Grade 11 Mathematics

Quadratic Equation [BCB Ex6.1]

Posted on August 16, 2023August 16, 2023 by MEAN

Determine the nature of roots of each of the following equations
1. $x^2-6x+5=0$
  
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  Solution :1a
2. $x^2-4x-3=0$
  
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  Solution :1b
3. $x^2-6x+9=0$
  
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  Solution :1c
4. $4x^2-4x+1=0$
  
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  Solution :1d
5. $2x^2-9x+35=0$
  
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  Solution :1e
6. $4x^2+8x-5=0$
  
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  Solution :1f
For what values of p will the equation $5x^2-px+45=0$ ?

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Solution :2
if the equation $x^2+2(k+2)x+9k=0$ has equal roots, find k.

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Solution :3
For what value of a will the equation $x^2-(3a-1)x+2(a^2-1)=0$ have equal roots?

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Solution :4
If the roots of the equation $(a^2+b^2)x^2-2(ac+bd)x+(c^2+d^2)=0$ are equal, then show that $\frac{a}{b}=\frac{c}{d}$

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Solution :5
Show that the roots of the equation $(a^2-bc)x^2+2(b^2-ca)x+(c^2-ab)=0$ will be equal, if either $b=0$ or $a^3+b^3+c^3-3abc=0$

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Solution :6
If $a,b,c$ are rational and $a+b+c=0$ , show that the roots $(b+c-a)x^2+(c+a-b)x+(a+b-c)=0$ are rational.

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Solution :7
Prove that the roots of the equation $(x-a)(x-b)=k^2$ are real for all values of k.

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Solution :8
Show that the roots of the equation $x^2-4abx+(a^2+2b^2)^2=0$ are imaginary.

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Solution :9
If the roots of the quadratic equation $qx^2+2px+2q=0$ are real and unequal, prove that the roots of the equation $(p+q)x^2+2qx+(p-q)=0$ are imaginary

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Solution :10

Quadratic Equation [BCB Ex6.2]

Posted on August 16, 2023August 16, 2023 by MEAN

From the equation whose roots are
1. 3,-2
  
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  Solution :1a
2. -5,4
  
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  Solution :1b
3. $\sqrt{3},-\sqrt{3}$
  
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  Solution :1c
4. $\frac{1}{2} (-1+\sqrt{5}),\frac{1}{2} (-1-\sqrt{5})$
  
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  Solution :1d
5. -3+5i,-i-5i
  
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  Solution :1e
6. a+ib,a-ib
  
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  Solution :1f
1. Find a quadratic equation whose roots are twice the roots of $4x^2+8x-5=0$
  
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  Solution :2a
2. Find a quadratic equation whose roots are reciprocals of the roots of $3x^2-5x-2=0$
  
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  Solution :2(b)
3. Find a quadratic equation whose roots are greater by h than the roots of $x^2-px+q=0$
  
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  Solution :2(c)
4. Find a quadratic equation whose roots are the squares of the roots of $3x^2-5x-2=0$
  
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  Solution :2(d)
Find a quadratic equation with rational coefficients one of whose roots is
1. 4+3i
  
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  Solution :3a
2. $\frac{1}{5+3i}$
  
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  Solution :3b
3. $2+\sqrt{3}$
  
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  Solution :3c
Find the value of k so that the equation
1. $2x^2+kx-15=0$ has one root 3
  
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  Solution :4a
2. $3x^2+kx-2=0$ has roots whose sum is equal to 6
  
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  Solution :4b
3. $2x^2+(4-k)x-17=0$ has roots equal but opposite in sign
  
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  Solution :4c
4. $3x^2+(5+k)x+8=0$ has roots numerically equal but opposite in sign
  
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  Solution :4d
5. $3x^2+7x+6-k=0$ has one root equal to zero
  
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  Solution :4e
6. $4x^2-17x+k=0$ has the reciprocal roots
  
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  Solution :4f
7. $4x^2+kx+5=0$ has roots whose difference is $\frac{1}{4}$
  
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  Solution :4g
Show that -1 is a root of the equation $a+b-2c$ x^2+(2a-b-c)x+(c+a-2b)=0\). Find the other root.

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Solution :5
Find the value of m for which the equation $(m+1)x^2+2(m+3)x+(2m+3)=0$ will have (a) reciprocal roots (b) one root zero.

