Complex Numbers and Quadratic Equations Test 50

Let, \(\alpha\) and \(\beta\) are the roots of \(x^{2}+6 x+8=0\), then

Sum of root \(= \alpha+\beta=-6\)

Arithmetic mean of root of equation \(=\frac{\alpha+\beta}{2}\) \(=\frac{-6}{2}\) \(=-3 \)

According to question, arithmetic mean is \(k\).

So, \(k=-3\)

Therefore, equation \(x^{2}+k x+2=0\) becomes

\(x^{2}-3 x+2=0 \) \(\Rightarrow(x-2)(x-1)=0 \)

\(\Rightarrow x=2\) and \(x=1\)

So, required roots of equation are \(1, 2\).

If \(\alpha ,\beta \) are the roots of quadratic equation \(a x^{2}+b x+c=0\), then

Sum of roots \(=\alpha+\beta=\frac{-b}{a}\)

Product of roots \(=\alpha \times \beta =\frac{c}{a}\)

Given,

\(3 a x^{2}+2 b x+c=0\)

Comparing with standard quadratic equation \(a x^{2}+b x+c=0\), we get

\(a=3 a, b=2 b\)

Roots are in the ratio \(3: 2\).

Let, roots are \(3 \alpha\) and \(2 \alpha\).

Sum of roots \(=3 \alpha+2 \alpha=\frac{-2 b }{3 a}\)

\(\Rightarrow 5 \alpha=\frac{-2 b }{ 3 a}\)

\(\Rightarrow \alpha=\frac{-2 b }{ 15 a}\)

\(\Rightarrow \alpha^{2}=\frac{4 b^{2}}{15 \times 15 a^{2}}\)

Now, product of roots \(=3 \alpha \times 2 \alpha=\frac{c }{ 3 a}\)

\(\therefore 6 \alpha^{2}=\frac{c }{ 3 a} \)

\(\Rightarrow 6\left(\frac{4 b^{2}}{15 \times 15 a^{2}}\right)=\frac{c}{3 a} \)

\(\Rightarrow 6 \times \frac{4 b^{2}\times 3a}{15 \times 5 a }=c \)

\(\Rightarrow 2 \times 4 b^{2}=25 ac\)

\(\Rightarrow 8 b ^{2}=25 ac\)

As we know,

\(i= \sqrt{-1}\)

\(i^2=-1\)

\(i^3=-i\)

\(i^4=1\)

\(i^{4n}=1\)

Given,

\((-1 \sqrt{-1})^{4 n+5}\) (where \(n\) is a positive integer)

\(=(-1\times i)^{4 n+5}\)

\(=(-i)^{4 n+5} \)

\(=(- i )^{4 n } \times(- i )^{5} \)

\(=(- i )^{4 n } \times (- i )^{4} \times(- i )^{1} \)

\(=1 \times 1 \times(- i ) \)

\(=- i\)

Given,

\(x^{2}+7 x-10=0\)

Comparing the given equation with \(a x^{2}+b x+c=0\), we get

\(a=1, b=7\) and \(c=-10\)

As we know,

Discriminant \(=b^{2}-4 a c\)

\(\therefore\)Discriminant \(=(7)^{2}-4 \times 1 \times (-10)\)

\(=49+40\)

\(=89\)

As we know,

If discriminant \((b^{2}-4 a c)>0\) then roots are real.

As we can see, discriminant \(=89 >0\)

\(\therefore\)The roots of the given equation are real.

