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Quadratic Equations Test - 35

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Quadratic Equations Test - 35
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  • Question 1
    1 / -0
    The number of values of aa for which (a23a+2)x2+(a25a+6)x+a24=0(a^2-3a+2)x^2+(a^2-5a+6)x+a^2-4=0 is an identity in xx is-
    Solution
    Equation is a identity if all its coefficients are equal to zero simultaneously

    a23a+2=0\Rightarrow a^{2}-3a+2=0

    a=2,1\Rightarrow a=2,1

    a25a+6=0\Rightarrow a^{2}-5a+6=0

    a24=0\Rightarrow a^{2}-4=0

    a=±2\Rightarrow a=\pm 2

    \Rightarrowcoefficients are equal to zero simultaneously when a=2a=2

    Therefore only one value of 'a' the equation is identity in  xx
  • Question 2
    1 / -0
    Solve the following quadratic equation by completing the square: x2+(3+1)x+3=0 x^2+(\sqrt{3}+1)x+\sqrt{3}=0
    Solution
    Given equation is x2(3+1)x+3=0x^2-(\sqrt{3}+1)x+\sqrt{3}=0

    \Rightarrow  x2(3+1)x=3x^2-(\sqrt{3}+1)x=-\sqrt{3}

    Now, adding (3+12)2\left(\dfrac{\sqrt{3}+1}{2}\right)^2 on both sides,

    \Rightarrow  x22×x×(3+12)+(3+12)2=3+(3+12)2x^2-2\times x\times \left(\dfrac{\sqrt{3}+1}{2}\right)+\left(\dfrac{\sqrt{3}+1}{2}\right)^2=-\sqrt{3}+\left(\dfrac{\sqrt{3}+1}{2}\right)^2

    \Rightarrow  [x(3+1)2]2=[(3+1)243]\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]^2=\left[\dfrac{(\sqrt{3}+1)^2}{4}-\sqrt{3}\right]

    \Rightarrow  [x(3+1)2]2=[3+23+1434]\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]^2=\left[\dfrac{3+2\sqrt{3}+1-4\sqrt{3}}{4}\right]

    \Rightarrow  [x(3+1)2]2=[323+14]\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]^2=\left[\dfrac{3-2\sqrt{3}+1}{4}\right]

    \Rightarrow  [x(3+1)2]2=[312]2\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]^2=\left[\dfrac{\sqrt{3}-1}{2}\right]^2

    Taking square root on both sides, 
      
    \Rightarrow  [x(3+1)2]=±312\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]=\pm\dfrac{\sqrt{3}-1}{2}

    \Rightarrow  [x(3+1)2]=312\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]=\dfrac{\sqrt{3}-1}{2} 
    and [x(3+1)2]=312\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]=-\dfrac{\sqrt{3}-1}{2}

    \therefore  x=232x=\dfrac{2\sqrt{3}}{2}  and x=22 x=\dfrac{2}{2}

    \therefore  x=3x=\sqrt{3} and x=1x=1

  • Question 3
    1 / -0
    Solve the following quadratic equation by completing the square: 4x2+4bx(a2b2)=0 4x^2+4bx-(a^2-b^2)=0
    Solution
    Given equation is 4x2+4bx(a2b2)=04x^2+4bx-(a^2-b^2)=0

    \Rightarrow  x2+bx(a2b2)4=0x^2+bx-\dfrac{(a^2-b^2)}{4}=0

    \Rightarrow  x2+bx=(a2b2)4x^2+bx=\dfrac{(a^2-b^2)}{4}

    Now, (b2)2=b24\left(\dfrac{b}{2}\right)^2=\dfrac{b^2}{4}

    Adding b24\dfrac{b^2}{4} on bothe sides,

    \Rightarrow  x2+bx+b24=a24b24+b24x^2+bx+\dfrac{b^2}{4}=\dfrac{a^2}{4}-\dfrac{b^2}{4}+\dfrac{b^2}{4}

    \Rightarrow  (x+b2)2=a24\left(x+\dfrac{b}{2}\right)^2=\dfrac{a^2}{4}
      
    Taking square root on both sides

    \Rightarrow  x+b2=±a2x+\dfrac{b}{2}=\pm\dfrac{a}{2}

    \Rightarrow  x+b2=a2x+\dfrac{b}{2}=\dfrac{a}{2}  and x+b2=a2x+\dfrac{b}{2}=-\dfrac{a}{2}

    \Rightarrow  x=ab2x=\dfrac{a-b}{2} and x=(a+b2)x=-\left(\dfrac{a+b}{2}\right)
  • Question 4
    1 / -0
    ............... is true for discriminate of quadratic equation x2+x+1=0x^2 + x + 1 = 0.
    Solution
    The polynomial is x2+x+1=0x^2 + x + 1 = 0.