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Solution :6
If the roots of the equation $x^2+ax+c=0$ differ by 1, prove that $a^2=4c+1$

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Solution :7
If are the roots of the equation , find the equation whose roots are
1. $\alpha ^2 \beta ^{-1}$ and $\beta ^2 \alpha ^{-1}$
  
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  Solution :8a
2. $\alpha + \frac{1}{\beta}$ and $\beta + \frac{1}{\alpha}$
  
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  Solution :8(b)
If are the roots of the equation , find the equation whose roots are
1. $\alpha ^2 \beta ^{-1}$ and $\beta ^2 \alpha ^{-1}$
  
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  Solution :9a
2. $\alpha ^3$ and $\beta ^3$
  
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  Solution :9b
3. $(\alpha-\beta)^2$ and $(\alpha+\beta)^2$
  
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  Solution :9c
4. the reciprocal of the roots of given equation
  
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  Solution :9d
1. If the roots of the equation $ax^2+bx+c=0$ be in the ratio of 3:4, prove that $12b^2=49ac$
  
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  Solution :10a
2. If one root of the equation $ax^2+bx+c=0$ be four times the other root, show that $4b^2=25ac$
  
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  Solution :10b
3. For what values of m, the equation $x^2-mx+m+1=0$ may have its root in the ratio 2:3
  
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  Solution :10c
1. If $\alpha, \beta$ are the roots of the equation $px^2+qx+q=0$ , prove that $\sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}+\sqrt{\frac{q}{p}}=0$
  
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  Solution :11a
2. If roots of the equation $lx^2+nx+n=0$ be in the ratio of p:q, prove that $\sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}}=0$
  
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  Solution :11b
If one root of the equation $ax^2+bx+c=0$ be square of the other root, prove that $b^3+a^2c+ac^2=3abc$

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Solution :12

Quadratic Equation[BCB Ex6.3]

Posted on August 16, 2023August 16, 2023 by MEAN

Show that each pair of following equations has a common root
1. $x^2-8x+15=0$ and $2x^2-x-15 =0$
  
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  Solution :1(a)
  Given quadratic equations are
  $x^2-8x+15=0$ and $2x^2-x-15 =0$
  Comparing with $a_1x^2+b_1x+c_1=0$ and $a_2x^2+b_2x+c_2=0$ , we get
  $a_1=1,b_1=-8,c_1=15,a_2=2,b_2=-1,c_2=-15$
  We know that, condition for two quadratic equations having a common root is
  $(a_2c_1-a_1c_2)^2 = (a_1b_2-a_2b_1) (b_1c_2-b_2c_1)$
  or $[2.(15)-1(-15)]^2 = [1(-1)-2(-8)] [(-8)(-15)-(-1)(15)]$
  or $[30+15]^2 = [-1+16] [120+15]$
  or $2035 =2035$
  Thus, $x^2-8x+15=0$ and $2x^2-x-15 =0$ has a common root.
2. $3x^2-8x+4=0$ and $4x^2-7x-2 =0$
  
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  Solution :1(b)
  Given quadratic equations are
  $3x^2-8x+4=0$ and $4x^2-7x-2 =0$
  Comparing with $a_1x^2+b_1x+c_1=0$ and $a_2x^2+b_2x+c_2=0$ , we get
  $a_1=3,b_1=-8,c_1=4,a_2=4,b_2=-7,c_2=-2$
  We know that, condition for two quadratic equations having a common root is
  $(a_2c_1-a_1c_2)^2 = (a_1b_2-a_2b_1) (b_1c_2-b_2c_1)$
  or $[(4)(4)-(3)(-2)]^2 = [(3)(-7)-(4)(-8)] [(-8)(-2)-(-7)(4)]$
  or $[16+6]^2 = [-21+32] [16+28]$
  or $484 =484$
  Thus, $3x^2-8x+4=0$ and $4x^2-7x-2 =0$ has a common root.
Find the value of p so that each pair of the equations may have one root common
1. $4x^2+px-12=0$ and $4x^2+3px-4 =0$
  
