A Given, \(p, q\) and \(r\) are in \(A P\) Then, \(\mathrm{q}=\frac{\mathrm{p}+\mathrm{r}}{2}\) And equation \(\mathrm{px}^2+\mathrm{qx}+\mathrm{r}=0\) have real roots So, \(\mathrm{D} \geq 0\) \(\mathrm{q}^2-4 \mathrm{pr} \geq 0\) On putting the value \(\mathrm{q}\) in above equation. \(\left(\frac{\mathrm{p}+\mathrm{r}}{2}\right)^2-4 \mathrm{pr} \geq 0\) \(\mathrm{p}^2+\mathrm{r}^2+2 \mathrm{pr}-16 \mathrm{pr} \geq 0\) \(\mathrm{p}^2+\mathrm{r}^2-14 \mathrm{pr} \geq 0\) On adding both side \(48 \mathrm{p}^2\) \(49 p^2+r^2-14 p r \geq 48 p^2\) \((r-7 p)^2 \geq(4 \sqrt{3} p)^2\) \(\left(\frac{r}{p}-7\right)^2 \geq(4 \sqrt{3})^2\) \(\because \quad p \neq 0\) \(\therefore \quad \left|\frac{r}{p}-7\right| \geq 4 \sqrt{3}\)
UPSEE-2013
Complex Numbers and Quadratic Equation
118460
If \(\alpha, \beta\) are the roots of the equation \({a x^2+b x}^2\) \(\mathbf{c}=\mathbf{0}\), then \(\frac{\alpha}{\mathrm{a} \beta+\mathrm{b}}+\frac{\beta}{\mathrm{a} \alpha+\mathrm{b}}\) is equal to
1 \(\frac{2}{\mathrm{a}}\)
2 \(\frac{2}{\mathrm{~b}}\)
3 \(\frac{2}{\mathrm{c}}\)
4 \(-\frac{2}{a}\)
Explanation:
D Given \(\alpha, \beta\) are the roots of equation \(a x^2+b x+c=0\) \(\alpha+\beta=\frac{-b}{a}, \alpha \cdot \beta=\frac{c}{a}\) \((\alpha+\beta)^2=\left(\frac{-b}{a}\right)^2\) \(\alpha^2+\beta^2=\frac{b^2}{a^2}-\frac{2 c}{a}\) \(\alpha^2+\beta^2=\frac{b^2-2 a c}{a^2}\) \(=\frac{\alpha}{a \beta+b}+\frac{\beta}{a \alpha+b}\) \(=\frac{a \alpha^2+b \alpha+a \beta^2+b \beta}{\alpha^2 \alpha \beta+a b \beta+b a \alpha+b^2}\) \(=\frac{a\left(\alpha^2+\beta^2\right)+b(\alpha+\beta)}{a^2 \alpha \beta+a b(\alpha+\beta)+b^2}\) \(=\frac{\left(\frac{b^2-2 a c}{a^2}\right)+b\left(\frac{-b}{a}\right)}{a^2 \times \frac{c}{a}+a b\left(\frac{-b}{a}\right)+b^2}\) \(=\frac{-2}{a}\)
UPSEE -2008
Complex Numbers and Quadratic Equation
118461
If \(\sin \alpha\) and \(\cos \alpha\) are the roots of the equation \(a x^2+b x+c=0\), then
1 \(\mathrm{a}^2-\mathrm{b}^2+2 \mathrm{ac}=0\)
2 \((a-c)^2=b^2+C^2\)
3 \(\mathrm{a}^2+\mathrm{b}^2-2 \mathrm{ac}=0\)
4 \(\mathrm{a}^2+\mathrm{b}^2+2 \mathrm{ac}=0\)
Explanation:
A It is given that, \(\sin \alpha\) and \(\cos \alpha\) are the roots of \(a x^2+b x+c=0\) \(\therefore \quad \text { Sum of the roots }=\sin \alpha+\cos \alpha=\frac{-b}{a}\) \(\text { and product of the roots }=\sin \alpha \cdot \cos \alpha=\frac{c}{a}\) \(\because \quad(\sin \alpha+\cos \alpha)^2=\sin ^2 \alpha+\cos ^2 \alpha+2 \sin \alpha \cos \alpha\) \(\because \quad\left(\frac{-b}{a}\right)^2=1+2\left(\frac{c}{a}\right)\) \(\Rightarrow \quad \frac{b^2}{a^2}=\frac{a+2 c}{a}\) \(\Rightarrow \quad b^2=a(a+2 c)\) \(\Rightarrow \quad b^2=a^2+2 a c\) \(\Rightarrow \quad a^2-b^2+2 a c=0\)
