Tag: vieta’s formula for quadratic equations

Questions Related to vieta’s formula for quadratic equations

If $\alpha$ and $\beta$ are the roots of the equation $ax^{2} \, + \, bx \, + \, c \, = \, 0$. The equation whose roots are as given below.
$\dfrac{\alpha }{\beta } \, ,\dfrac{\beta }{\alpha}$ is $acx^2 \, - \, (b^2 \, - \, 2ac) \, x \, + \, ac \, = \, 0$

maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

$\Rightarrow$  $\alpha$ and $\beta$ are roots of the equation $ax^2+bx+c=0$

$\Rightarrow$  $\alpha\beta=\dfrac{c}{a}$           ------ ( 1 )
$\Rightarrow$  $\alpha+\beta=\dfrac{-b}{a}$         ----- ( 2 )
$\Rightarrow$  $(\alpha+\beta)^2=\alpha^2+\beta^2+2\alpha\beta$
Using ( 1 ) and ( 2 ),
$\Rightarrow$  $\left(\dfrac{-b}{a}\right)^2=\alpha^2+\beta^2+2\times \dfrac{c}{a}$

$\Rightarrow$  $\dfrac{b^2}{a^2}=\alpha^2+\beta^2+\dfrac{2c}{a}$

$\therefore$   $\alpha^2+\beta^2=\dfrac{b^2}{a^2}-\dfrac{2c}{a}$

$\therefore$  $\alpha^2+\beta^2=\dfrac{b^2-2ac}{a^2}$          ------ ( 3 )
Now,
$\Rightarrow$  $\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}=\dfrac{\alpha^2+\beta^2}{\alpha\beta}$

                     $=\dfrac{\dfrac{b^2-2ac}{a^2}}{\dfrac{c}{a}}$           [ Using ( 1 ) and ( 3 ) ]

$\Rightarrow$  $\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}=\dfrac{b^2-2ac}{ac}$           ----- ( 4 )

$\Rightarrow$  $\dfrac{\alpha}{\beta}.\dfrac{\beta}{\alpha}=1$   ---- ( 5 )
New equation,
$\Rightarrow$  $x^2-\left(\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}\right)x+\left(\dfrac{\alpha}{\beta}.\dfrac{\beta}{\alpha}\right)=0$
By Using ( 4 ) and ( 5 ),
$\Rightarrow$  $x^2-\left(\dfrac{b^2-4ac}{ac}\right)x+1=0$

$\Rightarrow$  $acx^2-(b^2-4ac)x+1=0$
$\therefore$  We can see equation given in question is correct.

A quadratic polynomial $p(x)$ with $3$ and $\dfrac{-2}{5}$  as the sum and product of zeroes, respectively is $10x^2+30x-4$

maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

$10x^2+30x-4=0$

$\alpha+\beta=-\cfrac{30}{10}=-3$
$\alpha\beta=\cfrac{-4}{10}=\cfrac{-2}{5}$
So, a quadratic polynomial $p(x)$ with $3$ and $\cfrac{-2}{5}$ as the sum and product of zeroes, respectively is $10x^2+30x-4$ is false.

If the roots of a quadratic equation are reciprocals of the roots of $ax^2 + bx + c = 0$, then what will be the coefficient of $c$?

maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

  1. $x$

  2. $x^2$

  3. $-x$

  4. $x^3$

Show answer
Correct Option: B
Explanation:

We know to find the equation with reciprocal roots, we simply put $\dfrac {1}{x}$ in the place of $x$.
so, $a(\dfrac {1}{x})^2 + b(\dfrac {1}{x}) + c = 0$
On taking LCM we get the required equation
$a + bx + cx^2 = 0$
So the coefficient of $c$ is $x^2$

Find the Quadratic Equation whose roots are Reciprocal of $ax^2 + bx + c = 0$.

maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

  1. $ax^2 + bx + c = 0$

  2. $a(\dfrac {1}{x})^2 + b(\dfrac {1}{x}) + c = 0$

  3. $3ax^2 + 2bx + c = 0$

  4. None of the above

Show answer
Correct Option: B
Explanation:

We know that for reciprocal roots, we only need to replace $x$ by $\dfrac {1}{x}$, in the given equation.
So the above equation becomes:
$a(\dfrac {1}{x})^2 + b(\dfrac {1}{x}) + c = 0$, which is the required answer.