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  Solution :2(a)
  Given quadratic equations are
  $4x^2+px-12=0$ and $4x^2+3px-4 =0$
  Comparing with $a_1x^2+b_1x+c_1=0$ and $a_2x^2+b_2x+c_2=0$ , we get
  $a_1=4,b_1=p,c_1=-12,a_2=4,b_2=3p,c_2=-4$
  We know that, condition for two quadratic equations having a common root is
  $(c_1a_2-c_2a_1)^2 = (a_1b_2-a_2b_1) (b_1c_2-b_2c_1)$
  or $[(-12)(4)-(-4)(4)]^2 = [(4)(3p)-(4)(p)] [(p)(-4)-(3p)(-12)]$
  or $[-48+16]^2 = [12p-4p] [-4p+36p]$
  or $[-32]^2 = [8p] [32p]$
  or $p^2=4$
  or $p=\pm 2$
  Thus,fo $p=\pm 2$ , given pair of the equations may have one root common
2. $2x^2+px-1=0$ and $3x^2-2x-5 =0$
  
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  Solution :2(b)
  Given quadratic equations are
  $2x^2+px-1=0$ and $3x^2-2x-5=0$
  Comparing with $a_1x^2+b_1x+c_1=0$ and $a_2x^2+b_2x+c_2=0$ , we get
  $a_1=2,b_1=p,c_1=-1,a_2=3,b_2=-2,c_2=-5$
  We know that, condition for two quadratic equations having a common root is
  $(c_1a_2-c_2a_1)^2 = (a_1b_2-a_2b_1) (b_1c_2-b_2c_1)$
  or $[(-1)(3)-(-5)(2)]^2 = [(2)(-2)-(3)(p)] [(p)(-5)-(-2)(-1)]$
  or $[-3+10]^2 = [-4-3p] [-5p-2]$
  or $[7]^2 = [4+3p] [2+5p]$
  or $49 = 15p^2+26p+8$
  or $15p^2+26p-41=0$
  or $15p^2+(41-15)p-41=0$
  or $15p^2+41p-15p-41=0$
  or $p(15p+41)-1(15p+41)=0$
  or $p=1, -\frac{41}{15}$
  Thus,fo $p=1, -\frac{41}{15}$ , given pair of the equations may have one root common
If the quadratic equations $x^2+px+q=0$ and $x^2+p'x+q'=0$ have common roots show that it must be either $\frac{pq'-p'q}{q-q'}$ or $\frac{q-q'}{p'-p}$

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Solution :3
Given quadratic equations are
$x^2+px+q=0$ and $x^2+p'x+q'=0$
Comparing with $a_1x^2+b_1x+c_1=0$ and $a_2x^2+b_2x+c_2=0$ , we get
$a_1=1,b_1=p,c_1=q,a_2=1,b_2=p',c_2=q'$
We know that, condition for two quadratic equations having a common root is
$(c_1a_2-c_2a_1)^2 = (a_1b_2-a_2b_1) (b_1c_2-b_2c_1)$
or $[(q)(1)-(q')(1)]^2 = [(1)(p')-(1)(p)] [(p)(q')-(p')(q)]$
or $[q-q']^2 = [p'-p] [pq'-p'q]$
or $\frac{pq'-p'q}{q-q'} . \frac{p'-p}{q-q'} =0$
or $\frac{pq'-p'q}{q-q'} =0$ or $\frac{p'-p}{q-q'} =0$
or $\frac{pq'-p'q}{q-q'} =0$ or $\frac{q-q'}{p'-p} =0$
Thus, $x^2+px+q=0$ and $x^2+p'x+q'=0$ have one root common if $\frac{pq'-p'q}{q-q'} =0$ or $\frac{q-q'}{p'-p} =0$
If the quadratic equations $x^2+px+q=0$ and $x^2+qx+p=0$ have common roots show that it must be either $p=q$ or $p+q+1=0$

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Solution :4
Given quadratic equations are
$x^2+px+q=0$ and $x^2+qx+p=0$
Comparing with $a_1x^2+b_1x+c_1=0$ and $a_2x^2+b_2x+c_2=0$ , we get
$a_1=1,b_1=p,c_1=q,a_2=1,b_2=q,c_2=p$
We know that, condition for two quadratic equations having a common root is
$(c_1a_2-c_2a_1)^2 = (a_1b_2-a_2b_1) (b_1c_2-b_2c_1)$
or $[(q)(1)-(p)(1)]^2 = [(1)(q)-(1)(p)] [(p)(p)-(q)(q)]$
or $[q-p]^2 = [q-p] [p^2-q^2]$
or $(q-p) [(q-p)- (p^2-q^2)]=0$
or $q-p =0$ or $p+q-1 =0$
Thus, $x^2+px+q=0$ and $x^2+qx+p=0$ have one root common if $p=q$ or $p+q-1 =0$
If the quadratic equations $ax^2+bx+c=0$ and $bx^2+cx+a=0$ have common roots show that it must be either $a=b=c$ or $a+b+c=0$

