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Quadratic Equations Test - 26

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Quadratic Equations Test - 26
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  • Question 1
    1 / -0
    The equation $$(a+2){x}^{2}+(a-3)x=2a-1,a\neq -2$$ has rational roots for 
  • Question 2
    1 / -0
    If $$\displaystyle ax^{2}+bx+6=0$$ does not have two distinct real roots, where $$a\in R,b\in R$$, then the least value of $$3a+b$$ is
    Solution
    $$ax^{ 2 }+bx+6=0$$
    Substitute $$b=k-3a$$ to obtain $$ax^{ 2 }+\left( k-3a \right) x+6=0$$
    Since, the equation does not have real distinct roots.
    Therefore, $$D={ \left( k-3a \right)  }^{ 2 }-24a\le 0$$
    $$\Rightarrow 9a^{ 2 }-6a\left( 4+k \right) +{ k }^{ 2 }\le 0$$
    Above quadratic equation has real roots.
    Therefore, $$D=36\left[ { \left( 4+k \right)  }^{ 2 }-{ k }^{ 2 } \right] \ge 0$$
    $$\Rightarrow k\ge -2$$
    $$\Rightarrow 3a+b\ge -2$$.

    Ans: D
  • Question 3
    1 / -0
    If the roots of the equation $$x^{2} + a^{2} = 8x + 6a$$ are real, then:
    Solution
    The given equation can be written as
    $$ x^{2} - 8x + a^{2} -6a = 0$$
    Since the roots of the above equation are real,
    $$\therefore  B^{2} -4AC \geq  0 \Rightarrow  64-4(a^{2}-6a)\geq 0$$
    $$\Rightarrow a^{2} -6a -16 \leq  0$$
    $$\Rightarrow (a+2)(a-8) \leq 0$$
    $$\Rightarrow a\ \epsilon\  \left [ -2,8 \right ]$$
  • Question 4
    1 / -0
    Which of the following is not a quadratic equation?
    Solution
    It is not a quadratic equation because, here maximum power of x is 4.
  • Question 5
    1 / -0
    If both 'a' and 'b' belong to the set $$\left \{ 1, 2, 3, 4 \right \}$$, then the number of equations of the form $$ax^{2}+bx+1=0$$ having real roots is:
    Solution
    For a quadratic equation to have a real roots, discriminant must be greater than or equal to zero

    $$\Rightarrow b^{2}-4a\geq0$$

    Since, $$a,b \in \{1,2,3,4\}$$

    (1) When $$a=1$$

    $$\Rightarrow b^{2}\geq 4$$

    $$\Rightarrow b=2,3,4$$ 

    (2) When $$a=2$$

    $$\Rightarrow b^{2}\geq 8$$

    $$\Rightarrow b=3,4$$ 

    (3) When $$a=3$$

    $$\Rightarrow b^{2}\geq 12$$

    $$\Rightarrow b=4$$ 

    (3) When $$a=4$$

    $$\Rightarrow b^{2}\geq 16$$

    $$\Rightarrow b=4$$ 

    Therefore, total $$7$$ possibilities are there.

    Hence the answer is $$7$$.

  • Question 6
    1 / -0
    If the equation $$4x^{2} + x(p + 1) + 1 = 0$$ has exactly two equal roots, then one of the values of $$p$$ is
    Solution
    For the equation $$4x^{2} + x(p + 1) + 1 = 0$$ to have two equal roots, the condition is Discriminant $$=0$$
    $$b^{2}-4ac = 0$$
    $$(p+1)^{2}-4\times 4\times 1 = 0$$
    $$p^{2}+2p+1-16 = 0$$
    $$p^{2}+2p-15 = 0$$
    $$(p+5)(p-3) = 0$$
    $$\therefore p = -5$$ or $$p = 3$$
    Hence, one of the values of $$p$$ is $$3$$.
    Option D is correct.
  • Question 7
    1 / -0
    The value of $$x^{2} - 6x + 13$$ can never be less than
    Solution
    We know that a perfect square can never be less than $$0$$.
    $$\because x^{2}-6x+13 = x^{2}-6x+9+4$$
                             $$= (x-3)^{2}+4$$
    because $$(x-3)^{2}$$ can never be less than zero
    $$\therefore$$ least value of $$x^{2}-6x+13\ is \  4$$

  • Question 8
    1 / -0
    If $$\displaystyle ax^{2}+bx+1= 0, a \in R, b\in R,$$ does not have distinct real roots, then the maximum value of $$b^2$$ is
    Solution
    $$The\quad equation\quad a{ x }^{ 2 }-bx+1=0\quad has\quad no\quad distinct\quad
    real\quad roots,\\ This\quad implies\quad D=b^2-4ac\le 0,\\ b^2-4a\le 0 \\ b^2\le 4a$$.
  • Question 9
    1 / -0
    The number of values of $$k$$ for which $$\displaystyle \left \{x^{2}-(k-2)x+k^{2}\right\}+ \left \{x^{2}+kx+(2k-1)\right \}$$ is a perfect square is/are 
    Solution
    Consider $$p\left( x \right) =\left\{ x^{ 2 }-(k-2)x+k^{ 2 } \right\} +\left\{ x^{ 2 }+kx+(2k-1) \right\} $$
    $$\Rightarrow p\left( x \right) =2{ x }^{ 2 }+2x+{ k }^{ 2 }+2k-1$$
    $$p\left( x \right) $$ is perfect square when roots of the equation $$p\left( x \right) =0$$ are equal
    Therefore, $$D={ b }^{ 2 }-4ac=4-8\left( { k }^{ 2 }+2k-1 \right) =0$$
    $$\Rightarrow 2{ k }^{ 2 }-4k-3=0$$
    Therefore, number of values of $$k$$ are 2.

    Ans: B
  • Question 10
    1 / -0
    If at least one of the equations $$ { x^{ 2 } }+px+q=0$$ , $$ { x^{ 2 } }+rx+s=0 $$ has real roots, then
    Solution
    Given at least one of the equation $$x^{ 2 }+px+q=0$$ , $$x^{ 2 }+rx+s=0 $$ has real roots.
    Let $$D_1$$ and $$D_2$$ be the discriminant values.
    At least one of the equation has real roots when $$D_1+D_2 \geq 0$$.
    $$\Rightarrow p^2-4q+r^2-4s \geq 0$$
    $$\Rightarrow (p-r)^2+2pr-4(q+s) \geq 0$$
    Above inequality holds only if $$2pr-4(q+s)=0$$.
    $$\therefore pr=2(q+s)$$.
    Hence, option C.

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