Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients: \[{x^2} + 10x + 16\] A. 8 or 2 B. – 8 or 2 C. – 8 or – 2 D. 8 or – 2
ANSWER
Verified
Hint: To find the zeros of the given quadratic equation we have to only equate the given equation with zero and then split the middle term of the equation to get the roots of the equation. And then we can verify the relationship by using the formula for sum and product of the roots of the quadratic equation.
Complete step-by-step answer: Let f(x) be a quadratic equation and, f(x) = \[{x^2} + 10x + 16\]-----------(1) Now to find the zeros of equation 1. We had to equate that with zero. So, \[{x^2} + 10x + 16\] = 0-----------(2)
Now to find the zeros of the above equation. We had to split the middle term of the quadratic equation. So, splitting the middle term of the above equation. We get, \[{x^2} + 2x + 8x + 16 = 0\] On making factors of above equation. It becomes, \[x\left( {x + 2} \right) + 8\left( {x + 2} \right) = 0\]
Now taking (x + 2) as the common factor from the above equation. We get, \[\left( {x + 2} \right)\left( {x + 8} \right) = 0\] So, x = - 2 or – 8 So, the zeroes of a given quadratic equation will be x = - 2 or – 8. Now x = - 2 and x = - 8 are the roots of the quadratic equation. So, let \[\alpha = - 2\] and \[\beta = - 8\]
Now we know that if for any quadratic equation \[a{x^2} + bx + c\]. If \[\alpha \]and \[\beta \] are the zeros of this quadratic equation then, Sum of \[\alpha \]and \[\beta \] = \[\alpha \] + \[\beta \] = \[ - \dfrac{{{\text{Coefficient of }}x}}{{{\text{Coefficient of }}{x^2}}}\] = \[ - \dfrac{b}{a}\] And product of \[\alpha \]and \[\beta \] = \[\alpha \times \beta \] = \[\dfrac{{{\text{Constant term}}}}{{{\text{Coefficient of }}{x^2}}}\] = \[\dfrac{c}{a}\]
Now, we have to prove this result for equation 1. Proving sum of zeros So, – 2 + (– 8) = \[ - \dfrac{{{\text{Coefficient of }}x}}{{{\text{Coefficient of }}{x^2}}}\] = – 10 – 10 = – 10 Hence, the sum of zeros is equal to \[ - \dfrac{{{\text{Coefficient of }}x}}{{{\text{Coefficient of }}{x^2}}}\] in the given quadratic equation. Proving product of zeros So, – 2 * (– 8) = \[\dfrac{{{\text{Constant term}}}}{{{\text{Coefficient of }}{x^2}}}\] = – 16 – 16 = – 16
Hence, the product of zeros is equal to \[\dfrac{{{\text{Constant term}}}}{{{\text{Coefficient of }}{x^2}}}\] in the given quadratic equation. Hence the relation between zeros and coefficient of the given quadratic equation is verified. Hence, the correct option will be C.
Note: Whenever we come up with this type of problem then there is also another method to find the zeros of the quadratic equation. If \[a{x^2} + bx + c\] be a quadratic equation, then its zeros will be equal to \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]. And after that to verify the relation between zeros and coefficients of the quadratic equation we use the result that is if \[\alpha \]and \[\beta \] are the zeros of this quadratic equation then, \[\alpha \] + \[\beta \] = \[ - \dfrac{{{\text{Coefficient of }}x}}{{{\text{Coefficient of }}{x^2}}}\] and \[\alpha \times \beta \] = \[\dfrac{{{\text{Constant term}}}}{{{\text{Coefficient of }}{x^2}}}\].