Let,

\(\sqrt{(5+12 i )}= x + iy\)

On squaring both sides we get,

\(5+12 i=(x+i y)^{2}\)

\(\Rightarrow 5+12 i=x^2+(iy)^2+2(x)(iy)\)

\(\Rightarrow 5+12 i=x^2+i^2y^2+2xyi\)

As we know,

\(i^2=-1\)

\(\Rightarrow 5+12 i=x^2+(-1)y^2+2xyi\)

\(\Rightarrow 5+12 i=\left(x^{2}-y^{2}\right)+(2 x y)i\)

Comparing real and imaginary parts on both sides, we get

\(x^{2}-y^{2}=5 \) and \(2 x y=12\)

\(\therefore xy=6\)

\(\Rightarrow y=\frac{6}{x}\)

Now, \(x^{2}-y^{2})=5 \)

Putting value of \(y\) in above equation, we get

\(x^{2}-\left(\frac{6}{x}\right)^{2}=5\)

\(\Rightarrow x^{2}-\frac{36}{x^2}=5\)

\(\Rightarrow \frac{x^4-36}{x^2}=5\)

\(\Rightarrow x^4-36= 5x^2\)

\(\Rightarrow x^4-5x^2-36= 0\)

\(\Rightarrow x^4-9x^2+4x^2-36= 0\)

\(\Rightarrow x^2(x^2-9)+4(x^2-9)=0\)

\(\Rightarrow x^2-9=0\) or \(x^2+4=0\)

\(\Rightarrow x^{2}=9 \) or \(x^2=-4\)

\(\Rightarrow x=\pm 3\) (we know that \(x^2\) is always greater than zero so, we neglect \(x^2=−4\))

Now, \(x^2-y^{2}=5\)

\(\therefore 9- y^{2}=5 \)

\(\Rightarrow y^2=4\)

\(\Rightarrow y=\pm 2\)

So, \(\sqrt{(5+12 i )}=\pm(3+2 i )\)

As we know,

If \(z=x+iy\)...(1) be any complex number, then its modulus is given by,

\(|z|=\) \(\sqrt{ x ^{2}+ y ^{2}}\)

Let,

\(z=\frac{\sqrt{2}+ i }{\sqrt{2}- i } \)

Multiplying by \((\sqrt{2}+ i)\) in numerator and denominator, we get

\( z =\frac{\sqrt{2}+ i }{\sqrt{2}- i } \times \frac{\sqrt{2}+ i }{\sqrt{2}+ i } \)

\(\Rightarrow z =\frac{2+2 \sqrt{2} i + i ^{2}}{2- i ^{2}}\)

\(\Rightarrow z =\frac{2+2 \sqrt{2} i -1}{2-(-1)} \quad[\because i ^{2}=-1]\)

\(\Rightarrow z =\frac{1+2 \sqrt{2} i }{3}\)

\(\Rightarrow z=\frac{1}{3}+i \frac{2 \sqrt{2}}{3}\)...(2)

Comparing equation (1) and (2), we get

\(x =\frac{1}{3}\) and \(y =\frac{2 \sqrt{2}}{3}\)

As, \(|z|=\) \(\sqrt{ x ^{2}+ y ^{2}}\)

\(\therefore |z|=\sqrt{\left(\frac{1}{3}\right)^{2}+\left(\frac{2 \sqrt{2}}{3}\right)^{2}}\)

\(\Rightarrow | z |=\sqrt{\frac{1}{9}+\frac{8}{9}} \)

\(\Rightarrow| z |=1\)

So, the modulus of \(\frac{\sqrt{2}+ i }{\sqrt{2}- i }\) is \(1\).

Given,

\(a^{2} x^{2}-3 a b x+2 b^{2}=0\)

\(\Rightarrow a^{2} x^{2}-3 a b x+2 b^{2}=0 \)

\(\Rightarrow a^{2} x^{2}-2 a b x-a b x+2 b^{2}=0 \)

\(\Rightarrow a x(a x-2 b)-b(a x-2 b)=0 \)

\(\Rightarrow(a x-2 b)(a x-b)=0 \)

\(\Rightarrow(a x-2 b)=0\) or \((a x-b)=0 \)

\(\Rightarrow x=\frac{2 b }{ a}\) or \( x=\frac{b }{ a}\)

\(\therefore\) The roots are \(\frac{2 b }{ a}\) and \(\frac{b }{ a}\).

