \(f(x)=p x^2+q x+r\) \(\int_a^b\left(p x^2+q x+r\right) d x=(a-b)\left(a^2+b^2+a b\right)+2(a-b)\) \({\left[p \frac{x^3}{3}+q \frac{x^2}{2}+r x\right]_a^b=\left(a^3-b^3\right)+2(a-b)}\) \(\frac{p}{3}\left(b^3-a^3\right)+\frac{q}{2}\left(b^2-a^2\right)+r(b-a)=\left(a^3-b^3\right)+2(a-b)\) Comparing coefficient, we get- \(\frac{p}{3}=-1 \Rightarrow p=-3, r=-2\) \(\text { So, } f(x)=-3 x^2-2\)
SCRA-2015
Complex Numbers and Quadratic Equation
118106
If \(a\) and \(b\) are rational and \(\left(b^2+1\right)\) is not a perfect square, then the quadratic equation with rational coefficients whose one root is \(\frac{a}{2}\left(b+\sqrt{1+b^2}\right)\) is
1 \(x^2-2 a b x-a^2=0\) Roots are, \(\Rightarrow \frac{2 a b \pm \sqrt{4 a^2 b^2-4(1)\left(-a^2\right)}}{2(1)}\) \(\Rightarrow \frac{2 a b \pm \sqrt{4 a^2\left(b^2+1\right)}}{2}\) \(=\frac{2 a b \pm 2 a \sqrt{b^2+1}}{2}\) \(=a b \pm a \sqrt{b^2+1}\) So, it is not the required root . so, it is wrong option.
2 \(4 x^2-4 a b x-a^2=0\) Roots are, \(=\frac{4 a b \pm \sqrt{16 a^2 b^2+16 a^2}}{2(4)}\) \(=\frac{4 a b \pm 4 a \sqrt{b^2+1}}{8}\) \(=\frac{a}{2}\left[b \pm \sqrt{b^2+1}\right]\)So, this is the required option.
3 \(x^2-a b x-a^2=0\)
4 \(x^2-a b x+a^2=0\)
Explanation:
B a and b are rational \& \(\left(b^2+1\right)\) is a perfect square root is \(\frac{a}{2}\left(b+\sqrt{1+b^2}\right)\). (a.) \(x^2-2 a b x-a^2=0\) Roots are, \(\Rightarrow \frac{2 a b \pm \sqrt{4 a^2 b^2-4(1)\left(-a^2\right)}}{2(1)}\) \(\Rightarrow \frac{2 a b \pm \sqrt{4 a^2\left(b^2+1\right)}}{2}\) \(=\frac{2 a b \pm 2 a \sqrt{b^2+1}}{2}\) \(=a b \pm a \sqrt{b^2+1}\) So, it is not the required root . so, it is wrong option. (b.) \(4 x^2-4 a b x-a^2=0\) Roots are, \(=\frac{4 a b \pm \sqrt{16 a^2 b^2+16 a^2}}{2(4)}\) \(=\frac{4 a b \pm 4 a \sqrt{b^2+1}}{8}\) \(=\frac{a}{2}\left[b \pm \sqrt{b^2+1}\right]\)So, this is the required option.