If A.M. of the roots of a quadratic equation is $8/5$ and A.M. of their reciprocals is $8/7$, then the equation is?

maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

  1. $7x^2-16x+8=0$

  2. $3x^2-12x+7=0$

  3. $5x^2-16x+7=0$

  4. $7x^2-16x+5=0$

Show answer
Correct Option: C
Explanation:

$\Rightarrow$  Let $\alpha$ and $\beta$ are the roots of the equation.

According to the question,
$\Rightarrow$ $\dfrac{\alpha+\beta}{2}=\dfrac{8}{5}$

$\therefore$  $\alpha+\beta=\dfrac{16}{5}$                ----- ( 1 )

According to the equation,
$\Rightarrow$  $\dfrac{\dfrac{1}{\alpha}+\dfrac{1}{\beta}}{2}=\dfrac{8}{7}$
$\Rightarrow$  $\dfrac{\alpha+\beta}{2\alpha\beta}=\dfrac{8}{7}$
$\Rightarrow$  $7(\alpha+\beta)=16\alpha\beta$
$\Rightarrow$  $\dfrac{7\times \dfrac{16}{5}}{16}=\alpha\beta$
$\therefore$    $\alpha\beta=\dfrac{7}{5}$          - ------ ( 2 )
Now, new eqution,
$\Rightarrow$  $x^2-(\alpha+\beta)x+(\alpha.\beta)=0$
From ( 1 ) and ( 2 ),
$\Rightarrow$  $x^2-\dfrac{16}{5}x+\dfrac{7}{5}=0$
$\Rightarrow$  $5x^2-16x+7=0$   

If $\alpha, \beta$ are the root of a quadratic equation $x^2 - 3x+5=0$, then the equation whose roots are $(\alpha^2 - 3 \alpha +7)$ and $(\beta^2 -3\beta +7)$ is

maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

  1. $x^2 +4x+1=0$

  2. $x^2 -4x+4=0$

  3. $x^2 -4x-1=0$

  4. $x^2 +2x+3=0$

Show answer
Correct Option: B
Explanation:

Since $\alpha, \beta$ are the root of equation $x^2-3x+5=0$
So, $\alpha^2-3\alpha +5=0$
$\beta^2 -3\beta +5=0$
$\therefore \alpha^2 -3\alpha =-5$
$\beta^2 - 3 \beta =-5$
Putting in $(\alpha^2 - 3 \alpha +7) $  &  $(\beta^2 - 3\beta +7)$    ....... (1)
$-5 +7, -5 +7$
$\therefore$ 2 and 2 are the roots
$\therefore$ The required equation is $x^2 - 4x+4=0$

If the roots of ${a _1}{x^2}\, + \,{b _1}x\, + \,{c _1}\, = \,0$ are ${\alpha _1},\,{\beta _1},\,$ and those of ${a _2}{x^2}\, + \,{b _2}x\, + {c _2}\, = \,0$ are ${\alpha _2}\,,{\beta _2}$ such that ${\alpha _1}\,{\alpha _2} = \,{\beta _1}\,{\beta _2}\, = \,1$, then

maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

  1. $\dfrac{{{a _1}}}{{{a _2}}} = \,\dfrac{{{b _1}}}{{{b _2}}}\, = \,\dfrac{{{c _1}}}{{{c _2}}}$

  2. $\dfrac{{{a _1}}}{{{c _2}}} = \,\dfrac{{{b _1}}}{{{b _2}}}\, = \,\dfrac{{{c _1}}}{{{a _2}}}$

  3. ${a _1}\,{a _2}\, = \,{b _1}\,{b _2}\, = \,{c _1}\,{c _2}$

  4. None of these

Show answer
Correct Option: A
Explanation:

$a _1x^2+b _1x+c _1=0$

$\alpha _1+\beta _1=\cfrac{-b _1}{a _1}$
$\alpha _1\beta _1=\cfrac{c _1}{a _1}$

$a _2x^2+b _2x+c _2=0$
$\alpha _2+\beta _2=\cfrac{-b _2}{a _2}$
$\alpha _2\beta _2=\cfrac{c _2}{a _2}$

