Question

if the zeroes of a quadratic polynomial p(x) = ax^2 + x + a are equal then the value of a is

(a)
1
/2
(b)
−1
/2
(c) ±
1
/2
(d) ± 1​

Answer 1

Answer:

Given that the zeros of the quadratic polynomial ax

2

+bx+c,c



=0 are equal.

=> Value of the discriminant(D) has to be zero.

=>b

2

−4ac=0

=>b

2

=4ac

Since. L.H.S b

2

cannot be negative, thus, R.H.S. can also be never negative.

Therefore, a and c must be of the same sign.

Step-by-step explanation:

hope it helps☺

Answer 2

Answer:

Required value of a is $\pm \: \frac{1}{2}$

Step-by-step explanation:

Given polynomial is $p(x) = a {x}^{2} + x + a....(1)$

and it is also given this polynomial has all equal zeroes.

We know,

if $f(y) = q {y}^{2} + ry + s$

has equal roots then discriminant

${r}^{2} - 4qs = 0$

Now discriminant of polynomial p(x) is

${1}^{2} - 4 \times a \times a = 1 - 4 {a}^{2}$

According to rule,

$1 - 4 {a}^{2} = 0 \\ 4 {a}^{2} = 1 \\ {a}^{2} = \frac{1}{4} \\ a = \sqrt{ \frac{1}{4} } \\ a = \pm \: \frac{1}{2}$

Required value of a is $\pm \: \frac{1}{2}$

This is a problem of Algebra.

Some important Algebra formulas.

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab − b²

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)³ − 3ab(a + b)

a³ - b³ = (a -b)³ + 3ab(a - b)

a² − b² = (a + b)(a − b)

a² + b² = (a + b)² − 2ab

a² + b² = (a − b)² + 2ab

a³ − b³ = (a − b)(a² + ab + b²)

a³ + b³ = (a + b)(a² − ab + b²)

Know more about Algebra,

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