118102
The quadratic equation whose roots are \(\frac{1}{3+\sqrt{2}}\) and \(\frac{1}{3-\sqrt{2}}\), willbe
1 \(7 \mathrm{x}^2+6 \mathrm{x}+1=0\)
2 \(6 x^2-7 x+1=0\)
3 \(\mathrm{x}^2-6 \mathrm{x}+7=0\)
4 \(x^2-7 x+6=0\)
Explanation:
A Given, root are \(\frac{1}{3+\sqrt{2}}\) and \(\frac{1}{3-\sqrt{2}}\) \(\text { i.e. } \frac{1}{(3+\sqrt{2})} \times \frac{(3-\sqrt{2})}{(3-\sqrt{2})} \text { and } \frac{1}{(3-\sqrt{2})} \times \frac{(3+\sqrt{2})}{(3+\sqrt{2})}\) \(\Rightarrow \quad \frac{3-\sqrt{2}}{7} \text { and } \frac{3+\sqrt{2}}{7}\) So, the required quadratic equation is \(\mathrm{x}^2-\) (Sum of the roots) \(\mathrm{x}+\) Product of the roots \(=0\) \(\Rightarrow x^2-\left[\frac{3-\sqrt{2}}{7}+\frac{3+\sqrt{2}}{7}\right] x+\left(\frac{3-\sqrt{2}}{7}\right)\left(\frac{3+\sqrt{2}}{7}\right)=0\) \(\Rightarrow x^2+\left(\frac{6}{7}\right) x+\frac{9-2}{49}=0\) \(\Rightarrow x^2+\frac{6}{7} x+\frac{7}{49}=0\) \(\Rightarrow 7 x^2+6 x+1=0\)
CG PET- 2015
Complex Numbers and Quadratic Equation
118104
Let \(\alpha\) and \(\beta\) be the roots of the equation \(x^2+(2 \mathrm{i}\) - 1) \(=0\). Then, the value of \(\left|\alpha^8+\beta^8\right|\) is equal to :
1 50
2 250
3 1250
4 1500
Explanation:
A Given, \(x^2+(2 i-1)=0\) \(x^2=1-2 i\) \(\alpha\) and \(\beta\) are the roots. Therefore, \(\alpha\) and \(\beta\) will satisfy the equation, \(\therefore \quad \alpha^2=(1-2 \mathrm{i})\) \(\alpha^4= (1-2 \mathrm{i})^2\) \(=1+4 \mathrm{i}^2-4 \mathrm{i}\) \(=-3-4 \mathrm{i}\) \(\Rightarrow \quad \alpha^8=9+16 \mathrm{i}^2+24 \mathrm{i}\) \(=-7+24 \mathrm{i}\) Similarly \(\beta^2=(1-2 i)\) \(\therefore \beta^8=-7+24 i\) \(\therefore \alpha^8+\beta^8=-14+48 \mathrm{i}\) \(\therefore \left|\alpha^8+\beta^8\right|=|-14+48 \mathrm{i}|=50\)
JEE Main-29.06.2022
Complex Numbers and Quadratic Equation
118105
If \(f(x)\) is a second order polynomial (or quadratic expression in \(x\) ) and \(\int_a^b f(x) d x=(a-b)\left(a^2+b^2+a b+2\right)\), then \(f(x)\) with be of the form
1 \(3 x^2+x+2\)
2 \(2 x^2-x\)
3 \(-3 x^2-2\)
4 \(3 x^2+2\)
Explanation:
C Given, \(\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=(\mathrm{a}-\mathrm{b})\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{ab}+2\right)\) Let \(f(x)=p x^2+q x+r\) \(\int_a^b\left(p x^2+q x+r\right) d x=(a-b)\left(a^2+b^2+a b\right)+2(a-b)\) \({\left[p \frac{x^3}{3}+q \frac{x^2}{2}+r x\right]_a^b=\left(a^3-b^3\right)+2(a-b)}\) \(\frac{p}{3}\left(b^3-a^3\right)+\frac{q}{2}\left(b^2-a^2\right)+r(b-a)=\left(a^3-b^3\right)+2(a-b)\) Comparing coefficient, we get- \(\frac{p}{3}=-1 \Rightarrow p=-3, r=-2\) \(\text { So, } f(x)=-3 x^2-2\)
SCRA-2015
Complex Numbers and Quadratic Equation
118106
If \(a\) and \(b\) are rational and \(\left(b^2+1\right)\) is not a perfect square, then the quadratic equation with rational coefficients whose one root is \(\frac{a}{2}\left(b+\sqrt{1+b^2}\right)\) is
1 \(x^2-2 a b x-a^2=0\) Roots are, \(\Rightarrow \frac{2 a b \pm \sqrt{4 a^2 b^2-4(1)\left(-a^2\right)}}{2(1)}\) \(\Rightarrow \frac{2 a b \pm \sqrt{4 a^2\left(b^2+1\right)}}{2}\) \(=\frac{2 a b \pm 2 a \sqrt{b^2+1}}{2}\) \(=a b \pm a \sqrt{b^2+1}\) So, it is not the required root . so, it is wrong option.