$\alpha _1\alpha _2=\beta _1\beta _2=1$
$\therefore \alpha _1\beta _1\alpha _2\beta _2=(\alpha _1\alpha _1)\cdot (\beta _1\beta _1)=1\cdot 1=1$
$\therefore \alpha _1\beta _1\alpha _2\beta _2=(\alpha _1\alpha _1)\cdot (\beta _1\beta _1)=\cfrac{c _1}{a _1}\cdot \cfrac{c _2}{a _2}=1$
$\implies \cfrac{c _1}{a _2}=\cfrac{a _1}{c _2}$
Now, $\alpha _1+\beta _1=\cfrac{-b _1}{a _1}$
$\implies \cfrac{1}{\alpha _2}+\cfrac{1}{\beta _2}=\cfrac{-b _1}{a _1}$
$\implies \cfrac{\alpha _2+\beta _2}{\alpha _2\beta _2}=\cfrac{-b _1}{a _1}$
$\implies \cfrac{-b _2/a _2}{c _2/a _2}=\cfrac{-b _1}{a _1}$
$\implies \cfrac{b _2}{c _2}=\cfrac{b _1}{a _1}$
$\implies \cfrac{a _1}{c _2}=\cfrac{b _1}{b _2}$
$\therefore \cfrac{a _1}{a _2}=\cfrac{b _1}{b _2}=\cfrac{c _1}{c _2}$

If $alpha, beta$ are roots of $Ax^2 + Bx + C = 0$ and $\alpha^2, \beta^2$ are roots of $x^2 + px + q = 0$, the $p$ is equal to

maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

  1. $\dfrac{B^2 - 2AC}{A^2}$

  2. $\dfrac{2AC - B^2}{A^2}$

  3. $\dfrac{B^2 - 4AC}{A^2}$

  4. $\dfrac{4AC - B^2}{A^2}$

Show answer
Correct Option: B
Explanation:
$Ax^2+Bx+C=0$
$\alpha +\beta =\dfrac{-B}{A}$
$\alpha\beta =\dfrac{C}{A}$
Now it roots are $\alpha^2$ & $\beta^2$
Then equation will be
$(x-\alpha^2)(x-\beta^2)=0$
$x^2-x\beta^2-x\alpha^2+\alpha^2\beta^2=0$
$x^2-)(\alpha^2+\beta^2)x+\alpha^2\beta^2=0$
$x^2+[-(\alpha^2+\beta^2)]x+\alpha^2\beta^2=0$
$\alpha +\beta =\dfrac{-B}{-A}$
$\Rightarrow \alpha^2+\beta^2+2\alpha\beta =\dfrac{B^2}{A^2}$
$\alpha^2+\beta^2=\dfrac{B^2}{A^2}-\dfrac{2C}{A}$
$-(\alpha^2+\beta^2)=\dfrac{2C}{A}-\dfrac{B^2}{A^2}=\dfrac{2AC-B^2}{A^2}$
$p=\dfrac{2AC-B^2}{A^2}$.

If $\alpha+\beta$$=-2$ and ${\alpha}^{3}+{\beta}^{3}$$=-56$ then the quadratic equation whose roots are $\alpha,\beta$ is 

maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

  1. ${ x }^{ 2 }+2x-16$$=0$

  2. ${x}^{2}+2x-15$$=0$

  3. ${x}^2+2x-12$$=0$

  4. ${x}^{2}+2x-8$$=0$

Show answer
Correct Option: D
Explanation:

$\Rightarrow$  $\alpha+\beta=-2$             ------ ( 1 )


$\Rightarrow$  $\alpha^3+\beta^3=-56$


$\Rightarrow$  $(\alpha+\beta)^3=\alpha^3+\beta^3+3\alpha^2\beta+3\alpha\beta^2$

$\Rightarrow$  $(\alpha+\beta)^3=\alpha^3+\beta^3+3\alpha\beta(\alpha+\beta)$

$\Rightarrow$  $(-2)^3=-56+3\alpha\beta(-2)$            [ Using ( 1 ) and ( 2 ) ]

