Quadratic Equations - Class 10 Maths - Extra Questions

We have $$ x = \dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3}+\sqrt{2}} = \dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3}+\sqrt{2}} \times \dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3}- \sqrt{2}} = 5 – 2\sqrt{6}$$

$$3{ x }^{ 2 }-5x+2=0\\ { x }^{ 2 }-\dfrac { 5 }{ 3 } x+\dfrac { 2 }{ 3 } =0\\ { x }^{ 2 }-\dfrac { 5 }{ 3 } x+\dfrac { 25 }{ 36 } -\dfrac { 25 }{ 36 } +\dfrac { 2 }{ 3 } =0\\ { \left( x-\dfrac { 5 }{ 3 }  \right)  }^{ 2 }-\dfrac { 1 }{ 36 } =0\\ { \left( x-\dfrac { 5 }{ 3 }  \right)  }^{ 2 }=\dfrac { 1 }{ 36 } \\ \left( x-\dfrac { 5 }{ 3 }  \right) =\pm \dfrac { 1 }{ 6 } \\ x=\dfrac { 5 }{ 3 } \pm \dfrac { 1 }{ 6 } \\ x=\dfrac { 11 }{ 6 } ,\dfrac { 9 }{ 6 } $$

$$5{ x }^{ 2 }-6x-2\\ { x }^{ 2 }-\dfrac { 6 }{ 5 } x-\dfrac { 2 }{ 5 } =0\\ { x }^{ 2 }-\dfrac { 6 }{ 5 } x+\dfrac { 36 }{ 100 } -\dfrac { 36 }{ 100 } -\dfrac { 2 }{ 5 } =0\\ { \left( x-\dfrac { 6 }{ 5 }  \right)  }^{ 2 }=\dfrac { 76 }{ 100 } \\ x-\dfrac { 6 }{ 5 } =\pm \dfrac { \sqrt { 76 }  }{ 10 } \\ x=\dfrac { 6 }{ 5 } \pm \dfrac { \sqrt { 76 }  }{ 10 } $$

$$2{ x }^{ 2 }+x+4=0\\ { x }^{ 2 }+\dfrac { x }{ 2 } +2=0\\ { x }^{ 2 }+\dfrac { x }{ 2 } +\dfrac { 1 }{ 16 } -\dfrac { 1 }{ 16 } +2=0\\ { \left( x+\dfrac { 1 }{ 4 }  \right)  }^{ 2 }+\dfrac { 31 }{ 16 } =0\\ { \left( x+\dfrac { 1 }{ 4 }  \right)  }=\pm \dfrac { \sqrt { 31 } i }{ 4 } \\ x=-\dfrac { 1 }{ 4 } \pm \dfrac { \sqrt { 31 } i }{ 4 } $$

$$4{ x }^{ 2 }+3x+5=0\\ { x }^{ 2 }+\dfrac { 3 }{ 4 } x+\dfrac { 5 }{ 4 } =0\\ { x }^{ 2 }+\dfrac { 3 }{ 4 } x+\dfrac { 9 }{ 16 } -\dfrac { 9 }{ 16 } +\dfrac { 5 }{ 4 } =0\\ { \left( x+\dfrac { 3 }{ 4 }  \right)  }^{ 2 }=-\dfrac { 11 }{ 16 } \\ \left( x+\dfrac { 3 }{ 4 }  \right) =\pm \dfrac { \sqrt { 11 } i }{ 4 } \\ x=-\dfrac { 3 }{ 4 } \pm \dfrac { \sqrt { 11 } i }{ 4 } \\ $$

Let ,

  $$ y={{x}^{2}}+x+12=0 $$

It cut x-axis then, at x-axis $$y=0$$

$${{x}^{2}}+x+12=0$$

$$ x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$$

$$ x = \dfrac{-(1) \pm \sqrt{(1)^2-4*1*12}}{2*1}$$

$$ x = \dfrac{-1 \pm \sqrt{-47}}{2} $$

Hence required coordinates.

  $$ \left( x,y \right)=\left(\dfrac{-1+\sqrt{-47}}{2},\ 0\right) $$

 $$ \left( x,y \right)=\left(\dfrac{-1-\sqrt{-47}}{2},\ 0\right) $$

$$x^2+2x-5=0$$

$$x^2+2x+1-1-5=0$$

$$(x+1)^2=6$$

Given that,

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

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

 $$ \Rightarrow 2{{x}^{2}}-2{{x}^{2}}+5x-5x-2x+12=0 $$

 $$ \Rightarrow -2x+12=0 $$

 $$ \Rightarrow -2x=-12 $$

 $$ \Rightarrow x=6 $$

We have,

$$ 2{{y}^{2}}-y-1=0$$

$$ 2{{y}^{2}}-2y+y-1=0 $$

$$ 2y\left( y-1 \right)+1\left( y-1 \right)=0 $$

$$ \left( y-1 \right)\left( 2y+1 \right)=0 $$

$$ \Rightarrow y=1,-\dfrac{1}{2} $$

Hence, this is the answer.

