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Quadratic Equations Test - 27

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Quadratic Equations Test - 27
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  • Question 1
    1 / -0
    Find all values of parameter $$a$$ for which the quadratic equation $$  \left( a+1 \right) { x }^{ 2 }+2\left( a+1 \right) x+a-2=0 $$ has two distinct roots.

    Solution
    Discriminant of the given quadratic equation is,
    $$D =  4\left( a+1 \right) ^{ 2

    }-4\left( a+1 \right) \left( a-2 \right) =4\left( a+1 \right) \left(

    a+1-a+2 \right) =12\left( a+1 \right) $$
    Now for two distinct roots,
    $$D>0\Rightarrow a+1>0$$
    $$\Rightarrow a \in (-1,\infty)$$. 
  • Question 2
    1 / -0
    Find the discriminant for the given quadratic equation:
    $$x^2\,+\,x\,+\,1\,=\,0$$
    Solution
    Given equation is,
    $$x^2 + x + 1 = 0 $$
    We know, $$D = b^2 - 4ac$$
    $$a=1,b=1,c=1$$
    $$\therefore D = (1)^2 - 4(1)(1)$$
    $$\therefore D = -3$$
  • Question 3
    1 / -0
    If $$D$$ is the discriminant of $$x^2\,+\,4x\,+\,1\,=\,0$$, then the value of $$D^2$$, is
    Solution
    $$x^2 + 4x + 1 = 0 $$
    $$D = b^2 - 4ac$$
    $$D = (4)^2 - 4(1)(1)$$
    $$D = 16 - 4$$
    $$D = 12 $$
    $$D^2 = {12}^2 = 144$$
  • Question 4
    1 / -0
    Find the discriminant for the given quadratic equation:
    $$4x^2\,-\,kx\,+\,2\,=\,0$$
    Solution
    Given equation is,
    $$4x^2 - kx + 2 = 0 $$
    We know, $$D = b^2 - 4ac $$
    $$a=4, b=-k, c=2$$
    $$D = (-k)^2 - 4(4)(2)$$
    $$\therefore D = k^2 - 32$$
  • Question 5
    1 / -0
    Solve the following equations.
    $$x^4\,-\,3x^2\,+\,2\,=\,0$$, roots are 

    Solution
    Given equation is $$x^4 - 3x^2 +2 = 0$$
    $$x^4 - 2x^2 -x^2 + 2 =0$$
    $$(x^2 - 1)(x^2 - 2) = 0$$
    $$\therefore x^2 = 2 \Rightarrow x = \pm \sqrt{2}$$

    $$\therefore x^2 = 1 \Rightarrow x = \pm 1$$
  • Question 6
    1 / -0
    Find the roots of following quadratic equation 
    $$x^2\,+\,3x\,-\,2\,=\,0$$
    Solution
    Gven equation is
    $$x^2 + 3x - 2 = 0 $$
    Using quadratic formula,
    $$a=1, b=3, c=-2$$
    $$x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$$
    $$x = \frac{-3 \pm \sqrt{9 - 4(1)(-2)}}{2\times1}$$
    $$\therefore x = \frac{-3 \pm \sqrt{17}}{2}$$
    $$\therefore$$ the roots of the given equation are $$\displaystyle \frac {-3\pm \sqrt{17}}{2}$$
  • Question 7
    1 / -0
    Find the discriminant for the given quadratic equation:
    $$\sqrt{3}x^2\,+\,2\sqrt{2}x\,-\,2\sqrt{3}\,=\,0$$
    Solution
    Given equation is,
    $$\sqrt{3}x^2\,+\,2\sqrt{2}x\,-\,2\sqrt{3}\,=\,0$$
    We know, $$D = b^2 - 4ac$$
    $$a=\sqrt 3, b=2 \sqrt 2, c=-2\sqrt3$$
    $$D = (2\sqrt{2})^2 - 4\times\sqrt{3}\times(-2\sqrt{3})$$
    $$D = 8 + 24$$
    $$\therefore D = 32$$
  • Question 8
    1 / -0
    Find the discriminant for the given equation:
    $$3x^2\,+\,2x\,-\,1\,=\,0$$
    Solution
    Given equation is,
    $$3x^2 + 2x- 1 = 0$$
    We know, $$D = b^2 - 4c$$
    $$a=3,b=2,c=-1$$
    $$D = 2^2 - 4\times3\times(-1)$$
    $$D = 4 + 12$$
    $$\therefore D = 16$$
  • Question 9
    1 / -0
    Find the discriminant for the given quadratic equation: $$x^2\,+\,4x\,+\,k\,=\,0$$
    Solution
    Given equation is,
    $$x^2 + 4x + k = 0 $$
    We know,
    $$D = b^2 - 4ac $$
    $$a=1, b=4, c=k$$
    $$D = (4)^2 - 4(1)(k)$$
    $$\therefore D = 16 - 4k$$
  • Question 10
    1 / -0
    Find the value of discriminant for the following equation.
    $$x^{2}\, +\, 4x\, +\, k\, =\, 0$$
    Solution
    Given equation is:
    $$x^2+4x+k=0$$
    Discriminant = $$b^2 - 4ac$$ 
    $$a=1, b=4, c=k$$
    $$\therefore D= (4)^2 - 4(k)(1)$$
    $$= 16 - 4k$$
    $$\therefore \Delta =16-4k$$
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