
Roots of the equation \[{x^2} + bx - c = 0\], \[(b,c > 0)\] are
A) Both positive
B) Both negative
C) Of opposite sign
D) None of these
Answer
161.1k+ views
Hint: In this question, we are given a quadratic equation i.e., \[{x^2} + bx - c = 0\] where \[(b,c > 0)\]. We have to find the nature of the sign of its roots. Calculate the sum and roots of the given quadratic equation using the formula $\alpha + \beta = \dfrac{{ - B}}{A}$ and $\alpha \beta = \dfrac{C}{A}$. Then, apply the concept of the quadratic equation.
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:
Given that,
\[{x^2} + bx - c = 0\], a quadratic equation where \[(b,c > 0)\]
Compare the given equation \[{x^2} + bx - c = 0\] with the general quadratic equation.
We get, $A = 1,B = b,C = - c$
Now, let the root of the quadratic equation be \[\alpha \] and \[\beta \]
Using, the formula of the sum and the product of roots.
Sum, $\alpha + \beta = - b$ and Product, $\alpha \beta = - c$
Now, according to the question \[(b,c > 0)\]
Here, the product of the roots i.e., $pq$ is negative.
Therefore, it states that either of the roots will be negative.
Thus, the roots must be of opposite signs,
Hence, the correct option is (C).
Additional Information:Multiplication of the signs –
Neg. $( - )$ $ \times $ Neg. $( - )$ $ = $ Pos. $( + )$
Neg. $( - )$ $ \times $ Pos. $( + )$$ = $ Neg. $( - )$
Pos. $( + )$ $ \times $ Neg. $( - )$ $ = $ Neg. $( - )$
Pos. $( + )$ $ \times $ Pos. $( + )$ $ = $ Pos. $( + )$
Note: The values of $x$ that fulfil a given quadratic equation \[A{x^2} + Bx + C = 0\] are known as its roots. They are, in other words, the values of the variable $\left( x \right)$ that satisfy the equation. The roots of a quadratic function are the $x - $ coordinates of the function's $x - $ intercepts. Because the degree of a quadratic equation is $2$, it can only have two roots.
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:
Given that,
\[{x^2} + bx - c = 0\], a quadratic equation where \[(b,c > 0)\]
Compare the given equation \[{x^2} + bx - c = 0\] with the general quadratic equation.
We get, $A = 1,B = b,C = - c$
Now, let the root of the quadratic equation be \[\alpha \] and \[\beta \]
Using, the formula of the sum and the product of roots.
Sum, $\alpha + \beta = - b$ and Product, $\alpha \beta = - c$
Now, according to the question \[(b,c > 0)\]
Here, the product of the roots i.e., $pq$ is negative.
Therefore, it states that either of the roots will be negative.
Thus, the roots must be of opposite signs,
Hence, the correct option is (C).
Additional Information:Multiplication of the signs –
Neg. $( - )$ $ \times $ Neg. $( - )$ $ = $ Pos. $( + )$
Neg. $( - )$ $ \times $ Pos. $( + )$$ = $ Neg. $( - )$
Pos. $( + )$ $ \times $ Neg. $( - )$ $ = $ Neg. $( - )$
Pos. $( + )$ $ \times $ Pos. $( + )$ $ = $ Pos. $( + )$
Note: The values of $x$ that fulfil a given quadratic equation \[A{x^2} + Bx + C = 0\] are known as its roots. They are, in other words, the values of the variable $\left( x \right)$ that satisfy the equation. The roots of a quadratic function are the $x - $ coordinates of the function's $x - $ intercepts. Because the degree of a quadratic equation is $2$, it can only have two roots.
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