Question

3. Find the zeros of the polynomial 2x²+5x-12 and verify the relation
between its zeros and co-efficients



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Answer 1

Solution

Given :-

• Equation, 2x² + 5x - 12 = 0

Find :-

• Zeroes of this equation, & Relation between its zeros and coefficients

Explanation

Important Formula

★Sum of Zeroes = -(Coefficient of x)/(Coefficient of x²)

★ Product of zeroes = (Constant part)/(Coefficient of x²)

Let,

Zeroes of this equation be p & q

So, Now Calculate relation

relationbetween its zeros and co-efficients

==> Sum of Zeroes = -(5)/2

==> p + q = -5/2 ----------(1)

And,

==> Product of zeroes = -12/2

==> P . q = -6 -------------(2)

Now, Calculate Zeroes of this equation,

==> , 2x² + 5x - 12 = 0

==> 2x² + 8x - 3x - 12 = 0

==> 2x(x+4) - 3(x+4) = 0

==> (x+4)(2x-3) = 0

==> (x +4) = 0 Or, (2x-3) = 0

==> x = -4 Or, x = 3/2

Here, Zeroes of this equation be

• p = -4
• q = 3/2

Now, check relation

==> Sum of zeroes = -4 + 3/2

==> p + q = (-4*2+3)/2

==> p + q = (-8+3)/2

==> p + q -5/3 -----------(3)

And,

==> Product of zeroes = -4 * 3/2

==> P . q = -2 * 3

==> P . q = -6 ---------------(4)

Equ(1) , (2) & equ(3) , (4) are equal

Hence , we can say that Relation is satisfied .

_________________

Answer 2

✯✯ QUESTION ✯✯

Find the zeros of the polynomial 2x²+5x-12 and verify the relation between its zeros and co-efficients..

━━━━━━━━━━━━━━━━━━━━

✰✰ ANSWER ✰✰

$\implies\tt{{2x}^{2}+5x-12}$

By Splitting Middle Term : -

$\implies\tt{{2x}^{2}+(8x-3x)-12=0}$

$\implies\tt{2x(x+4)-3(x+4)=0}$

$\implies\tt{(2x-3)(2x+4)=0}$

• x = 3/2
• x = -4

☛So , 3/2 and -4 are the zeroes of polynomial 2x²+5x-12..

_______________________

➮HERE : -

• a = 2
• b = 5
• c = -12

★Sum of Zeroes : -

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

$\implies\tt{\dfrac{3}{2}+\dfrac{(-4)}{1}=\dfrac{-5}{2}}$

$\implies\tt{\dfrac{3}{2}-\dfrac{4}{1}=\dfrac{-5}{2}}$

$\implies\tt{\dfrac{3-8}{2}=\dfrac{-5}{2}}$

$\implies\tt{\dfrac{-5}{2}=\dfrac{-5}{2}}$

$\orange\longmapsto\:\large\underline{\boxed{\bf\green{L.H.S}\pink{=}\red{R.H.S}}}$

★Product of Zeroes : -

$\implies\tt{\alpha\beta=\dfrac{c}{a}}$

$\implies\tt{\dfrac{3}{2}\times{-4}=\dfrac{-12}{2}}$

$\implies\tt{\cancel\dfrac{-12}{2}=\cancel\dfrac{-12}{2}}$

$\implies\tt{-6=-6}$

$\red\longmapsto\:\large\underline{\boxed{\bf\pink{L.H.S}\orange{=}\purple{R.H.S}}}$

✺ HENCE VERIFIED ✺