CG PET- 2015
Complex Numbers and Quadratic Equation
118462
If \(\alpha\) and \(\beta\) are the roots of the equation \(x^2-\) \(4 x+5=0\), then the quadratic equation whose roots are \(\alpha^2+\beta\) and \(\alpha+\beta^2\) is
1 \(x^2+10 x+34=0\)
2 \(x^2-10 x+34=0\)
3 \(x^2-10 x-34=0\)
4 \(x^2+10 x-34=0\)
Explanation:
B Given, \(x^2-4 x+5=0\) \(\alpha+\beta=4\) \(\alpha \cdot \beta=5\) Now, \(\left(\alpha^2+\beta\right)+\left(\alpha+\beta^2\right)\) \(=\alpha^2+\beta+\alpha+\beta^2\) \(=\left(\alpha^2+\beta^2\right)+(\alpha+\beta)\) \(=(\alpha+\beta)^2-2 \alpha \beta+(\alpha+\beta)\) \(=(4)^2-(2 \times 5)+4\) \(=16-10+4\) \(=10\) Now, \(\left(\alpha^2+\beta\right)\left(\alpha+\beta^2\right)=\alpha^3+\alpha^2 \beta^2+\beta \alpha+\beta^3\) \(=\alpha^3+\beta^3+\alpha \beta(\alpha \beta+1)\) \(=(\alpha+\beta)\left(\alpha^2+\beta^2-\alpha \beta\right)+\alpha \beta(\alpha \beta+1)\) \(=(\alpha+\beta)\left[(\alpha+\beta)^2-3 \alpha \beta\right]+\alpha \beta(\alpha \beta+1)\) \(=4(16-15)+5(5+1)\) \(=4+30=34\) So, the quadratic equation whose roots are \(\left(\alpha^2+\beta\right) \text { and }\left(\alpha+\beta^2\right) \text { is }\) \(x^2-\left(\alpha^2+\beta+\alpha+\beta^2\right) x+\left(\alpha^2+\beta\right)\left(\alpha+\beta^2\right)=0\) \(x^2-10 x+34=0\)
A Given, \(p, q\) and \(r\) are in \(A P\) Then, \(\mathrm{q}=\frac{\mathrm{p}+\mathrm{r}}{2}\) And equation \(\mathrm{px}^2+\mathrm{qx}+\mathrm{r}=0\) have real roots So, \(\mathrm{D} \geq 0\) \(\mathrm{q}^2-4 \mathrm{pr} \geq 0\) On putting the value \(\mathrm{q}\) in above equation. \(\left(\frac{\mathrm{p}+\mathrm{r}}{2}\right)^2-4 \mathrm{pr} \geq 0\) \(\mathrm{p}^2+\mathrm{r}^2+2 \mathrm{pr}-16 \mathrm{pr} \geq 0\) \(\mathrm{p}^2+\mathrm{r}^2-14 \mathrm{pr} \geq 0\) On adding both side \(48 \mathrm{p}^2\) \(49 p^2+r^2-14 p r \geq 48 p^2\) \((r-7 p)^2 \geq(4 \sqrt{3} p)^2\) \(\left(\frac{r}{p}-7\right)^2 \geq(4 \sqrt{3})^2\) \(\because \quad p \neq 0\) \(\therefore \quad \left|\frac{r}{p}-7\right| \geq 4 \sqrt{3}\)
UPSEE-2013
Complex Numbers and Quadratic Equation
118460
If \(\alpha, \beta\) are the roots of the equation \({a x^2+b x}^2\) \(\mathbf{c}=\mathbf{0}\), then \(\frac{\alpha}{\mathrm{a} \beta+\mathrm{b}}+\frac{\beta}{\mathrm{a} \alpha+\mathrm{b}}\) is equal to
1 \(\frac{2}{\mathrm{a}}\)
2 \(\frac{2}{\mathrm{~b}}\)
3 \(\frac{2}{\mathrm{c}}\)
4 \(-\frac{2}{a}\)
Explanation:
