मराठी

The Values of X Satisfying Log3 ( X 2 + 4 X + 12 ) = 2 Are - Mathematics

Advertisements
Advertisements

प्रश्न

The values of x satisfying log3 \[( x^2 + 4x + 12) = 2\] are

पर्याय

  • 2, −4

  • 1, −3

  • −1, 3

  • −1, −3

MCQ

उत्तर

−1, −3 The given equation is \[\log_3 ( x^2 + 4x + 12) = 2\] .

\[\Rightarrow x^2 + 4x + 12 = 3^2 = 9\]

\[ \Rightarrow x^2 + 4x + 3 = 0\]

\[ \Rightarrow \left( x + 1 \right)\left( x + 3 \right) = 0\]

\[ \Rightarrow x = - 1, - 3\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 14: Quadratic Equations - Exercise 14.4 [पृष्ठ १६]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 14 Quadratic Equations
Exercise 14.4 | Q 5 | पृष्ठ १६

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Solve the equation x2 + 3 = 0


Solve the equation –x2 + x – 2 = 0


Solve the equation x2 – x + 2 = 0


Solve the equation  `sqrt3 x^2 - sqrt2x + 3sqrt3 = 0`


Solve the equation 27x2 – 10x + 1 = 0


Solve the equation 21x2 – 28x + 10 = 0


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


x2 + 1 = 0


x2 + 2x + 5 = 0


x2 + x + 1 = 0


\[4 x^2 + 1 = 0\]


\[x^2 - 4x + 7 = 0\]


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


\[x^2 + x + 1 = 0\]


\[17 x^2 - 8x + 1 = 0\]


\[27 x^2 - 10 + 1 = 0\]


\[8 x^2 - 9x + 3 = 0\]


\[13 x^2 + 7x + 1 = 0\]


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


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


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


Solving the following quadratic equation by factorization method:

\[x^2 - \left( 2\sqrt{3} + 3i \right) x + 6\sqrt{3}i = 0\]


Solve the following quadratic equation:

\[2 x^2 + \sqrt{15}ix - i = 0\]


Solve the following quadratic equation:

\[i x^2 - x + 12i = 0\]


If a and b are roots of the equation \[x^2 - px + q = 0\], than write the value of \[\frac{1}{a} + \frac{1}{b}\].


If α, β are roots of the equation \[x^2 + lx + m = 0\] , write an equation whose roots are \[- \frac{1}{\alpha}\text { and } - \frac{1}{\beta}\].


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 \[x^2 + px + 1 = 0; \gamma, \delta\] 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 values of k for which the quadratic equation \[k x^2 + 1 = kx + 3x - 11 x^2\] has real and equal roots are


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


The least value of which makes the roots of the equation  \[x^2 + 5x + k = 0\]  imaginary 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.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×