2 \(4 x^2-4 a b x-a^2=0\) Roots are, \(=\frac{4 a b \pm \sqrt{16 a^2 b^2+16 a^2}}{2(4)}\) \(=\frac{4 a b \pm 4 a \sqrt{b^2+1}}{8}\) \(=\frac{a}{2}\left[b \pm \sqrt{b^2+1}\right]\)So, this is the required option.
3 \(x^2-a b x-a^2=0\)
4 \(x^2-a b x+a^2=0\)
Explanation:
B a and b are rational \& \(\left(b^2+1\right)\) is a perfect square root is \(\frac{a}{2}\left(b+\sqrt{1+b^2}\right)\). (a.) \(x^2-2 a b x-a^2=0\) Roots are, \(\Rightarrow \frac{2 a b \pm \sqrt{4 a^2 b^2-4(1)\left(-a^2\right)}}{2(1)}\) \(\Rightarrow \frac{2 a b \pm \sqrt{4 a^2\left(b^2+1\right)}}{2}\) \(=\frac{2 a b \pm 2 a \sqrt{b^2+1}}{2}\) \(=a b \pm a \sqrt{b^2+1}\) So, it is not the required root . so, it is wrong option. (b.) \(4 x^2-4 a b x-a^2=0\) Roots are, \(=\frac{4 a b \pm \sqrt{16 a^2 b^2+16 a^2}}{2(4)}\) \(=\frac{4 a b \pm 4 a \sqrt{b^2+1}}{8}\) \(=\frac{a}{2}\left[b \pm \sqrt{b^2+1}\right]\)So, this is the required option.
118102
The quadratic equation whose roots are \(\frac{1}{3+\sqrt{2}}\) and \(\frac{1}{3-\sqrt{2}}\), willbe
1 \(7 \mathrm{x}^2+6 \mathrm{x}+1=0\)
2 \(6 x^2-7 x+1=0\)
3 \(\mathrm{x}^2-6 \mathrm{x}+7=0\)
4 \(x^2-7 x+6=0\)
Explanation:
A Given, root are \(\frac{1}{3+\sqrt{2}}\) and \(\frac{1}{3-\sqrt{2}}\) \(\text { i.e. } \frac{1}{(3+\sqrt{2})} \times \frac{(3-\sqrt{2})}{(3-\sqrt{2})} \text { and } \frac{1}{(3-\sqrt{2})} \times \frac{(3+\sqrt{2})}{(3+\sqrt{2})}\) \(\Rightarrow \quad \frac{3-\sqrt{2}}{7} \text { and } \frac{3+\sqrt{2}}{7}\) So, the required quadratic equation is \(\mathrm{x}^2-\) (Sum of the roots) \(\mathrm{x}+\) Product of the roots \(=0\) \(\Rightarrow x^2-\left[\frac{3-\sqrt{2}}{7}+\frac{3+\sqrt{2}}{7}\right] x+\left(\frac{3-\sqrt{2}}{7}\right)\left(\frac{3+\sqrt{2}}{7}\right)=0\) \(\Rightarrow x^2+\left(\frac{6}{7}\right) x+\frac{9-2}{49}=0\) \(\Rightarrow x^2+\frac{6}{7} x+\frac{7}{49}=0\) \(\Rightarrow 7 x^2+6 x+1=0\)
CG PET- 2015
Complex Numbers and Quadratic Equation
118104
Let \(\alpha\) and \(\beta\) be the roots of the equation \(x^2+(2 \mathrm{i}\) - 1) \(=0\). Then, the value of \(\left|\alpha^8+\beta^8\right|\) is equal to :
1 50
2 250
3 1250
4 1500
Explanation:
