Question

find the zeros of the quadratic polynomial 3√2x^2+13^x+6√2 and verify the relation between zeros and coefficient​

Answer 1

f(x)=3√2x^2+13x+6√2=0

splitting middle term

3√2x^2+9x+4x+6√2=0

3x(√2x+3)2√2(√2x+3)=0

(3x+2√2)(√2x+3)=0

case(1)

3x+2√2=0

x=-2√2/3 ,α=-2√2/3

case(2)

√2x+3=0

x=-3/√2 ,β=-3/√2

relation

a=3√2 ,b=13,c=6√2

α*β=c/a

-2√2/3*-3/√2 = 6√2/3√2

2=2

prooved

α+β=-b/a

-2√2/3+-3/√2=-13/3√2

-9-4/3√2=-13/3√2

-13/3√2=-13/3√2

prooved

Answer 2

3√2x² + 13x + 6√2 = 0

___________ [GIVEN]

• We have to find the zeros of the quadratic polynomial and verify the relation between zeros and coefficient.

_____________________________

3√2x² + 13x + 6√2 = 0

It's like that ax² + bx + c = 0

Here..

a = 3√2

b = 13

c = 6√2

We have to find the numbers whose sum is b and product is c.

=> 3√2x² + 9x + 4x + 6√2 = 0

=> 3x (√2x + 3) + 2√2 (√2x + 3) = 0

=> (3x + 2√2) (√2x + 3) = 0

• 3x + 2√2 = 0

=> 3x = - 2√2

=> x = $\dfrac{ - 2 \sqrt{2} }{3}$

• √2x + 3 = 0

=> √2x = - 3

=> x = $\dfrac{ - 3}{ \sqrt{2} }$

______________________________

Let $\alpha$ = $\dfrac{ - 2 \sqrt{2} }{3}$ and $\beta$ = $\dfrac{ - 3}{ \sqrt{2} }$

Now..

• $\alpha$ + $\beta$ = $\dfrac{-b}{a}$

=> $\dfrac{ - 2 \sqrt{2} }{3}$ + $\dfrac{ (- 3)}{ \sqrt{2} }$ = $\dfrac{-13}{3 \sqrt{2} }$

=> $\dfrac{ - 2 \sqrt{2} \: \times \: \sqrt{2} \: - \: 9}{3 \sqrt{2} }$ = $\dfrac{-13}{3 \sqrt{2} }$

=> $\dfrac{-13}{3 \sqrt{2} }$ = $\dfrac{-13}{3 \sqrt{2} }$

• $\alpha$ $\beta$ = $\dfrac{c}{a}$

=> $\dfrac{ - 2 \sqrt{2} }{3}$ × $\dfrac{ (- 3)}{ \sqrt{2} }$ = $\dfrac{6 \sqrt{2} }{3 \sqrt{2} }$

=> $\dfrac{6 \sqrt{2} }{3 \sqrt{2} }$ = $\dfrac{6 \sqrt{2} }{3 \sqrt{2} }$

=> $\dfrac{6}{3}$ = $\dfrac{6}{3}$

=> 2 = 2

_____________ [ANSWER]

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