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Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and their coefficients:
$p(x)\ =\ x^2\ +\ 2\sqrt{2}x\ –\ 6$
Given:
$f(x) = x^2 + 2\sqrt{2}x – 6$
To find:
Here, we have to find the zeros of f(x).
Solution:
To find the zeros of f(x), we have to put $f(x)=0$.
This implies,
$x^2 +2\sqrt{2}x – 6 = 0$
$x^2 +3\sqrt{2}x -\sqrt{2}x – 6= 0$
$x(x +3\sqrt{2}) -\sqrt{2}(x - 3\sqrt{2}) = 0$
$(x +3\sqrt{2})(x -\sqrt{2}) = 0$
$x+3\sqrt{2}=0$ and $x-\sqrt{2}=0$
$x =-3\sqrt{2}$ and $x = \sqrt{2}$
Therefore, the zeros of the quadratic equation $f(x) = x^2 +2\sqrt{2}x – 6$ are $\sqrt{2}$ and $-3\sqrt{2}$.
Verification:
We know that,
Sum of zeros $= -\frac{coefficient of x}{coefficient of x^2}$
$= –\frac{2\sqrt{2}}{1}$
$=-2\sqrt{2}$
Sum of the zeros of $f(x)=\sqrt{2}+(-3\sqrt{2})=-2\sqrt{2}$
Product of roots $= \frac{constant}{coefficient of x^2}$
$= \frac{(-6)}{1}$
$= -6$
Product of the roots of $f(x)=\sqrt{2}\times(-3\sqrt{2})=-6$
Hence, the relationship between the zeros and their coefficients is verified.
Updated on: 10-Oct-2022
5K+ Views
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