A Given, \(x^2+(2 i-1)=0\) \(x^2=1-2 i\) \(\alpha\) and \(\beta\) are the roots. Therefore, \(\alpha\) and \(\beta\) will satisfy the equation, \(\therefore \quad \alpha^2=(1-2 \mathrm{i})\) \(\alpha^4= (1-2 \mathrm{i})^2\) \(=1+4 \mathrm{i}^2-4 \mathrm{i}\) \(=-3-4 \mathrm{i}\) \(\Rightarrow \quad \alpha^8=9+16 \mathrm{i}^2+24 \mathrm{i}\) \(=-7+24 \mathrm{i}\) Similarly \(\beta^2=(1-2 i)\) \(\therefore \beta^8=-7+24 i\) \(\therefore \alpha^8+\beta^8=-14+48 \mathrm{i}\) \(\therefore \left|\alpha^8+\beta^8\right|=|-14+48 \mathrm{i}|=50\)
JEE Main-29.06.2022
Complex Numbers and Quadratic Equation
118105
If \(f(x)\) is a second order polynomial (or quadratic expression in \(x\) ) and \(\int_a^b f(x) d x=(a-b)\left(a^2+b^2+a b+2\right)\), then \(f(x)\) with be of the form
1 \(3 x^2+x+2\)
2 \(2 x^2-x\)
3 \(-3 x^2-2\)
4 \(3 x^2+2\)
Explanation:
C Given, \(\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=(\mathrm{a}-\mathrm{b})\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{ab}+2\right)\) Let \(f(x)=p x^2+q x+r\) \(\int_a^b\left(p x^2+q x+r\right) d x=(a-b)\left(a^2+b^2+a b\right)+2(a-b)\) \({\left[p \frac{x^3}{3}+q \frac{x^2}{2}+r x\right]_a^b=\left(a^3-b^3\right)+2(a-b)}\) \(\frac{p}{3}\left(b^3-a^3\right)+\frac{q}{2}\left(b^2-a^2\right)+r(b-a)=\left(a^3-b^3\right)+2(a-b)\) Comparing coefficient, we get- \(\frac{p}{3}=-1 \Rightarrow p=-3, r=-2\) \(\text { So, } f(x)=-3 x^2-2\)
SCRA-2015
Complex Numbers and Quadratic Equation
118106
If \(a\) and \(b\) are rational and \(\left(b^2+1\right)\) is not a perfect square, then the quadratic equation with rational coefficients whose one root is \(\frac{a}{2}\left(b+\sqrt{1+b^2}\right)\) is
1 \(x^2-2 a b x-a^2=0\) Roots are, \(\Rightarrow \frac{2 a b \pm \sqrt{4 a^2 b^2-4(1)\left(-a^2\right)}}{2(1)}\) \(\Rightarrow \frac{2 a b \pm \sqrt{4 a^2\left(b^2+1\right)}}{2}\) \(=\frac{2 a b \pm 2 a \sqrt{b^2+1}}{2}\) \(=a b \pm a \sqrt{b^2+1}\) So, it is not the required root . so, it is wrong option.
2 \(4 x^2-4 a b x-a^2=0\) Roots are, \(=\frac{4 a b \pm \sqrt{16 a^2 b^2+16 a^2}}{2(4)}\) \(=\frac{4 a b \pm 4 a \sqrt{b^2+1}}{8}\) \(=\frac{a}{2}\left[b \pm \sqrt{b^2+1}\right]\)So, this is the required option.
