Find the roots of
\[{x^2} - 4ax + 4{a^2} - {b^2} = 0\]
Answer
565.5k+ views
Hint: To solve this we have to use the formula that is used to find the value of roots of a quadratic equation. the formula is \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] put the values of \[a,b,c\] in order to get the values of the roots of that equation. while taking out from the root check sign of that term also.
Complete answer: Given,
A quadratic equation
\[{x^2} - 4ax + 4{a^2} - {b^2} = 0\]
To find,
The roots of the quadratic equation \[{x^2} - 4ax + 4{a^2} - {b^2} = 0\]
To find the roots of this quadratic equation we have to use the formula that is used to find the roots of the general quadratic equation.\[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
Here, \[b\] is the coefficient of \[x\]
\[a\] is the coefficient of \[{x^2}\]
\[c\] is the constant term of the quadratic equation.
So in the given equation values of \[a,b,c\] are as follows.
\[ \Rightarrow a = 1\],
\[ \Rightarrow b = - 4a\] and
\[ \Rightarrow c = 4{a^2} - {b^2}\]
On putting all these values in the formula of finding the roots of the quadratic equation.
\[roots = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
\[roots = \dfrac{{ - \left( { - 4} \right) \pm \sqrt {{{\left( { - 4a} \right)}^2} - 4\left( 1 \right)\left( {4{a^2} - {b^2}} \right)} }}{{2\left( 1 \right)}}\]
On further solving
\[roots = \dfrac{{4 \pm \sqrt {16{a^2} - 16{a^2} + 4{b^2}} }}{2}\]
On simplifying
\[roots = \dfrac{{4 \pm \sqrt {4{b^2}} }}{2}\]
Now taking \[4{b^2}\] outside of bracket
\[roots = \dfrac{{4 \pm 2b}}{2}\]
On further solving
\[roots = 2 \pm b\]
First taking a positive sign we get the first root of the equation.
\[roots = 2 + b\]
Now taking the negative sign we get the second root of the equation.
\[roots = 2 - b\]
Final answer:
The roots of the quadratic equation:
\[ \Rightarrow root = 2 + b\] and
\[ \Rightarrow root = 2 - b\]
Probability of getting both boys if the elder of them is boy \[P(Y/X) = \dfrac{1}{2}\].
Note:
To solve this type of question we have to use the formula that is used to find the roots of the quadratic equation. then assign the value of \[a,b,c\] according to the equation. then put all those values in the formula of roots. You may commit a mistake in assigning the value of \[a,b,c\] and solving the root part of the formula.
Complete answer: Given,
A quadratic equation
\[{x^2} - 4ax + 4{a^2} - {b^2} = 0\]
To find,
The roots of the quadratic equation \[{x^2} - 4ax + 4{a^2} - {b^2} = 0\]
To find the roots of this quadratic equation we have to use the formula that is used to find the roots of the general quadratic equation.\[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
Here, \[b\] is the coefficient of \[x\]
\[a\] is the coefficient of \[{x^2}\]
\[c\] is the constant term of the quadratic equation.
So in the given equation values of \[a,b,c\] are as follows.
\[ \Rightarrow a = 1\],
\[ \Rightarrow b = - 4a\] and
\[ \Rightarrow c = 4{a^2} - {b^2}\]
On putting all these values in the formula of finding the roots of the quadratic equation.
\[roots = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
\[roots = \dfrac{{ - \left( { - 4} \right) \pm \sqrt {{{\left( { - 4a} \right)}^2} - 4\left( 1 \right)\left( {4{a^2} - {b^2}} \right)} }}{{2\left( 1 \right)}}\]
On further solving
\[roots = \dfrac{{4 \pm \sqrt {16{a^2} - 16{a^2} + 4{b^2}} }}{2}\]
On simplifying
\[roots = \dfrac{{4 \pm \sqrt {4{b^2}} }}{2}\]
Now taking \[4{b^2}\] outside of bracket
\[roots = \dfrac{{4 \pm 2b}}{2}\]
On further solving
\[roots = 2 \pm b\]
First taking a positive sign we get the first root of the equation.
\[roots = 2 + b\]
Now taking the negative sign we get the second root of the equation.
\[roots = 2 - b\]
Final answer:
The roots of the quadratic equation:
\[ \Rightarrow root = 2 + b\] and
\[ \Rightarrow root = 2 - b\]
Probability of getting both boys if the elder of them is boy \[P(Y/X) = \dfrac{1}{2}\].
Note:
To solve this type of question we have to use the formula that is used to find the roots of the quadratic equation. then assign the value of \[a,b,c\] according to the equation. then put all those values in the formula of roots. You may commit a mistake in assigning the value of \[a,b,c\] and solving the root part of the formula.
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