    Comparing with ax2+bx+c=0ax^2 + bx + c = 0, we get

    a=1,b=1,c=1a = 1, b = 1, c = 1

    D=b24ac=14(1)(1)=3<0D = b^2 - 4ac = 1 - 4(1) (1) = -3 < 0
  • Question 5
    1 / -0
    The discriminant value of equation 5x26x+1=05{x}^{2}-6x+1=0 is ............... 
    Solution
    The given equation is 5x26x+1=05{x}^{2}-6x+1=0 ........... (i)(i)

    General form of quadratic equation is given by ax2+bx+c=0ax^2+bx+c=0 ....... (ii)(ii)

    Comparing equation (i)(i) with (ii)(ii) we have,

    a=5,b=6,c=1a=5,b=-6,c=1

    The discriminant DD is given by 

    D=b24ac=(6)24(5)(1)=3620=16D={b}^{2}-4ac={(-6)}^{2}-4(5)(1)=36-20=16

    \therefore Discriminant value is D=16D=16
  • Question 6
    1 / -0
    Solve the following quadratic equation by completing the square: 5x+7x1=3x+2\dfrac{5x+7}{x-1}=3x+2
    Solution
    Given 5x+7x1=3x+2\dfrac{5x+7}{x-1}=3x+2

    \Rightarrow  5x+7=(3x+2)(x1)5x+7=(3x+2)(x-1)

    \Rightarrow  5x+7=3x23x+2x25x+7=3x^2-3x+2x-2

    \Rightarrow  3x26x9=03x^2-6x-9=0

    \Rightarrow  x22x3=0x^2-2x-3=0

    \Rightarrow  Here, a=1,b=2,c=3a=1,\,b=-2,\,c=-3

    \Rightarrow  x22x=3x^2-2x=3

    Now, (b2)2=(22)2=1\left(\dfrac{b}{2}\right)^2=\left (\dfrac{-2}{2} \right )^2=1

    Adding 11 on both sides,

    \Rightarrow  x22x+1=3+1x^2-2x+1=3+1

    \Rightarrow  (x1)2=4(x-1)^2=4
      
    Taking square root on both sides

    \Rightarrow  x1=±2x-1=\pm2

    \Rightarrow  x1=2x-1=2 and x1=2x-1=-2

    \therefore  x=3x=3 and x=1x=-1
  • Question 7
    1 / -0
    If the roots of the equation x2+px+c=0x^{2} + px + c = 0 are (2,2)(2, -2) and the roots of the equation x2+bx+q=0x^{2} + bx + q = 0 are (1,2)(-1, -2), then the roots of the equation x2+bx+c=0x^{2} + bx + c = 0 are
    Solution
    Given, roots of the equation x2+px+c=0x^{2} + px + c = 0 are 22 and 2-2.

    p=22p=0\therefore -p = 2 - 2 \Rightarrow p = 0

    and (2)(2)=cc=4(2)\cdot (-2) = c\Rightarrow c = -4

    Again, roots of the equation x2+bx+q=0x^{2} + bx + q = 0 are 1-1 and 2-2.

    b=11b=3\therefore -b = -1 -1 \Rightarrow b = 3

    and (1)(2)=qq=2(1-) \cdot (-2) = q\Rightarrow q = 2

    x2+bc+cx3+3x4=0\therefore x^{2} + bc + c \equiv x^{3} + 3x - 4 =0

    (x1)(x+4)=0\Rightarrow (x - 1)(x + 4) = 0

    So, the required roots are 11 and 4-4.
  • Question 8
    1 / -0
    Let α\alpha be the root of the equation 25cos2θ+5cosθ12=C25\cos^{2}\theta + 5\cos \theta - 12 = C, where π2<α<π\dfrac {\pi}{2} < \alpha < \pi.
    What is tanα\tan \alpha equal to?
    Solution
    Given equation is 25cos2θ+5cosθ12=025\cos^2\theta+5\cos\theta-12=0

