
If the difference of the roots of the equation $x^{2}-2px+8= 0$ is 2 then p equals
A. $\pm 6$
B. $\pm 2$
C. $2, -6$
D. $2, 6$
Answer
164.4k+ views
Hint:The values of the variable that satisfy a quadratic equation are known as its roots. The solutions or zeros of the quadratic equation are other names for them.
We have to find the sum of the roots and the product of the roots. Then, applying their values in the given condition will give the value of p.
Formula Used:
General quadratic equation: – \[A{x^2} + Bx + C = 0\]
Let, the roots of the above quadratic equation be $\alpha $ and $\beta $
Therefore,
Sum of roots, $\alpha + \beta = \dfrac{{ - B}}{A}$
Product of roots, $\alpha \beta = \dfrac{C}{A}$
Complete step by step Solution:
We have the equation $x^{2}-2px+8= 0$
Let, $\alpha$and $\beta$ be the roots of the above equation.
Comparing the above equation with $ax^{2}+bx+c=0$, we can say that b
is the coefficient of x and a is the coefficient of $x^{2}$where c is the constant term.
Thus, $a=1$, $b=-p$, $c=8$
According to the question, the difference of roots is 2 which is $\alpha -\beta =2$
Now, taking square on both sides of above equation we get $\left ( \alpha -\beta \right )^{2}=4$
Here, $\alpha +\beta =p$ and $\alpha \beta = 8$ which is found by using the sum of the roots and the product of the roots formula.
We know $\left ( a-b \right )^{2}=a^{2}-2ab+b^{2}$
So, by expanding the square of the difference between the roots by using the above identity we get $\alpha ^{2}-2\alpha \beta +\beta ^{2}=4$
We know, $\alpha ^{2}+\beta ^{2}=\left ( \alpha +\beta \right )^{2}-2\alpha \beta$. So substituting it in the above equation we get
$\left ( \alpha +\beta \right )^{2}-4\alpha \beta =4$
Now substitute the values of the sum of roots and product of roots in the above equation
$p^{2}-4\times 8=4\Rightarrow p^{2}=4+32$
$\rightarrow p=\pm 6$
Hence, the correct option is (A).
Note: Many students make mistakes opening up $\alpha ^{2}+\beta ^{2}$. They directly write it as $\left ( \alpha +\beta \right )^{2}$, which is wrong. The correct way is by using its correct identity and then expanding it. Also, after finding the value of p, we always have to check with the given equation if its value satisfies the equation or not.
We have to find the sum of the roots and the product of the roots. Then, applying their values in the given condition will give the value of p.
Formula Used:
General quadratic equation: – \[A{x^2} + Bx + C = 0\]
Let, the roots of the above quadratic equation be $\alpha $ and $\beta $
Therefore,
Sum of roots, $\alpha + \beta = \dfrac{{ - B}}{A}$
Product of roots, $\alpha \beta = \dfrac{C}{A}$
Complete step by step Solution:
We have the equation $x^{2}-2px+8= 0$
Let, $\alpha$and $\beta$ be the roots of the above equation.
Comparing the above equation with $ax^{2}+bx+c=0$, we can say that b
is the coefficient of x and a is the coefficient of $x^{2}$where c is the constant term.
Thus, $a=1$, $b=-p$, $c=8$
According to the question, the difference of roots is 2 which is $\alpha -\beta =2$
Now, taking square on both sides of above equation we get $\left ( \alpha -\beta \right )^{2}=4$
Here, $\alpha +\beta =p$ and $\alpha \beta = 8$ which is found by using the sum of the roots and the product of the roots formula.
We know $\left ( a-b \right )^{2}=a^{2}-2ab+b^{2}$
So, by expanding the square of the difference between the roots by using the above identity we get $\alpha ^{2}-2\alpha \beta +\beta ^{2}=4$
We know, $\alpha ^{2}+\beta ^{2}=\left ( \alpha +\beta \right )^{2}-2\alpha \beta$. So substituting it in the above equation we get
$\left ( \alpha +\beta \right )^{2}-4\alpha \beta =4$
Now substitute the values of the sum of roots and product of roots in the above equation
$p^{2}-4\times 8=4\Rightarrow p^{2}=4+32$
$\rightarrow p=\pm 6$
Hence, the correct option is (A).
Note: Many students make mistakes opening up $\alpha ^{2}+\beta ^{2}$. They directly write it as $\left ( \alpha +\beta \right )^{2}$, which is wrong. The correct way is by using its correct identity and then expanding it. Also, after finding the value of p, we always have to check with the given equation if its value satisfies the equation or not.
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