हिंदी

√ 2 X 2 + X + √ 2 = 0 - Mathematics

Advertisements
Advertisements

प्रश्न

\[\sqrt{2} x^2 + x + \sqrt{2} = 0\]

उत्तर

Given:

\[\sqrt{2} x^2 + x + \sqrt{2} = 0\]

Comparing the given equation with the general form of the quadratic equation 

\[a x^2 + bx + c = 0\], we get
\[a = \sqrt{2}, b = 1\] and \[c = \sqrt{2}\] .
Substituting these values in
\[\alpha = \frac{- b + \sqrt{b^2 - 4ac}}{2a}\] and \[\beta = \frac{- b - \sqrt{b^2 - 4ac}}{2a}\] , we get:
\[\alpha = \frac{- 1 + \sqrt{1 - 4 \times \sqrt{2} \times \sqrt{2}}}{2\sqrt{2}}\] and   \[\beta = \frac{- 1 - \sqrt{1 - 4 \times \sqrt{2} \times \sqrt{2}}}{2\sqrt{2}}\]
\[\Rightarrow \alpha = \frac{- 1 + \sqrt{- 7}}{2\sqrt{2}}\] and \[\beta = \frac{- 1 - \sqrt{- 7}}{2\sqrt{2}}\]
\[\Rightarrow \alpha = \frac{- 1 + i\sqrt{7}}{2\sqrt{2}}\] and  \[\beta = \frac{- 1 - i\sqrt{7}}{2\sqrt{2}}\]
Hence, the roots of the equation are \[\frac{- 1 \pm i\sqrt{7}}{2\sqrt{2}}\].    
shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 14: Quadratic Equations - Exercise 14.1 [पृष्ठ ६]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
अध्याय 14 Quadratic Equations
Exercise 14.1 | Q 21 | पृष्ठ ६

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Solve the equation `x^2 + x + 1/sqrt2 = 0`


Solve the equation  `x^2 + x/sqrt2 + 1 = 0`


Solve the equation   `x^2 -2x + 3/2 = 0`  


If z1 = 2 – i,  z2 = 1 + i, find `|(z_1 + z_2 + 1)/(z_1 - z_2 + 1)|`


9x2 + 4 = 0


x2 + 2x + 5 = 0


4x2 − 12x + 25 = 0


x2 + x + 1 = 0


\[5 x^2 - 6x + 2 = 0\]


\[17 x^2 + 28x + 12 = 0\]


\[21 x^2 - 28x + 10 = 0\]


\[\sqrt{3} x^2 - \sqrt{2}x + 3\sqrt{3} = 0\]


\[x^2 + x + \frac{1}{\sqrt{2}} = 0\]


\[x^2 + \frac{x}{\sqrt{2}} + 1 = 0\]


\[\sqrt{5} x^2 + x + \sqrt{5} = 0\]


\[3 x^2 - 4x + \frac{20}{3} = 0\]


Solving the following quadratic equation by factorization method:

\[6 x^2 - 17ix - 12 = 0\]

 

Solve the following quadratic equation:

\[\left( 2 + i \right) x^2 - \left( 5 - i \right) x + 2 \left( 1 - i \right) = 0\]


Write the number of real roots of the equation \[(x - 1 )^2 + (x - 2 )^2 + (x - 3 )^2 = 0\].


If roots α, β of the equation \[x^2 - px + 16 = 0\] satisfy the relation α2 + β2 = 9, then write the value P.


If α, β are roots of the equation \[x^2 - a(x + 1) - c = 0\] then write the value of (1 + α) (1 + β).


The complete set of values of k, for which the quadratic equation  \[x^2 - kx + k + 2 = 0\] has equal roots, consists of


For the equation \[\left| x \right|^2 + \left| x \right| - 6 = 0\] ,the sum of the real roots is


If α, β are roots of the equation \[4 x^2 + 3x + 7 = 0, \text { then } 1/\alpha + 1/\beta\] is equal to


If α, β are the roots of the equation \[a x^2 + bx + c = 0, \text { then } \frac{1}{a\alpha + b} + \frac{1}{a\beta + b} =\]


If α, β are the roots of the equation \[x^2 + px + 1 = 0; \gamma, \delta\] the roots of the equation \[x^2 + qx + 1 = 0, \text { then } (\alpha - \gamma)(\alpha + \delta)(\beta - \gamma)(\beta + \delta) =\]


If x is real and \[k = \frac{x^2 - x + 1}{x^2 + x + 1}\], then


The value of a such that  \[x^2 - 11x + a = 0 \text { and } x^2 - 14x + 2a = 0\] may have a common root is


If one root of the equation \[x^2 + px + 12 = 0\] while the equation \[x^2 + px + q = 0\] has equal roots, the value of q is


The number of roots of the equation \[\frac{(x + 2)(x - 5)}{(x - 3)(x + 6)} = \frac{x - 2}{x + 4}\] is 


The equation of the smallest degree with real coefficients having 1 + i as one of the roots is


If 1 – i, is a root of the equation x2 + ax + b = 0, where a, b ∈ R, then find the values of a and b.


Show that `|(z - 2)/(z - 3)|` = 2 represents a circle. Find its centre and radius.


If `|(z - 2)/(z + 2)| = pi/6`, then the locus of z is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×