$\Rightarrow$  $-8+56=-6\alpha\beta$

$\Rightarrow$  $48=-6\alpha\beta$

$\Rightarrow$  $\alpha\beta=-8$                      ----- ( 2 )

The required quadratic equation,

$x^2-(\alpha+\beta)x+(\alpha\beta)=0$

Using ( 1 ) and ( 3 ) we get,
$\Rightarrow$  $x^2+2x-8=0$

If $\alpha \neq \beta$ but $\alpha^2 = 5 \alpha -3$ and $\beta^2 = 5\beta -3$, then the equation whose roots are $\dfrac{\alpha}{\beta}$ and $\dfrac{\beta}{\alpha}$is

maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

  1. $3x^2 - 25x+3=0$

  2. $x^2 +5x-3=0$

  3. $x^2 -5x+3=0$

  4. $3x^2 - 19x+3=0$

Show answer
Correct Option: D
Explanation:
${\alpha}^{2}=5\alpha-3$

${\alpha}^{2}-5\alpha+3=0$

$\alpha=\dfrac{5\pm\sqrt{{5}^{2}-4\times 1\times 3}}{2}$

$\alpha=\dfrac{5\pm\sqrt{25-12}}{2}$

$\alpha=\dfrac{5\pm\sqrt{13}}{2}$

${\beta}^{2}=5\beta-3$

${\beta}^{2}-5\beta+3=0$

$\beta=\dfrac{5\pm\sqrt{{5}^{2}-4\times 1\times 3}}{2}$

$\beta=\dfrac{5\pm\sqrt{25-12}}{2}$

$\beta=\dfrac{5\pm\sqrt{13}}{2}$

Given:$\alpha\neq\,\beta$

Let $\alpha=\dfrac{5+\sqrt{13}}{2}$ and $\beta=\dfrac{5-\sqrt{13}}{2}$

$\Rightarrow\,\dfrac{\alpha}{\beta}=\dfrac{\dfrac{5+\sqrt{13}}{2}}{\dfrac{5-\sqrt{13}}{2}}$

$=\dfrac{5+\sqrt{13}}{5-\sqrt{13}}$

$=\dfrac{5+\sqrt{13}}{5-\sqrt{13}}\times \dfrac{5+\sqrt{13}}{5+\sqrt{13}}$

$=\dfrac{25+13+10\sqrt{13}}{25-13}$

$=\dfrac{38+10\sqrt{13}}{12}$

$=\dfrac{19+5\sqrt{13}}{6}$

$\Rightarrow\,\dfrac{\beta}{\alpha}=\dfrac{\dfrac{5-\sqrt{13}}{2}}{\dfrac{5+\sqrt{13}}{2}}$

$=\dfrac{5-\sqrt{13}}{5+\sqrt{13}}$

$=\dfrac{5-\sqrt{13}}{5+\sqrt{13}}\times \dfrac{5-\sqrt{13}}{5-\sqrt{13}}$

$=\dfrac{25+13-10\sqrt{13}}{25-13}$

$=\dfrac{19-10\sqrt{13}}{12}$

$=\dfrac{19-5\sqrt{13}}{6}$

Sum of the zeroes$=\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}$

$=\dfrac{19+5\sqrt{13}}{6}+\dfrac{19-5\sqrt{13}}{6}$

$=\dfrac{19+5\sqrt{13}+19-5\sqrt{13}}{6}$

$=\dfrac{2\times 19}{6}=\dfrac{19}{3}$

Product of the zeroes$=\dfrac{\alpha}{\beta}\times\dfrac{\beta}{\alpha}$

$=\dfrac{19+5\sqrt{13}}{6}\times\dfrac{19-5\sqrt{13}}{6}$

$=\dfrac{361-25\times 13}{36}=\dfrac{361-325}{36}=\dfrac{36}{36}=1$

Now,we know the foumula for finding quadratic equations,
${x}^{2}-\left(sum\,of \,the \,zeroes\right)x+product\,of\,the \,zeroes=0$

${x}^{2}-\dfrac{19}{3}x+1=0$

Hence the equation is $3{x}^{2}-19x+3=0$