Question

in the adjoining figure , abcd is a square and triangle edc is an equilateral triangle ,prove that 1) ae= be 2) angle Dae= 15 degree

Answer 1

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The zeroes of quadratic polynomial are $-\dfrac{3}{\sqrt{2}}$ and $\dfrac{1}{2\sqrt{2}}$

Step-by-step explanation:

Given: $4x^2+5\sqrt{2}x-3$

Factor the trinomial:-

$\Rightarrow 4x^2+5\sqrt{2}x-3$

$\Rightarrow 4x^2+6\sqrt{2}x-\sqrt{2}x-3$

$\Rightarrow 2\sqrt{2}x(\sqrt{2}x+3)-1(\sqrt{2}x-3)$

$\Rightarrow (\sqrt{2}x+3)(2\sqrt{2}x-1)$

Now we will set each factor to 0 ans solve for x

$\sqrt{2}x+3=0\Rightarrow x=-\dfrac{3}{\sqrt{2}}$

$2\sqrt{2}x-1=0\Rightarrow x=\dfrac{1}{2\sqrt{2}}$

Hence, The zeroes of quadratic polynomial are $-\dfrac{3}{\sqrt{2}}$ and $\dfrac{1}{2\sqrt{2}}$

Answer 2

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The zeroes of quadratic polynomial are $-\dfrac{3}{\sqrt{2}}$ and $\dfrac{1}{2\sqrt{2}}$

Step-by-step explanation:

Given: $4x^2+5\sqrt{2}x-3$

Factor the trinomial:-

$\Rightarrow 4x^2+5\sqrt{2}x-3$

$\Rightarrow 4x^2+6\sqrt{2}x-\sqrt{2}x-3$

$\Rightarrow 2\sqrt{2}x(\sqrt{2}x+3)-1(\sqrt{2}x-3)$

$\Rightarrow (\sqrt{2}x+3)(2\sqrt{2}x-1)$

Now we will set each factor to 0 ans solve for x

$\sqrt{2}x+3=0\Rightarrow x=-\dfrac{3}{\sqrt{2}}$

$2\sqrt{2}x-1=0\Rightarrow x=\dfrac{1}{2\sqrt{2}}$

Hence, The zeroes of quadratic polynomial are $-\dfrac{3}{\sqrt{2}}$ and $\dfrac{1}{2\sqrt{2}}$

HOPE SO IT WILL HELP....