D Given \(\alpha, \beta\) are the roots of equation \(a x^2+b x+c=0\) \(\alpha+\beta=\frac{-b}{a}, \alpha \cdot \beta=\frac{c}{a}\) \((\alpha+\beta)^2=\left(\frac{-b}{a}\right)^2\) \(\alpha^2+\beta^2=\frac{b^2}{a^2}-\frac{2 c}{a}\) \(\alpha^2+\beta^2=\frac{b^2-2 a c}{a^2}\) \(=\frac{\alpha}{a \beta+b}+\frac{\beta}{a \alpha+b}\) \(=\frac{a \alpha^2+b \alpha+a \beta^2+b \beta}{\alpha^2 \alpha \beta+a b \beta+b a \alpha+b^2}\) \(=\frac{a\left(\alpha^2+\beta^2\right)+b(\alpha+\beta)}{a^2 \alpha \beta+a b(\alpha+\beta)+b^2}\) \(=\frac{\left(\frac{b^2-2 a c}{a^2}\right)+b\left(\frac{-b}{a}\right)}{a^2 \times \frac{c}{a}+a b\left(\frac{-b}{a}\right)+b^2}\) \(=\frac{-2}{a}\)
UPSEE -2008
Complex Numbers and Quadratic Equation
118461
If \(\sin \alpha\) and \(\cos \alpha\) are the roots of the equation \(a x^2+b x+c=0\), then
1 \(\mathrm{a}^2-\mathrm{b}^2+2 \mathrm{ac}=0\)
2 \((a-c)^2=b^2+C^2\)
3 \(\mathrm{a}^2+\mathrm{b}^2-2 \mathrm{ac}=0\)
4 \(\mathrm{a}^2+\mathrm{b}^2+2 \mathrm{ac}=0\)
Explanation:
A It is given that, \(\sin \alpha\) and \(\cos \alpha\) are the roots of \(a x^2+b x+c=0\) \(\therefore \quad \text { Sum of the roots }=\sin \alpha+\cos \alpha=\frac{-b}{a}\) \(\text { and product of the roots }=\sin \alpha \cdot \cos \alpha=\frac{c}{a}\) \(\because \quad(\sin \alpha+\cos \alpha)^2=\sin ^2 \alpha+\cos ^2 \alpha+2 \sin \alpha \cos \alpha\) \(\because \quad\left(\frac{-b}{a}\right)^2=1+2\left(\frac{c}{a}\right)\) \(\Rightarrow \quad \frac{b^2}{a^2}=\frac{a+2 c}{a}\) \(\Rightarrow \quad b^2=a(a+2 c)\) \(\Rightarrow \quad b^2=a^2+2 a c\) \(\Rightarrow \quad a^2-b^2+2 a c=0\)
CG PET- 2015
Complex Numbers and Quadratic Equation
118462
If \(\alpha\) and \(\beta\) are the roots of the equation \(x^2-\) \(4 x+5=0\), then the quadratic equation whose roots are \(\alpha^2+\beta\) and \(\alpha+\beta^2\) is
1 \(x^2+10 x+34=0\)
2 \(x^2-10 x+34=0\)
3 \(x^2-10 x-34=0\)
4 \(x^2+10 x-34=0\)
Explanation:
B Given, \(x^2-4 x+5=0\) \(\alpha+\beta=4\) \(\alpha \cdot \beta=5\) Now, \(\left(\alpha^2+\beta\right)+\left(\alpha+\beta^2\right)\) \(=\alpha^2+\beta+\alpha+\beta^2\) \(=\left(\alpha^2+\beta^2\right)+(\alpha+\beta)\) \(=(\alpha+\beta)^2-2 \alpha \beta+(\alpha+\beta)\) \(=(4)^2-(2 \times 5)+4\) \(=16-10+4\) \(=10\) Now, \(\left(\alpha^2+\beta\right)\left(\alpha+\beta^2\right)=\alpha^3+\alpha^2 \beta^2+\beta \alpha+\beta^3\) \(=\alpha^3+\beta^3+\alpha \beta(\alpha \beta+1)\) \(=(\alpha+\beta)\left(\alpha^2+\beta^2-\alpha \beta\right)+\alpha \beta(\alpha \beta+1)\) \(=(\alpha+\beta)\left[(\alpha+\beta)^2-3 \alpha \beta\right]+\alpha \beta(\alpha \beta+1)\) \(=4(16-15)+5(5+1)\) \(=4+30=34\) So, the quadratic equation whose roots are \(\left(\alpha^2+\beta\right) \text { and }\left(\alpha+\beta^2\right) \text { is }\) \(x^2-\left(\alpha^2+\beta+\alpha+\beta^2\right) x+\left(\alpha^2+\beta\right)\left(\alpha+\beta^2\right)=0\) \(x^2-10 x+34=0\)
