Question

Find the zeroes of the quadratic polynomial and verify the relationship between the zeroes and the coefficient x square-10x+24.

Answer 1

Answer:

p(x)=x²-10x+24

x²-4x-6x+24

x(x-4)-6(x-4)

(x-6) (x-4)

x=6 & x=4

let alpha be '¶' and

beta be '£'

Step-by-step explanation:

¶=6. £=4

a=1, b=-10, ©=24

sum of the zeroes =¶+£

6+4=10

-b/a=-(-10)

=10

product of the zeroes=¶×£=6×4=24

c/a=24/1=24

hence verified..

Answer 2

Question:

Find the zeroes of the quadratic polynomial and verify the relationship between the zeroes and the coefficient.

${x}^{2} - 10x + 24 \: .$

Solution:

${x}^{2} - 10x + 24 = 0 \\ \\ {x}^{2} - x(6 + 4) + 24 = 0 \\ \\ {x}^{2} - 6x - 4x + 24 = 0 \\ \\ x(x - 6) - 4(x - 6) = 0 \\ \\ (x - 4)(x - 6) = 0 \\ \\ taking \: x - 4 = 0 \: \: we \: will \: get \\ x = 4 \\ \\ and \\ taking \: x - 6 = 0 \: we \: will \: get \\ x = 6$

Hence we have two zeroes of given quadratic polynomial as

$\alpha = 6 \: \: and \: \beta = 4$

NOW,

comparing the given quadratic polynomial with the standard form of quadratic polynomial

$a {x}^{2} + bx + c \\ \\ we \: will \: get \: coefficients \: as$

a = 1 ; b = - 10 ; c = 24

Now,

as we know,

$\alpha + \beta = \frac{ - b}{a}$

for verification firstly putting values in LHS

$lhs \: = \alpha + \beta = 6 + 4 = 10$

now putting values in RHS

$rhs = \frac{ - b}{a} = \frac{ - ( - 10)}{1} = 10$

since,

LHS = RHS

hence verified.

also we know,

$\alpha \beta = \frac{c}{a}$

putting values in LHS

$lhs \: = \alpha \beta = 6 \times 4 = 24$

putting values in RHS

$rhs = \frac{c}{a} = \frac{24}{1} = 24$

since,

LHS = RHS

hence verified.