3 \(x^2-a b x-a^2=0\)
4 \(x^2-a b x+a^2=0\)
Explanation:
B a and b are rational \& \(\left(b^2+1\right)\) is a perfect square root is \(\frac{a}{2}\left(b+\sqrt{1+b^2}\right)\). (a.) \(x^2-2 a b x-a^2=0\) Roots are, \(\Rightarrow \frac{2 a b \pm \sqrt{4 a^2 b^2-4(1)\left(-a^2\right)}}{2(1)}\) \(\Rightarrow \frac{2 a b \pm \sqrt{4 a^2\left(b^2+1\right)}}{2}\) \(=\frac{2 a b \pm 2 a \sqrt{b^2+1}}{2}\) \(=a b \pm a \sqrt{b^2+1}\) So, it is not the required root . so, it is wrong option. (b.) \(4 x^2-4 a b x-a^2=0\) Roots are, \(=\frac{4 a b \pm \sqrt{16 a^2 b^2+16 a^2}}{2(4)}\) \(=\frac{4 a b \pm 4 a \sqrt{b^2+1}}{8}\) \(=\frac{a}{2}\left[b \pm \sqrt{b^2+1}\right]\)So, this is the required option.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Complex Numbers and Quadratic Equation
118102
The quadratic equation whose roots are \(\frac{1}{3+\sqrt{2}}\) and \(\frac{1}{3-\sqrt{2}}\), willbe
1 \(7 \mathrm{x}^2+6 \mathrm{x}+1=0\)
2 \(6 x^2-7 x+1=0\)
3 \(\mathrm{x}^2-6 \mathrm{x}+7=0\)
4 \(x^2-7 x+6=0\)
Explanation:
A Given, root are \(\frac{1}{3+\sqrt{2}}\) and \(\frac{1}{3-\sqrt{2}}\) \(\text { i.e. } \frac{1}{(3+\sqrt{2})} \times \frac{(3-\sqrt{2})}{(3-\sqrt{2})} \text { and } \frac{1}{(3-\sqrt{2})} \times \frac{(3+\sqrt{2})}{(3+\sqrt{2})}\) \(\Rightarrow \quad \frac{3-\sqrt{2}}{7} \text { and } \frac{3+\sqrt{2}}{7}\) So, the required quadratic equation is \(\mathrm{x}^2-\) (Sum of the roots) \(\mathrm{x}+\) Product of the roots \(=0\) \(\Rightarrow x^2-\left[\frac{3-\sqrt{2}}{7}+\frac{3+\sqrt{2}}{7}\right] x+\left(\frac{3-\sqrt{2}}{7}\right)\left(\frac{3+\sqrt{2}}{7}\right)=0\) \(\Rightarrow x^2+\left(\frac{6}{7}\right) x+\frac{9-2}{49}=0\) \(\Rightarrow x^2+\frac{6}{7} x+\frac{7}{49}=0\) \(\Rightarrow 7 x^2+6 x+1=0\)
CG PET- 2015
Complex Numbers and Quadratic Equation
118104
Let \(\alpha\) and \(\beta\) be the roots of the equation \(x^2+(2 \mathrm{i}\) - 1) \(=0\). Then, the value of \(\left|\alpha^8+\beta^8\right|\) is equal to :
1 50
2 250
3 1250
4 1500
Explanation:
A Given, \(x^2+(2 i-1)=0\) \(x^2=1-2 i\) \(\alpha\) and \(\beta\) are the roots. Therefore, \(\alpha\) and \(\beta\) will satisfy the equation, \(\therefore \quad \alpha^2=(1-2 \mathrm{i})\) \(\alpha^4= (1-2 \mathrm{i})^2\) \(=1+4 \mathrm{i}^2-4 \mathrm{i}\) \(=-3-4 \mathrm{i}\) \(\Rightarrow \quad \alpha^8=9+16 \mathrm{i}^2+24 \mathrm{i}\) \(=-7+24 \mathrm{i}\) Similarly \(\beta^2=(1-2 i)\) \(\therefore \beta^8=-7+24 i\) \(\therefore \alpha^8+\beta^8=-14+48 \mathrm{i}\) \(\therefore \left|\alpha^8+\beta^8\right|=|-14+48 \mathrm{i}|=50\)