Given that quadetric equation is $$\sqrt{5}x^2+x+\sqrt{5}=0$$

[If given qudratics equation as $$ ax^2+bx+c=0$$, then $$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$$]

Then we compare given quadratic equation with standrad form, we get $$a=\sqrt5, b=1, c=\sqrt5$$, then

$$x=\dfrac{-1\pm\sqrt{1-\left(4\times\sqrt{5}\times\sqrt{5}\right)}}{2\sqrt{5}}$$

$$x=\dfrac{-1\pm\sqrt{-19}}{2\sqrt5}$$

$$x=\dfrac{-1+\sqrt{19}i}{2\sqrt5} $$and $$ x=\dfrac{-1-\sqrt{19}i}{2\sqrt5}$$

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

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

$$\\(x-2)^2=3$$

$$\\(x-2)=+\sqrt3\>or\>-\sqrt3$$

$$\\x=2+\sqrt3\>or\>x=2-\sqrt3$$

We have,

$$p=99$$

 Since,

$$ \Rightarrow {{p}^{3}}+3{{p}^{2}}+3p+1-1 $$

$$ \Rightarrow {{\left( p+1 \right)}^{3}}-1 $$

$$ \Rightarrow {{\left( 99+1 \right)}^{3}}-1 $$

$$ \Rightarrow {{\left( 100 \right)}^{3}}-1 $$

$$ \Rightarrow 1000000-1 $$

$$ \Rightarrow 999999 $$

 Hence, this is the answer.

Consider given equation,

$$ {{x}^{2}}-12x+36=0 $$

$$ {{x}^{2}}-6x-6x+36=0 $$

$$ x\left( x-6 \right)-6\left( x-6 \right)=0 $$

$$ \left( x-6 \right)\left( x-6 \right)=0 $$

Hence, $$x=6,6$$

Hence, this is the answer.

$${\textbf{Step -1: Simplify the given equation.}}$$

                $${\text{Consider,}}$$

                $${\left( {3x - 1} \right)^2} = 5\left( {x + 8} \right)$$

                $$ \Rightarrow {\left( {3x} \right)^2} - 2\left( {3x} \right)\left( 1 \right) + {\left( 1 \right)^2} = 5x + 40$$   $$\left( {{\textbf{Using formula: }}{{\left(\mathbf{ {a + b}} \right)}^\mathbf2} = \mathbf{{a^2} + 2ab + {b^2}}} \right)$$

                $$ \Rightarrow 9{x^2} - 6x + 1 = 5x + 40$$

                $$ \Rightarrow 9{x^2} - 6x - 5x - 40 + 1 = 0$$ 

                $$ \Rightarrow 9{x^2} - 11x - 39 = 0$$ $$ \ldots \left( 1 \right)$$

$${\textbf{Step -2: Compare equation }}\left(\mathbf 1 \right)$$ $${\textbf{with standard form of quadratic equation.}}$$

                $${\text{A quadratic equation is an equation of the form }}$$ $$a{x^2} + bx + c = 0$$ 

                $${\text{Where}}$$ $$a,b,c$$ $${\text{are knowns numbers}}$$ $${\text{and}}$$ $$a \ne 0$$ $${\text{and}}$$ $$x$$ $${\text{is unknown variable}}{\text{.}}$$

                $${\text{Here, }}$$ $${\left( {3x - 1} \right)^2} = 5\left( {x + 8} \right)$$ $${\text{can be written as}}$$ $$9{x^2} - 11x - 39 = 0.$$ 

                $${\text{Which is a quadratic equation}}{\text{.}}$$

$${\textbf{ Hence, }}$$ $$\mathbf{{\left( {3x - 1} \right)^2} = 5\left( {x + 8} \right)}$$ $${\textbf{is a quadratic equation.}}$$

$$\textbf{Step-1: Let}$$ $$x=\sqrt{a+\sqrt{a+\sqrt{a+...........\infty }}}$$

                 $$\text{On squaring both sides we get,}$$

                 $${{x}^{2}}=a+\sqrt{a+\sqrt{a+\sqrt{a+............\infty }}}$$

                 $$\Rightarrow {{x}^{2}}=a+x$$

                 $$\Rightarrow {{x}^{2}}-x-a=0$$

$$\textbf{Step-2: Solving the quadratic equation}$$

                 $$\Rightarrow x=\dfrac{1\pm \sqrt{1+4a}}{2}$$               $$\mathbf{[\because x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}]}$$

$$\textbf{ Therefore, the value of}$$ $$\mathbf{\sqrt{a+\sqrt{a+\sqrt{a+...........\infty }}}}$$ $$\textbf{ is }$$ $$\mathbf{{=\dfrac{1\pm \sqrt{1+4a}}{2}}}$$

The given mathematical statement is $$\left( {a + 4} \right)\left( {a + 2} \right) = {a^2} + 8$$.