Given,

\(z =\frac{3-2 i \sin \theta}{2- i \sin \theta}\)

Multiplying the numerator and denominator by \((2+i \sin \theta )\), we get

\(z =\frac{3-2 i\sin \theta}{2-i\sin \theta} \times \frac{2+i \sin \theta}{2+i \sin \theta} \)

\(\Rightarrow z=\frac{6+3i \sin \theta-4i \sin \theta-2 i^{2} \sin ^{2} \theta}{(2)^2-(i \sin \theta)^2}\)

\(\Rightarrow z=\frac{6-i \sin \theta+2 i^{2}\sin ^{2} \theta}{4- i ^{2} \sin ^{2} \theta} \)

\(\Rightarrow z=\frac{6-i \sin \theta+2 \times (-1) \sin ^{2} \theta}{4-(-1)\sin ^{2} \theta}\quad [\because i^2=-1] \)

\(\Rightarrow z=\frac{6-i \sin \theta+2 \times (-1) \sin ^{2} \theta}{4-(-1)\sin ^{2} \theta}\)

\(\Rightarrow z =\frac{6-2 \sin ^{2} \theta}{4+\sin ^{2} \theta}+i\left(\frac{- \sin \theta}{4+\sin ^{2} \theta}\right)\)

Imaginary part of z,

\( \operatorname{Im}( z )=\frac{-\sin \theta}{4+\sin ^{2} \theta}\)

As we know,

For \(z\) to be purely real, \(\operatorname{Im}(z)=0\)

\(\therefore \frac{-\sin \theta}{4+\sin ^{2} \theta}=0 \)

\(\Rightarrow \sin \theta=0\)

\(\Rightarrow \theta= n \pi\), where \(n\) belongs to an integer

As we know,

If \(\alpha\) and \(\beta\) are the roots of the equation, \(a x^{2}+b x+c=0\)

Sum of roots \((\alpha+\beta)=\frac{-b}{a}\)

Product of roots \((\alpha \beta)=\frac{c}{a}\)

Given,

\(x^{2}-6 x+3=0\)

\(\alpha+\beta=-\left(\frac{-6}{1}\right)=6\)

\( \alpha \beta=\frac{3}{1}=3 \)

Now,

\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^{2}+\beta^{2}}{\alpha \cdot \beta}\)...(1)

\((\alpha+\beta)^{2}=\alpha^{2}+\beta^{2}+2 \alpha \beta\)

\(\Rightarrow(6)^{2}=\alpha^{2}+\beta^{2}+2 \times 3 \)

\(\Rightarrow \alpha^{2}+\beta^{2}=36-6=30\)

Putting the value of \((\alpha^{2}+\beta^{2})\) in equation (1), we get

\(\frac{\alpha^{2}+\beta^{2}}{\alpha \cdot \beta}=\frac{30}{3}\)

\(\therefore \frac{\alpha^{2}+\beta^{2}}{\alpha \cdot \beta}=10\)

So, the value of \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^{2}+\beta^{2}}{\alpha \cdot \beta}=10\)

As we know,

The sum of the roots of a quadratic equation \(a x^{2}+b x+c\) \(=0, a \neq 0\) is given by,

\(\frac{ \text{coefficient of} x }{\text{ coefficient of }x^{2}}=-(\frac{b }{ a})\)

Let us verify the options,

(A) \(2 x^{2}-3 x+6=0\)

Sum of the roots \(=\frac{-b }{ a}=-(\frac{-3 }{ 2})=\frac{3 }{ 2}\)

(B) \(-x^{2}+3 x-3=0\)

Sum of the roots \(=\frac{-b }{ a}=-(\frac{3 }{-1})=3\)

(C) \(\sqrt{2} x^{2}-\frac{3 }{ \sqrt{2 x}}+1=0\)

\(2 x^{2}-3 x+\sqrt{2}=0\)

Sum of the roots \(=\frac{-b }{ a}=-(\frac{-3 }{ 2})=\frac{3 }{ 2}\)

(D) \(3 x^{2}-3 x+3=0\)

Sum of the roots \(=\frac{-b }{ a}=-(\frac{-3 }{ 3})=1\)

So, the equation which has the sum of its roots as 3 is –x² + 3x – 3 = 0.