A Given, \(p, q\) and \(r\) are in \(A P\) Then, \(\mathrm{q}=\frac{\mathrm{p}+\mathrm{r}}{2}\) And equation \(\mathrm{px}^2+\mathrm{qx}+\mathrm{r}=0\) have real roots So, \(\mathrm{D} \geq 0\) \(\mathrm{q}^2-4 \mathrm{pr} \geq 0\) On putting the value \(\mathrm{q}\) in above equation. \(\left(\frac{\mathrm{p}+\mathrm{r}}{2}\right)^2-4 \mathrm{pr} \geq 0\) \(\mathrm{p}^2+\mathrm{r}^2+2 \mathrm{pr}-16 \mathrm{pr} \geq 0\) \(\mathrm{p}^2+\mathrm{r}^2-14 \mathrm{pr} \geq 0\) On adding both side \(48 \mathrm{p}^2\) \(49 p^2+r^2-14 p r \geq 48 p^2\) \((r-7 p)^2 \geq(4 \sqrt{3} p)^2\) \(\left(\frac{r}{p}-7\right)^2 \geq(4 \sqrt{3})^2\) \(\because \quad p \neq 0\) \(\therefore \quad \left|\frac{r}{p}-7\right| \geq 4 \sqrt{3}\)
UPSEE-2013
Complex Numbers and Quadratic Equation
118460
If \(\alpha, \beta\) are the roots of the equation \({a x^2+b x}^2\) \(\mathbf{c}=\mathbf{0}\), then \(\frac{\alpha}{\mathrm{a} \beta+\mathrm{b}}+\frac{\beta}{\mathrm{a} \alpha+\mathrm{b}}\) is equal to
1 \(\frac{2}{\mathrm{a}}\)
2 \(\frac{2}{\mathrm{~b}}\)
3 \(\frac{2}{\mathrm{c}}\)
4 \(-\frac{2}{a}\)
Explanation:
D Given \(\alpha, \beta\) are the roots of equation \(a x^2+b x+c=0\) \(\alpha+\beta=\frac{-b}{a}, \alpha \cdot \beta=\frac{c}{a}\) \((\alpha+\beta)^2=\left(\frac{-b}{a}\right)^2\) \(\alpha^2+\beta^2=\frac{b^2}{a^2}-\frac{2 c}{a}\) \(\alpha^2+\beta^2=\frac{b^2-2 a c}{a^2}\) \(=\frac{\alpha}{a \beta+b}+\frac{\beta}{a \alpha+b}\) \(=\frac{a \alpha^2+b \alpha+a \beta^2+b \beta}{\alpha^2 \alpha \beta+a b \beta+b a \alpha+b^2}\) \(=\frac{a\left(\alpha^2+\beta^2\right)+b(\alpha+\beta)}{a^2 \alpha \beta+a b(\alpha+\beta)+b^2}\) \(=\frac{\left(\frac{b^2-2 a c}{a^2}\right)+b\left(\frac{-b}{a}\right)}{a^2 \times \frac{c}{a}+a b\left(\frac{-b}{a}\right)+b^2}\) \(=\frac{-2}{a}\)
UPSEE -2008
Complex Numbers and Quadratic Equation
118461
If \(\sin \alpha\) and \(\cos \alpha\) are the roots of the equation \(a x^2+b x+c=0\), then
1 \(\mathrm{a}^2-\mathrm{b}^2+2 \mathrm{ac}=0\)
2 \((a-c)^2=b^2+C^2\)
3 \(\mathrm{a}^2+\mathrm{b}^2-2 \mathrm{ac}=0\)
4 \(\mathrm{a}^2+\mathrm{b}^2+2 \mathrm{ac}=0\)
Explanation:
A It is given that, \(\sin \alpha\) and \(\cos \alpha\) are the roots of \(a x^2+b x+c=0\) \(\therefore \quad \text { Sum of the roots }=\sin \alpha+\cos \alpha=\frac{-b}{a}\) \(\text { and product of the roots }=\sin \alpha \cdot \cos \alpha=\frac{c}{a}\) \(\because \quad(\sin \alpha+\cos \alpha)^2=\sin ^2 \alpha+\cos ^2 \alpha+2 \sin \alpha \cos \alpha\) \(\because \quad\left(\frac{-b}{a}\right)^2=1+2\left(\frac{c}{a}\right)\) \(\Rightarrow \quad \frac{b^2}{a^2}=\frac{a+2 c}{a}\) \(\Rightarrow \quad b^2=a(a+2 c)\) \(\Rightarrow \quad b^2=a^2+2 a c\) \(\Rightarrow \quad a^2-b^2+2 a c=0\)