Simplifying the LHS of $$\left( {a + 4} \right)\left( {a + 2} \right) = {a^2} + 8$$

$$\left( {a + 4} \right)\left( {a + 2} \right) = {a^2} + 2a + 4a + 8$$

$$ = {a^2} + 6a + 8$$

$$ \ne {\rm{RHS}}$$

Therefore, the given mathematical statement is not correct.

Consider the given equation,

$$2{{\cos }^{2}}\theta -\sqrt{3}\sin \theta +1=0$$

$$ 2\left( 1-{{\sin }^{2}}\theta  \right)-\sqrt{3}\sin \theta +1=0 $$   $$[\because \cos ^2 \theta=1-\sin^2 \theta]$$

$$ 2-2{{\sin }^{2}}\theta -\sqrt{3}\sin \theta +1=0 $$

$$ 2{{\sin }^{2}}\theta +\sqrt{3}\sin \theta -3=0 $$

$$ \sin \theta =\dfrac{-\sqrt{3}\pm \sqrt{3-4\times 2\times -3}}{2\times 2} $$  (use quadratic formula)

$$ \sin \theta =\dfrac{-\sqrt{3}\pm \sqrt{3+24}}{2\times 2} $$

$$ \sin \theta =\dfrac{-\sqrt{3}\pm 3\sqrt{3}}{4} $$

$$\therefore \sin \theta =\dfrac{\sqrt{3}}{2},-\sqrt{3}$$

Given that,

$$p{{x}^{2}}+qx+r=0$$    $$p,q,r$$$$\in R$$

We know that if a quadratic equation has no real root then,

$$ D<0 $$

$$ {{B}^{2}}-4AC=0 $$

$$ \Rightarrow {{q}^{2}}-4pr<0 $$

$$\Rightarrow {{q}^{2}}<4pr$$

$$((\frac{a}{4})-(\frac{b}{2})+1)^2\\=(\frac{a}{4})^2+(\frac{-b}{2})^2+1^2+2\times(\frac{a}{4})(\frac{-b}{2})+2(\frac{-b}{2})\times1+2\times1\times(\frac{a}{4})\\=(\frac{a^2}{16})+(\frac{b^2}{4})+1-(\frac{ab}{2})-b+(\frac{a}{2})$$

Let the roots of the required equation be $$M$$ and $$N$$

let the roots of the equation $$2x²+2(p+q)x+p²+q²=0$$ be $$a$$ and $$b$$

$$\Rightarrow a + b = -(p+q)$$

$$\Rightarrow ab = \dfrac{(p^2 + q^2)}{ 2}$$

$$\Rightarrow (a+b)^2 = (p+q)^2$$

$$\Rightarrow (a-b)^2 = (a+b)^2 - 4ab$$

$$\Rightarrow (a-b)^2 = -(p - q)^2$$

we wanted the values of square of sum of the roots and square of difference of the roots

Now$$\Rightarrow M = (a+b)^2 = (p+q)^2$$ and

$$\Rightarrow N = (a-b)^2 = -(p - q)^2$$

$$\Rightarrow M + N = 4pq$$

$$\Rightarrow MN = (p+q)^2 [-(p - q)^2]$$

$$\Rightarrow MN= -(p^2 - q^2)^2$$

hence the required equation is

$$\Rightarrow x^2 - (4pq)x - (p^2 - q^2)^2 = 0$$

Given: $$27-125a^2-135a+225a^2$$

$$\Rightarrow 100a^2-135a+27=0$$

Here, $$a=100, b = -135, c=27$$

$$a =\dfrac{-(b)\pm \sqrt{(b)^2-4\times a\times c}}{2\times a}$$

Using quadratic formula to find roots we get,

$$a =\dfrac{-(-135)\pm \sqrt{(-135)^2-4\times 100\times 27}}{2\times 100}$$

$$=\dfrac{135\pm \sqrt{18225-10800}}{200}$$

$$ = \dfrac{135\pm \sqrt{7425}}{200}$$

$$=\dfrac{135\pm 5\sqrt{297}}{200}$$

$$=\dfrac{27\pm \sqrt{297}}{40}$$

Consider the given equation,

$$ 4{{x}^{2}}+12x+8=0 $$

$$ {{x}^{2}}+3x+2=0 $$     ..... (1)

Add and subtract $${{\left( \dfrac{3}{2} \right)}^{2}}$$ in equation (1), we get 

$$ {{x}^{2}}+3x+2+{{\left( \dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}}=0 $$

$$ {{\left( x+\dfrac{3}{2} \right)}^{2}}+2-\dfrac{9}{4}=0 $$

$$ {{\left( x+\dfrac{3}{2} \right)}^{2}}-\dfrac{1}{4}=0 $$

Hence, need to add $$\dfrac{1}{4}$$ so that $$ {{x}^{2}}+3x+2=0 $$  become a perfect square.