Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

This is one of our many quizzes on Equations. Click on this to access more such competitive quizzes on Algebra.

Equations Quiz 2

mathematics

 A quadratic equation $3{x^2} - 2x + 3 = 0$ has _____ roots.

 Determine the discriminant of the equation which is  $D = {b^2} - 4ac$ and according to its value, determine the nature of the roots.

  W.K.T the roots of a quadratic equation is given by, $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$, where $D = {b^2} - 4ac$. $ \\ $ In the given equation $3{x^2} - 2x + 3 = 0$,  $ \\ a = 3   \\  b =  - 2   \\  c = 3   \\ \Rightarrow D = {\left( { - 2} \right)^2} - 4\left( {3 \times 3} \right) = 4 - 36 =  - 32 < 0   \\ $ As this value is negative, the given equation has non-real roots.

 If the equation ${x^2} + ax - \dfrac{3}{4} = 0$ has a root $\dfrac{1}{2}$, then what is the value of $a$?

 Substitute the given root in the equation, and then solve for $a$.

  Given the equation ${x^2} + ax - \dfrac{3}{4} = 0$. $ \\ $ It has a root $x = \dfrac{1}{2}$. $ \\ $ Now, we will substitute this value in the given equation. $ \\ {\left( {\dfrac{1}{2}} \right)^2} + a\left( {\dfrac{1}{2}} \right) - \dfrac{3}{4} = 0   \\   \Rightarrow \dfrac{1}{4} + \dfrac{a}{2} - \dfrac{3}{4} = 0   \\   \Rightarrow 1 + 2a - 3 = 0   \\   \Rightarrow 2a - 2 = 0   \\   \Rightarrow 2a = 2   \\   \Rightarrow a = 1   \\$

 If two roots of the equation $8{x^2} - 4x + k - 8 = 0$ are reciprocals, then find $k$.

 Use the formula for multiplication of two roots of quadratic equation and then solve for $k$.

Given the equation $8{x^2} - 4x + k - 8 = 0$. $ \\ $ Here, $a = 8,b =  - 4,c = k - 8$ $ \\ $ Let us say one of its roots is $\alpha $, therefore its other root will be $\dfrac{1}{\alpha }$. $ \\ $ The product of two roots of the quadratic equation: $ \\ \alpha  \times \dfrac{1}{\alpha } = \dfrac{c}{a}   \\   \Rightarrow 1 = \dfrac{{k - 8}}{8}   \\   \Rightarrow k - 8 = 8   \\   \Rightarrow k = 8 + 8 = 16   \\ $

 If the equation $6{x^2} - x - 2 = 0$ has two roots $m$ and $n$, What will be the value of $\dfrac{1}{m} + \dfrac{1}{n}$?

 Use the formulas for addition and multiplication of two roots of quadratic equations.

 $\dfrac{1}{m} + \dfrac{1}{n} = \dfrac{{m + n}}{{mn}}$ $ \\ $ The sum of two roots of the quadratic equation $m + n = \dfrac{{ - b}}{a} = \dfrac{{ - \left( { - 1} \right)}}{6} = \dfrac{1}{6}$. $ \\ $ The multiplication of two roots of quadratic equation $mn = \dfrac{c}{a} = \dfrac{{ - 2}}{6} =  - \dfrac{1}{3}$. $ \\ \dfrac{1}{m} + \dfrac{1}{n} = \dfrac{{m + n}}{{mn}} = \dfrac{{\dfrac{1}{6}}}{{ - \dfrac{1}{3}}} =  - \dfrac{3}{6} =  - \dfrac{1}{2}$

 $\sqrt {12 + \sqrt {12 + \sqrt {12 + ...} } }  = $_______ 

 Consider a variable for the given term, make a quadratic equation and solve it.

 Let $\sqrt {12 + \sqrt {12 + \sqrt {12 + ...} } }  = k$ $ \\ \sqrt {12 + k}  = k   \\   \Rightarrow 12 + k = {k^2}   \\   \Rightarrow {k^2} - k - 12 = 0   \\   \Rightarrow {k^2} - 4k + 3k - 12 = 0   \\   \Rightarrow k\left( {k - 4} \right) + 3\left( {k - 4} \right) = 0   \\   \Rightarrow \left( {k - 4} \right)\left( {k + 3} \right) = 0   \\   \Rightarrow k - 4 = 0   \\   \Rightarrow k = 4   \\  $ or $   \Rightarrow k + 3 = 0   \\   \Rightarrow k =  - 3   \\  $ Since the square root of a negative number is not a real number, $k =  - 3$ is not possible. Hence, the right answer is $k = 4$.