CG PET- 2015
Complex Numbers and Quadratic Equation
118462
If \(\alpha\) and \(\beta\) are the roots of the equation \(x^2-\) \(4 x+5=0\), then the quadratic equation whose roots are \(\alpha^2+\beta\) and \(\alpha+\beta^2\) is
1 \(x^2+10 x+34=0\)
2 \(x^2-10 x+34=0\)
3 \(x^2-10 x-34=0\)
4 \(x^2+10 x-34=0\)
Explanation:
B Given, \(x^2-4 x+5=0\) \(\alpha+\beta=4\) \(\alpha \cdot \beta=5\) Now, \(\left(\alpha^2+\beta\right)+\left(\alpha+\beta^2\right)\) \(=\alpha^2+\beta+\alpha+\beta^2\) \(=\left(\alpha^2+\beta^2\right)+(\alpha+\beta)\) \(=(\alpha+\beta)^2-2 \alpha \beta+(\alpha+\beta)\) \(=(4)^2-(2 \times 5)+4\) \(=16-10+4\) \(=10\) Now, \(\left(\alpha^2+\beta\right)\left(\alpha+\beta^2\right)=\alpha^3+\alpha^2 \beta^2+\beta \alpha+\beta^3\) \(=\alpha^3+\beta^3+\alpha \beta(\alpha \beta+1)\) \(=(\alpha+\beta)\left(\alpha^2+\beta^2-\alpha \beta\right)+\alpha \beta(\alpha \beta+1)\) \(=(\alpha+\beta)\left[(\alpha+\beta)^2-3 \alpha \beta\right]+\alpha \beta(\alpha \beta+1)\) \(=4(16-15)+5(5+1)\) \(=4+30=34\) So, the quadratic equation whose roots are \(\left(\alpha^2+\beta\right) \text { and }\left(\alpha+\beta^2\right) \text { is }\) \(x^2-\left(\alpha^2+\beta+\alpha+\beta^2\right) x+\left(\alpha^2+\beta\right)\left(\alpha+\beta^2\right)=0\) \(x^2-10 x+34=0\)
A Given, \(p, q\) and \(r\) are in \(A P\) Then, \(\mathrm{q}=\frac{\mathrm{p}+\mathrm{r}}{2}\) And equation \(\mathrm{px}^2+\mathrm{qx}+\mathrm{r}=0\) have real roots So, \(\mathrm{D} \geq 0\) \(\mathrm{q}^2-4 \mathrm{pr} \geq 0\) On putting the value \(\mathrm{q}\) in above equation. \(\left(\frac{\mathrm{p}+\mathrm{r}}{2}\right)^2-4 \mathrm{pr} \geq 0\) \(\mathrm{p}^2+\mathrm{r}^2+2 \mathrm{pr}-16 \mathrm{pr} \geq 0\) \(\mathrm{p}^2+\mathrm{r}^2-14 \mathrm{pr} \geq 0\) On adding both side \(48 \mathrm{p}^2\) \(49 p^2+r^2-14 p r \geq 48 p^2\) \((r-7 p)^2 \geq(4 \sqrt{3} p)^2\) \(\left(\frac{r}{p}-7\right)^2 \geq(4 \sqrt{3})^2\) \(\because \quad p \neq 0\) \(\therefore \quad \left|\frac{r}{p}-7\right| \geq 4 \sqrt{3}\)
UPSEE-2013
Complex Numbers and Quadratic Equation
118460
If \(\alpha, \beta\) are the roots of the equation \({a x^2+b x}^2\) \(\mathbf{c}=\mathbf{0}\), then \(\frac{\alpha}{\mathrm{a} \beta+\mathrm{b}}+\frac{\beta}{\mathrm{a} \alpha+\mathrm{b}}\) is equal to
1 \(\frac{2}{\mathrm{a}}\)
2 \(\frac{2}{\mathrm{~b}}\)
3 \(\frac{2}{\mathrm{c}}\)
4 \(-\frac{2}{a}\)
Explanation:
