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This is the second of our many quizzes on Finding zeros of polynomials. Click on this to access more such competitive quizzes on Algebra.

Finding zeros of polynomials Quiz 2

mathematics

 If ${{\left( \dfrac{1}{8} \right)}^{3x - 8}} = 1$, then find the value of $x$.

 First write the RHS as $\dfrac{1}{8}$ to a power of 0, then equate the bases of the sides, solving for $x$.

  $\text{Here, }{{\left( \dfrac{1}{8} \right)}^{3x - 8}} = {{\left( \dfrac{1}{8} \right)}^{0}} \\ \text{Bases are equal, therefore: } 3x - 8 = 0 \\ x = \dfrac{8}{3}$ 

  If $\dfrac{\left( 5x - 2y \right)}{\left( 6{{x}^{2}} - 5xy + {{y}^{2}} \right)} =  \dfrac{4}{3}$ and $\dfrac{y}{\left( 6{{x}^{2}} - 5xy + {{y}^{2}} \right)} = 1$ Then $y$ is equal to: 

 First split the middle terms of both the denominators, then solve for $x$ and $y$.

  Here, first we split the denominator then equation becomes $\\ \dfrac{\left( 5x - 2y \right)}{\left( \left( 2x - y \right)\left( 3x - y \right) \right)} = \dfrac{4}{3} \text{ and } \\ \dfrac{y}{\left( \left( 2x - y \right)\left( 3x - y \right) \right)} = 1 \\ \dfrac{1}{2x - y}+\dfrac{1}{3x - y}=\dfrac{4}{3} \text{ and } \\ \dfrac{2}{2x - y}+\dfrac{3}{3x - y}=1 \\ \text{Now let } a=\dfrac{1}{2x-y} \text{ and } b=\dfrac{1}{3x-y} \\ \text{So equation becomes} \\ a + b = \dfrac{4}{3} \text{ and } 2a - 3b = 1 \\ \text{Solving these two equation we get a=1 and } b=\dfrac{1}{3} \\ \text{So now equation becomes } 2x - y = 1 \text{ and } 3x - y = 3 \\ \text{Solving these two equation we get x=2 and y=3}$

 ${{x}^{2}}+  \left( 3p + 2 \right) x + 3{{p}^{2}} - 14p + 15 = 0$ is a quadratic equation with respect to x. If the product of the roots equal to $7$, find the sum of the roots.

 First by the formula of product of roots, find the value of p, then use summation of roots formula.

$\text{Product of roots }= 3{{p}^{2}} - 14p + 15 \\ 3{{p}^{2}} - 14p + 15 = 7 \\ p = 4 or \dfrac{2}{3} \\ \text{Sum of the roots }= 3p + 2 \\ = 14 \text{ or } 4$

 The equation  $\sqrt{\left( 4x + 9 \right)} - \sqrt{\left( 11x + 1 \right)} - \sqrt{\left( 7x + 4 \right)}=0$ has

 By squaring on both sides of the equation, then splitting the middle term we will get the values for $x$, one real root and one imaginary.

 $\text{Here, } \sqrt{\left( 4x + 9 \right)} - \sqrt{\left( 11x + 1 \right)}\text{ = }\sqrt{\left( 7x + 4 \right)} \\  \left( 4x + 9 \right) + \left( 7x + 4 \right) - 2\sqrt{\left( 4x + 9 \right)\left( 7x + 4 \right)} = \left( 11x + 1 \right) \\ 28 {{x}^{2}} + 79 x +36 = 36 \\  x = 0 or -\dfrac{79}{28} \\ \text{So } x= -\dfrac{79}{28}$ is not a real root. There is one real solution and one imaginary solution.

  ${{x}^{2}} - \left( {{k}^{2}} + 2 \right)x + {{k}^{2}} + 3k = 1$ is a quadratic equation in $x$, where $k$ is natural number. If the roots of the equation are identical and real, find the value of $x$.

 By putting the values from the given options in the summation of the root formula, we will get $k$, then from there we will get $x$.

 $\text{Sum of the roots }= {{k}^{2}} + 2 \\ \text{Product of the roots }= {{k}^{2}} + 3k - 1 \\ \text{Let ‘a’ be the root;} \\ \text{Sum of the roots = 2a} \\ \text{Product of the roots= a2} \\ \text{From option (c), we get,} \\ {{k}^{2}} + 2 =\text{ 6} \\ k = 2 or - 2 \\$ Other options do not satisfy the value of ‘k’ as it is a natural number. Product of the roots $ \\= {{k}^{2}} + 3k - 1  \\ = 2^{2} + 3\left( 2 \right) - 1  \\ = 9  \\ \Rightarrow x = 3$ 