Complex Numbers and Quadratic Equations Test 48

Let \(\alpha\) and \(\beta\) be roots of the equation \(x^{2}-n x+m=0\).

Given,

The roots of the equation \(x^{2}-n x+m=0\) differ by 1.

\(\Rightarrow \alpha-\beta=1 \)

\(\Rightarrow \alpha=1+\beta\)

Now,

The sum of the roots \(=\alpha+\beta=-(-n)=n\)

\(\therefore \beta+1+\beta=n \)

\(\Rightarrow 2 \beta+1=n \)

\(\Rightarrow \beta=\frac{(n-1)}{2}\)

The product of the roots \(=\alpha \beta=m\)

\(\therefore \beta(1+\beta)=m \)

\(\Rightarrow\left(\frac{n-1}{2}\right)\left(1+\frac{n-1}{2}\right)=m \)

\(\Rightarrow\left(\frac{n-1}{2}\right)\left(\frac{2+n-1}{2}\right)=m \)

\(\Rightarrow(n-1)(n+1)=4 m \)

\(\Rightarrow n^{2}-1=4 m \)

\(\Rightarrow n^{2}-4 m-1=0\)

Given,

Quadratic equation \(7 x^{2}-28 x+21\)

\( 7 x^{2}-28 x+21=0 \)

\(\Rightarrow x^{2}-4 x+3=0 \)

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

\(\Rightarrow(x-1)(x-3)=0 \)

\(\Rightarrow x=1,3\)

Now,

\( \frac{1}{\alpha}+\frac{1}{\beta}=\frac{k}{\alpha \beta} \)

\(\Rightarrow \frac{1}{1}+\frac{1}{3}=\frac{k}{1 \times 3} \)

\(\Rightarrow \frac{4}{3}=\frac{k}{3}\)

\(\Rightarrow k=4\)

So, the value of \(k\) is \(4\).

A quadratic equation ax² + bx + c = 0 has no real roots, if b² – 4ac < 0. That means, the quadratic equation contains imaginary roots.

Given,

Complex number \(z=\frac{1-i}{1+i}\)

Multiplyingby \((1 - i)\) inthe numerator and denominator, we get

\( z=\frac{1-i}{1+i} \times \frac{1-i}{1-i} \)

\(\Rightarrow z=\frac{(1-i)(1-i)}{1-i^2} \)

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

\(\Rightarrow z=\frac{1-1-2 i}{2} \)

\(\Rightarrow z=\frac{0-2 i}{2} \)

\(\Rightarrow z=0-1i\)

\(\therefore\) Re(z) \(=0\) and Im(z)\(=-1\)

Given,

\(\alpha\) and \(\beta\) are the roots of the equation \(x^{2}-q(1+x)-r=0\)

\(\Rightarrow x^{2}-q-q x-r=0 \)

\(\Rightarrow x^{2}-q x-(q+r)=0\)

Sum of roots \(=\alpha+\beta=q\)

Product of roots \(=\alpha \beta=-(q+r)=-q-r\)

Now,

\((1+\alpha)(1+\beta)=1+\alpha+\beta+\alpha \beta \)

\(=1+q-q-r \)

\(=1-r\)

As we know,

If discriminant \( b ^{2}-4 ac <0\), then it has no real roots.

Given,

\(2 x^{2}-\sqrt{5 x}+1=0\)

Comparing with the standard form of a quadratic equation, we get

\(a=2, b=-\sqrt{5}, c=1\)

Now,

\( b ^{2}-4 ac =(-\sqrt{5})^{2}-4\times (2) \times (1) \)

\(=5-8 \)

\(=-3\)

\( b ^{2}-4 ac <0\)

Therefore, the given equation has no real roots.

Let,

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

\(\Rightarrow z=\frac{2+4 i-i-2 i^{2}}{6-9 i+2 i-3 i^{2}}\)

\(\Rightarrow z=\frac{2+4 i-i+2}{6-9 i+2 i+3} \quad[\because i^2=-1]\)

\(\Rightarrow z=\frac{4+3 i}{9-7 i} \)

Multiplying by \((9+7i)\) in numerator and denominator, we get

\( z=\frac{4+3 i}{9-7 i} \times \frac{9+7 i}{9+7 i} \)

\(\Rightarrow z=\frac{36+28 i+27 i+21 i^{2}}{81-49 i^{2}}\)

\(\Rightarrow z=\frac{36+28 i+27 i-21}{81+49} \quad[\because i^2=-1]\)

\(\Rightarrow z=\frac{15+55 i }{130} \)

\(\Rightarrow z=\frac{15}{130}+i \frac{55}{130}\)

As we know,

Conjugate of \(z =\bar{ z }= x - iy\)

\(\therefore \bar{ z }=\frac{15}{130}- i \frac{55}{130}\)

Given,

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

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

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

Now,

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

\(=a^{2} \times\left(\frac{c}{a}\right)+a b(\alpha+\beta)+b^{2}\)

\(=a c+a b \times\left(\frac{-b}{a}\right)+b^{2}\)

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

\(=a c\)

As we know,

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

\( i ^{2}=-1\)

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

Given,

\( i^{1325}\)

\(=i^{(1324+1)}\)

\(=\left(i^{4}\right)^{331} \times i\)

\(=1 \times i \quad\left[\because i^{4 n}=1\right]\)

\(=i\)

\(\therefore i^{1325}=i\)

For quadratic 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, \((5+\sqrt{2}) x^{2}-(4+\sqrt{5}) x+(8+2 \sqrt{5})=0\)

\(\therefore  \alpha \beta=\frac{(8+2 \sqrt{5}) }{(5+\sqrt{2})}\) and \(\alpha+\beta=\frac{(4+\sqrt{5}) }{(5+\sqrt{2})}\)

Now, \(\frac{2 \alpha \beta }{(a+\beta)}= \frac{\frac{2(8+2 \sqrt{5}) }{(5+\sqrt{2})} }{\frac{(4+\sqrt{5}) }{(5+\sqrt{2})}}\)

\(=\frac{2(8+2 \sqrt{5}) }{(5+\sqrt{2})} \times \frac{(5+\sqrt{2})}{(4+\sqrt{5})} \)

\(=\frac{2(8+2 \sqrt{5})}{(4+\sqrt{5})} \) \(=\frac{2×2(4+ \sqrt{5})}{(4+\sqrt{5})} \) \(=4\)