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

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Question:1

              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

Question:2

         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 = \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:

find the value of k for which the pair of equation

rahul-had-some-bananas-and-he-divided-them-into-two-lots

Ruhi invested a certain amount of money in two schemes

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