Find the zeros of the quadratic polynomial x² + 7x + 12

Question1:-Find the zeros of the quadratic polynomial x² + 7x + 12, and verify the relation between the zeros and its coefficients.

Solution:- We have, f(x) = x² + 7x + 12

f(x) = x² + 7x + 12

⇒ f(x) = x² + 4x + 3x + 12

⇒ f(x) = x(x + 4) + 3(x + 4)

⇒ f(x) = (x + 4)(x + 3)

The zeros of f(x) are given by

f(x) = 0

(x + 4)(x + 3) = 0

⇒ x + 4 = 0 or x + 3 = 0

⇒ x = -4 or x = -3

Thus, the zeros of f(x) = x² + 7x + 12 are α = -4 and β = -3

Now, Sum of the zeros = α + β = (- 4) + (-3) = -7

Again, sum of the zeros = -b/a = -(7)/1 = -7

And

Product of the zeros = α×β = (-4)(-3) = 12

Again product of the zeros = c/a = 12/1 = 12

Hence, the relation verified

Question2:-Find the zeros of the quadratic polynomial f(x) = 6x² – 3, and verify the relation between the zeros and its coefficients.

Solution:We have, f(x) = 6x² – 3

$\Rightarrow f(x) =(\sqrt{6}x)^2-\sqrt{3})^2$

$\Rightarrow f(x) = (\sqrt{6}x-\sqrt{3})(\sqrt{6}x+\sqrt{3})$

The zeros of f(x) are given by

f(x) = 0

$(\sqrt{6}x-\sqrt{3})(\sqrt{6}x+\sqrt{3})=0$

$\Rightarrow \sqrt{6}x-\sqrt{3} = 0$ or $\sqrt{6}x+\sqrt{3}=0$

$\Rightarrow x = \dfrac{\sqrt{3}}{\sqrt{6}}$ or $x = -\dfrac{\sqrt{3}}{\sqrt{6}}$

$\Rightarrow x = \dfrac{1}{\sqrt{2}}$ or $x = -\dfrac{1}{\sqrt{2}}$

Hence, the zeros of f(x) = 6x² – 3 are : $\alpha=\dfrac{1}{\sqrt{2}}$ and $\beta = -\dfrac{1}{\sqrt{2}}$

Now, sum of the zeros = $\alpha + \beta = \dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}=0$

Again, sum of the zeros $= -\dfrac{b}{a} = -\dfrac{0}{6}=0$

And

Product of the zeros = $t\alpha \times \beta = \dfrac{1}{\sqrt{2}}\times (-\dfrac{1}{\sqrt{2}})=-\dfrac{1}{2}$

Again product of the zeros = $\dfrac{c}{a} = \dfrac{-3}{6} = -\dfrac{1}{2}$

Hence, the relation verified

Some other question: