Question

Find the zeroes of the following quadratic polynomial and verify the relationship between the zeroes and the coefficients.
1)$2y\cos x=y^2+2c$
2) $4x^2-4x+1$
3) $6x^2-3-7x$
4) $4u^2+8u$
5) $t^2-15$
6) $3x^2-x-4$

Answer 1

Part (1)

$2y\cos x=y^2+2c$

$\Rightarrow y^2+2c-2y\cos x=0$

$\Rightarrow y^2-2y\cos x+2c=0$

Comparing this equation $A{y}^2+By+C=0$

Then,$A=1,B=-2\cos x,C=2c$

Using, quadratic equation and we get,

$y=\frac{-B\pm \sqrt{B^2-4AC}}{2A}$

$y=\frac{-(-2\cos x)\pm \sqrt{4{\cos }^{2}x-4\times 1\times 2c}}{2}$

$y=\frac{2(\cos x\pm \sqrt{{\cos }^{2}x-2c})}{2}$

$y=\cos x\pm \sqrt{{\cos }^{2}x-2c}$

$y=\cos x+\sqrt{{\cos }^{2}x-2c}\&y=\cos x-\sqrt{{\cos }^{2}x-2c}$

Verify that,

Comparing equation (1) $a{y}^2+by+c=0$

Then,$A=1,B=-2\cos x,C=2c$

Sums of zeros$=-\frac{cofficentofx}{cofficentof{x}^2}$

$\cos x+\sqrt{{\cos }^{2}x-2c}+\cos x-\sqrt{{\cos }^{2}x-2c}=-\frac{-2\cos x}{1}$

$2\cos x=2\cos x$

Product of zeros=$=\frac{constantterm}{cofficentofx}$

$(\cos x+\sqrt{{\cos }^{2}x-2c})\times (\cos x-\sqrt{{\cos }^{2}x-2c})=\frac{2c}{1}$

$\Rightarrow {\cos }^{2}x-{\left(\sqrt{{\cos }^{2}x-2c}\right)}^2=\frac{2c}{1}$

$\Rightarrow {\cos }^{2}x-{\cos }^{2}x+2c=2c$

$\Rightarrow 2c=2c$

Part (2)

$4x^2-4x+1=0$

$\Rightarrow 4x^2-(2+2)x+1=0$

$\Rightarrow 4x^2-2x-2x+1=0$

$\Rightarrow 2x(2x-1)-1(2x-1)=0$

$\Rightarrow (2x-1)(2x-1)=0$

$\Rightarrow 2x=1$ $2x=1$

$\Rightarrow x=\frac{1}{2},\frac{1}{2}$ ans.

Verify that,

Sums of zeros$=-\frac{cofficentofx}{cofficentof{x}^2}$

$\frac{1}{2}+\frac{1}{2}=-\frac{-4}{4}$

$1=1$

Product of zeros$=\frac{constantterm}{cofficentofx}$

$\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$

$\frac{1}{4}=\frac{1}{4}$

Part (3)

$6x^2-3-7x=0$

$\Rightarrow 6x^2-7x-3=0$

$\Rightarrow 6x^2-(9-2)x-3=0$

$\Rightarrow 6x^2-9x+2x-3=0$

$\Rightarrow 3x(2x-3)+1(2x-3)=0$

$\Rightarrow (2x-3)(3x+1)=0$

$If,$ $2x-3=0$ $3x+1=0$

$x=\frac{3}{2}$,$\frac{-1}{3}$

Verify that,

Sums of zeros$=-\frac{cofficentofx}{cofficentof{x}^2}$

$\frac{3}{2}+\left(\frac{-1}{3}\right)=-\frac{-7}{6}$

$\frac{7}{6}=\frac{7}{6}$

Product of zeros$=\frac{constantterm}{cofficentofx}$

$\frac{3}{2}\times -\frac{1}{3}=\frac{-3}{6}$

$\frac{-1}{2}=\frac{-1}{2}$

Part (4)

$4u^2+8u=0$

$\Rightarrow 4u(u+2)=0$

$\Rightarrow If,4u=0$ $u=0$

If,$u+2=0$ ……$u=-2$

Verify that,

Sums of zeros$=-\frac{cofficentofx}{cofficentof{x}^2}$

$0+(-2)=-\frac{8}{4}$

$-2=-2$

Product of zeros$=\frac{constantterm}{cofficentofx}$

$0\times (-2)=\frac{0}{4}$

$0=0$

Part (5)

$t^2-15=0$

$\Rightarrow t^2=15$

$\Rightarrow t=\pm \sqrt{15}$

Verify that,

Sums of zeros$=-\frac{cofficentofx}{cofficentof{x}^2}$

$\sqrt{15}-\sqrt{15}=-\frac{0}{1}$

$0=0$

Product of zeros$=\frac{constantterm}{cofficentofx}$

$\sqrt{15}\times -\sqrt{15}=-\frac{15}{1}$

$-15=-15$

Part (6)

$3x^2-x-4=0$

$\Rightarrow 3x^2-(4-3)x-4=0$

$\Rightarrow 3x^2-4x+3x-4=0$

$\Rightarrow x(3x-4)+1(3x-4)=0$

$\Rightarrow (3x-4)(x+1)=0$

If,$3x-4=0$ $x=\frac{4}{3}$

If,$x+1=0$ $x=-1$

Verify that,

Sums of zeros$=-\frac{cofficentofx}{cofficentof{x}^2}$

$\frac{4}{3}+(-1)=-\frac{-1}{3}$

$\frac{1}{3}=\frac{1}{3}$

Product of zeros$=\frac{constantterm}{cofficentofx}$

$\frac{4}{3}\times (-1)=\frac{-4}{3}$

$=-\frac{4}{3}$