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Solution :5
Given quadratic equations are
$ax^2+bx+c=0$ and $bx^2+cx+a=0$
Comparing with $a_1x^2+b_1x+c_1=0$ and $a_2x^2+b_2x+c_2=0$ , we get
$a_1=a,b_1=b,c_1=c,a_2=b,b_2=c,c_2=a$
We know that, condition for two quadratic equations having a common root is
$(c_1a_2-c_2a_1)^2 = (a_1b_2-a_2b_1) (b_1c_2-b_2c_1)$
or $[(c)(b)-(a)(a)]^2 = [(a)(c)-(b)(b)] [(b)(a)-(c)(c)]$
or $[bc-a^2]^2 = [ca-b^2] [ab-c^2]$
or $b^2c^2-2a^2bc+a^4=a^2bc-ac^3-ab^3+b^2c^2$
or $a^4=3a^2bc-ac^3-ab^3$
or $a^3=3abc-c^3-b^3$
or $a^3+b^3+c^3-3abc=0$
or $a^3+b^3+c^3-3abc=0$
or $(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0$
or Either $(a+b+c)=0$ or $(a^2+b^2+c^2-ab-bc-ca)=0$
or Either $a+b+c=0$ or $2a^2+2b^2+2c^2-2ab-2bc-2ca=0$
or Either $a+b+c=0$ or $(a-b)^2+(b-c)^2+(c-a)^2=0$
or Either $a+b+c=0$ or $a-b=0, b-c=0,c-a$
or Either $a+b+c=0$ or $a=b,b=c,c=a$
or Either $a+b+c=0$ or $a=b=c$
Thus, $ax^2+bx+c=0$ and $bx^2+cx+a=0$ have one root common if $a+b+c=0$ or $a=b=c$
Prove that if the equations $x^2+bx+ca=0$ and $x^2+cx+ab=0$ have a common root, their other root will satisfy $x^2+ax+bc=0$

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Solution :6
Given quadratic equations are
$x^2+bx+ca=0$ and $x^2+cx+ab=0$
Let $x$ be the common solution of the equations $x^2+bx+ca=0$ and $x^2+cx+ab=0$ then
$\frac{x^2}{ \begin{vmatrix} b & ca \\ c & ab \end{vmatrix} }=\frac{x}{\begin{vmatrix} ca & 1 \\ ab & 1 \end{vmatrix}}=\frac{1}{\begin{vmatrix} 1 & b \\ 1 & c \end{vmatrix}}$
Comparing first two, we get
$x=\frac{ \begin{vmatrix} b & ca \\ c & ab \end{vmatrix} } {\begin{vmatrix} ca & 1 \\ ab & 1 \end{vmatrix} }$
or $x=\frac{ab^2-ac^2}{ac-ab}$
or $x=-(b+c)$
Again, comparing last two, we get
$x= \frac{\begin{vmatrix} ca & 1 \\ ab & 1 \end{vmatrix}}{\begin{vmatrix} 1 & b \\ 1 & c \end{vmatrix}}$
or $x=\frac{ac-ab}{c-b}$
or $x=a$
Using both values of x, we get
$a=-(b+c)$
or $a+b+c=0$
Now, suppose $\alpha$ be the other root of $x^2+bx+ca=0$ , then
$x+\alpha=-b$
or $a+\alpha=-b$
or $\alpha=-a-b$
Next, suppose $\beta$ be the other root of $x^2+cx+ab=0$ , then
$x+\beta=-c$
or $a+\beta=-c$
or $\beta=-a-c$
Now, new quadratic equation containing other roots $\alpha, \beta$ is given by
$x^2-(\alpha+ \beta)x+ (\alpha \beta)=0$
or $x^2-(-a-b-a-c)x+ (a-b)(a-c)=0$
or $x^2-(-2a-b-c)x+ (a^2-ab-ac+bc)=0$
or $x^2-(-2a+a)x+ (0+bc)=0$
or $x^2+ax+ bc=0$