JEE Main-29.06.2022
Complex Numbers and Quadratic Equation
118105
If \(f(x)\) is a second order polynomial (or quadratic expression in \(x\) ) and \(\int_a^b f(x) d x=(a-b)\left(a^2+b^2+a b+2\right)\), then \(f(x)\) with be of the form
1 \(3 x^2+x+2\)
2 \(2 x^2-x\)
3 \(-3 x^2-2\)
4 \(3 x^2+2\)
Explanation:
C Given, \(\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=(\mathrm{a}-\mathrm{b})\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{ab}+2\right)\) Let \(f(x)=p x^2+q x+r\) \(\int_a^b\left(p x^2+q x+r\right) d x=(a-b)\left(a^2+b^2+a b\right)+2(a-b)\) \({\left[p \frac{x^3}{3}+q \frac{x^2}{2}+r x\right]_a^b=\left(a^3-b^3\right)+2(a-b)}\) \(\frac{p}{3}\left(b^3-a^3\right)+\frac{q}{2}\left(b^2-a^2\right)+r(b-a)=\left(a^3-b^3\right)+2(a-b)\) Comparing coefficient, we get- \(\frac{p}{3}=-1 \Rightarrow p=-3, r=-2\) \(\text { So, } f(x)=-3 x^2-2\)
SCRA-2015
Complex Numbers and Quadratic Equation
118106
If \(a\) and \(b\) are rational and \(\left(b^2+1\right)\) is not a perfect square, then the quadratic equation with rational coefficients whose one root is \(\frac{a}{2}\left(b+\sqrt{1+b^2}\right)\) is
1 \(x^2-2 a b x-a^2=0\) Roots are, \(\Rightarrow \frac{2 a b \pm \sqrt{4 a^2 b^2-4(1)\left(-a^2\right)}}{2(1)}\) \(\Rightarrow \frac{2 a b \pm \sqrt{4 a^2\left(b^2+1\right)}}{2}\) \(=\frac{2 a b \pm 2 a \sqrt{b^2+1}}{2}\) \(=a b \pm a \sqrt{b^2+1}\) So, it is not the required root . so, it is wrong option.
2 \(4 x^2-4 a b x-a^2=0\) Roots are, \(=\frac{4 a b \pm \sqrt{16 a^2 b^2+16 a^2}}{2(4)}\) \(=\frac{4 a b \pm 4 a \sqrt{b^2+1}}{8}\) \(=\frac{a}{2}\left[b \pm \sqrt{b^2+1}\right]\)So, this is the required option.
3 \(x^2-a b x-a^2=0\)
4 \(x^2-a b x+a^2=0\)
Explanation:
B a and b are rational \& \(\left(b^2+1\right)\) is a perfect square root is \(\frac{a}{2}\left(b+\sqrt{1+b^2}\right)\). (a.) \(x^2-2 a b x-a^2=0\) Roots are, \(\Rightarrow \frac{2 a b \pm \sqrt{4 a^2 b^2-4(1)\left(-a^2\right)}}{2(1)}\) \(\Rightarrow \frac{2 a b \pm \sqrt{4 a^2\left(b^2+1\right)}}{2}\) \(=\frac{2 a b \pm 2 a \sqrt{b^2+1}}{2}\) \(=a b \pm a \sqrt{b^2+1}\) So, it is not the required root . so, it is wrong option. (b.) \(4 x^2-4 a b x-a^2=0\) Roots are, \(=\frac{4 a b \pm \sqrt{16 a^2 b^2+16 a^2}}{2(4)}\) \(=\frac{4 a b \pm 4 a \sqrt{b^2+1}}{8}\) \(=\frac{a}{2}\left[b \pm \sqrt{b^2+1}\right]\)So, this is the required option.