    25cos2θ+20cosθ15cosθ12=0\Rightarrow 25\cos^2\theta+20\cos\theta-15\cos\theta-12=0

    5cosθ(5cosθ+4)3(5cosθ+4)=0\Rightarrow 5\cos\theta(5\cos\theta+4)-3(5\cos\theta+4)=0

    (5cosθ3)(5cosθ+4)=0\Rightarrow (5\cos\theta-3)(5\cos\theta+4)=0

    So, cosθ=35\cos\theta=\dfrac{3}{5} or cosθ=45\cos\theta=\dfrac{-4}{5}

    Since α\alpha is the root of the equation and it lies between $$\dfrac{\pi}{2}<\alpha<\pi\Rightarrow \cos \alpha\  $$ is '-ve'.

    Thus α=cos1(45)\alpha=\cos^{-1}\left (\dfrac{-4}{5}\right) 

    Since it lies in the range π2<α<π\dfrac{\pi}{2}<\alpha<\pi, so tanα\tan \alpha is '-ve'

    tanα=tan[cos1(45)]\tan\alpha=\tan\left [\cos^{-1}\left (\dfrac{-4}{5}\right)\right]

    =tan[tan1(34)]=\tan \left [\tan^{-1}\left (-\dfrac{3}{4}\right)\right]

    =34=-\dfrac{3}{4}
  • Question 9
    1 / -0
    Sum of the roots of the equation x3 2+x32=0{ \left| x-3 \right|  }^{ 2 }+\left| x-3 \right| -2=0 is
    Solution
    Given equation is x3 2+x32=0{ \left| x-3 \right|  }^{ 2 }+\left| x-3 \right| -2=0

    x3 2+2x3x32=0\Rightarrow { \left| x-3 \right|  }^{ 2 }+2\left| x-3 \right| -\left| x-3 \right| -2=0

    x3(x3+2)1(x3+2)=0\Rightarrow \left| x-3 \right| \left( \left| x-3 \right| +2 \right) -1\left( \left| x-3 \right| +2 \right) =0

    (x3+2)(x31)=0\Rightarrow \left( \left| x-3 \right| +2 \right) \left( \left| x-3 \right| -1 \right) =0

    Therefore, x3=2,1 \left| x-3 \right| =-2,1

    But x32\left| x-3 \right| \neq -2

    Hence, x3=1\left| x-3 \right| =1

    x3=±1\Rightarrow x-3=\pm 1

    x=3±1\Rightarrow x=3\pm 1

    x=4,2\Rightarrow x=4, 2

    Now, sum of the roots =4+2=6=4+2=6.
  • Question 10
    1 / -0
    The equation esinx esinx 4=0{ e }^{ \sin { x }  }-{ e }^{ -\sin { x }  }-4=0 has
    Solution
    Given, esinx esinx 4=0{ e }^{ \sin { x }  }-{ e }^{ -\sin { x }  }-4=0 

    multiply throughout by esinxe^{sinx}

    e2sinx 4esinx 1=0\Rightarrow { e }^{ 2\sin { x }  }-4{ e }^{ \sin { x }  }-1=0

    this is quadratc equation in esinxe^{sinx}

    converting from exponential to logarithmic form.

    esinx =4±16+4 2=2±5{ e }^{ \sin { x }  }=\dfrac { 4\pm \sqrt { 16+4 }  }{ 2 } =2\pm \sqrt { 5 }

    sinx=log(2+5 ) \Rightarrow \sin { x } =\log { \left( 2+\sqrt { 5 }  \right)  }               [log(25 )  \because \log { \left( 2-\sqrt { 5 }  \right)  } is not defined ] 

    Since, 2+5>elog(2+5 ) >12+\sqrt { 5 } > e \Rightarrow \log { \left( 2+\sqrt { 5 }  \right)  } > 1

    sinx>1\Rightarrow \sin { x } > 1, which is not possible.
    Hence, no solution exist.
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