D Given \(\alpha, \beta\) are the roots of equation \(a x^2+b x+c=0\) \(\alpha+\beta=\frac{-b}{a}, \alpha \cdot \beta=\frac{c}{a}\) \((\alpha+\beta)^2=\left(\frac{-b}{a}\right)^2\) \(\alpha^2+\beta^2=\frac{b^2}{a^2}-\frac{2 c}{a}\) \(\alpha^2+\beta^2=\frac{b^2-2 a c}{a^2}\) \(=\frac{\alpha}{a \beta+b}+\frac{\beta}{a \alpha+b}\) \(=\frac{a \alpha^2+b \alpha+a \beta^2+b \beta}{\alpha^2 \alpha \beta+a b \beta+b a \alpha+b^2}\) \(=\frac{a\left(\alpha^2+\beta^2\right)+b(\alpha+\beta)}{a^2 \alpha \beta+a b(\alpha+\beta)+b^2}\) \(=\frac{\left(\frac{b^2-2 a c}{a^2}\right)+b\left(\frac{-b}{a}\right)}{a^2 \times \frac{c}{a}+a b\left(\frac{-b}{a}\right)+b^2}\) \(=\frac{-2}{a}\)
UPSEE -2008
Complex Numbers and Quadratic Equation
118461
If \(\sin \alpha\) and \(\cos \alpha\) are the roots of the equation \(a x^2+b x+c=0\), then
1 \(\mathrm{a}^2-\mathrm{b}^2+2 \mathrm{ac}=0\)
2 \((a-c)^2=b^2+C^2\)
3 \(\mathrm{a}^2+\mathrm{b}^2-2 \mathrm{ac}=0\)
4 \(\mathrm{a}^2+\mathrm{b}^2+2 \mathrm{ac}=0\)
Explanation:
A It is given that, \(\sin \alpha\) and \(\cos \alpha\) are the roots of \(a x^2+b x+c=0\) \(\therefore \quad \text { Sum of the roots }=\sin \alpha+\cos \alpha=\frac{-b}{a}\) \(\text { and product of the roots }=\sin \alpha \cdot \cos \alpha=\frac{c}{a}\) \(\because \quad(\sin \alpha+\cos \alpha)^2=\sin ^2 \alpha+\cos ^2 \alpha+2 \sin \alpha \cos \alpha\) \(\because \quad\left(\frac{-b}{a}\right)^2=1+2\left(\frac{c}{a}\right)\) \(\Rightarrow \quad \frac{b^2}{a^2}=\frac{a+2 c}{a}\) \(\Rightarrow \quad b^2=a(a+2 c)\) \(\Rightarrow \quad b^2=a^2+2 a c\) \(\Rightarrow \quad a^2-b^2+2 a c=0\)
CG PET- 2015
Complex Numbers and Quadratic Equation
118462
If \(\alpha\) and \(\beta\) are the roots of the equation \(x^2-\) \(4 x+5=0\), then the quadratic equation whose roots are \(\alpha^2+\beta\) and \(\alpha+\beta^2\) is
1 \(x^2+10 x+34=0\)
2 \(x^2-10 x+34=0\)
3 \(x^2-10 x-34=0\)
4 \(x^2+10 x-34=0\)
Explanation:
B Given, \(x^2-4 x+5=0\) \(\alpha+\beta=4\) \(\alpha \cdot \beta=5\) Now, \(\left(\alpha^2+\beta\right)+\left(\alpha+\beta^2\right)\) \(=\alpha^2+\beta+\alpha+\beta^2\) \(=\left(\alpha^2+\beta^2\right)+(\alpha+\beta)\) \(=(\alpha+\beta)^2-2 \alpha \beta+(\alpha+\beta)\) \(=(4)^2-(2 \times 5)+4\) \(=16-10+4\) \(=10\) Now, \(\left(\alpha^2+\beta\right)\left(\alpha+\beta^2\right)=\alpha^3+\alpha^2 \beta^2+\beta \alpha+\beta^3\) \(=\alpha^3+\beta^3+\alpha \beta(\alpha \beta+1)\) \(=(\alpha+\beta)\left(\alpha^2+\beta^2-\alpha \beta\right)+\alpha \beta(\alpha \beta+1)\) \(=(\alpha+\beta)\left[(\alpha+\beta)^2-3 \alpha \beta\right]+\alpha \beta(\alpha \beta+1)\) \(=4(16-15)+5(5+1)\) \(=4+30=34\) So, the quadratic equation whose roots are \(\left(\alpha^2+\beta\right) \text { and }\left(\alpha+\beta^2\right) \text { is }\) \(x^2-\left(\alpha^2+\beta+\alpha+\beta^2\right) x+\left(\alpha^2+\beta\right)\left(\alpha+\beta^2\right)=0\) \(x^2-10 x+34=0\)