Question

8. Find the zeroes of the quadratic polynomial 2x2 – 7x -15 and verify the relationship between the zeroes and the coefficients.
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Answer 1

$\large\red{\underline{\underline{\bf{\blue{Given}}}}}$

• $2 {x}^{2} - 7x - 15 = 0$

$\large\red{\underline{\underline{\bf{\blue{Find}}}}}$

• Relationship between zeroes and coefficients.

$\large\red{\underline{\underline{\bf{\blue{Solution}}}}}$

First we have to find zeroes of the polynomial

$= > 2 {x}^{2} - 7x - 15 = 0 \\ \\ = > 2 {x}^{2} - 10x + 3x - 15 = 0 \\ \\ = > 2x(x - 5) + 3(x - 5) = 0 \\ \\ = > (2x + 3)(x - 5) = 0 \\ \\ = > (2x + 3) = 0 \: or \: (x - 5) = 0 \\ \\ = > x = \frac{ - 3}{2} or \: x = 5$

• Zeroes are -3/2 , 5

α = -3/2 , β = 5

On comparing this equation with

$= > a {x}^{2} + bx + c = 0$

We get,

• a = 2 , b = -7 , c = -15

Now,

We know that

Sum of zeroes = (-Coefficient of x) /Coefficient of x^2

=> α + β = -b/a

$= > \frac{ - 3}{2} + 5 = \frac{ - ( - 7)}{2} \\ = > \frac{7}{2} = \frac{7}{2}$

Hence verified.

Product of zeroes = Constant term / Coefficient of x^2

=> αβ = c/a

$= > \frac{ - 3}{2} \times 5 = \frac{ - 15}{2} \\ = > \frac{ - 15}{2} = \frac{ - 15}{2}$

Hence verified

Answer 2

Given:

2x²- 7x -15= 0

To verify relationship between zeroes and coefficients.

Answer:

By splitting the middle term,

2x²-7x-15=0

2x²-10x +3x -15=0

2x(x-5)+3(x-5)=0

(2x+3)(x-5)=0

2x+3=0 , x-5 =0

x= (-3/2) , x = 5

Hence the Zeroes are -3/2 , 5

Let

α = -3/2 , β = 5

On comparing this equation with

= > ax²+ bx + c = 0

We get,

a = 2 , b = -7 , c = -15

Now,

We know that

Sum of zeroes = (-Coefficient of x) /Coefficient of x²)

= (-b / a)

substitute the values from the given polynomial

=> -(-7) / 2

Also the sum of zeroes =

=> (-3 / 2) + 5 = 7/ 2

7/2 = 7/2

Also,

Product of zeroes = (Constant term / Coefficient of x²)

= (c/a)

substitute the values from given polynomial

=> (-15/2)

also product of zeroes =

=> (-3)/2 ×5

=> -15/2

-15/2 = -15/2

Hence verified