Question

Find the zeroes of the quadratic polynomial abx2 (b2-ac)x-bc

Answer 1

Question:

Find the zeros of the quadratic polynomial

abx² + (b² - ac)x - bc.

Answer:

x = -b/a , c/b

Note:

• The values of x for which the polynomial p(x) becomes zero are called the zeros of the polynomial p(x).

• To find the zeros of the given polynomial p(x) , equate it to zero and then find the possible values of x , ie; operate on p(x) = 0.

Solution:

Here,

The given polynomial is ;

abx² + (b² - ac)x - bc = p(x) {say}

Now,

To get the zeros of the given polynomial p(x) ;

we have;

=> p(x) = 0

=> abx² + (b² - ac)x - bc = 0

=> abx^2 + b²x - acx - bc = 0

=> bx(ax + b) - c(ax + b) = 0

=> (ax + b)(bx - c) = 0

=> ax + b = 0 OR bx - c = 0

=> ax = - b OR bx = c

=> x = - b/a OR x = c/b

Hence,

The zeros of the given polynomial are :

x = -b/a and x = c/b

Verification:

Case1 : When x = -b/a

=> p(x) = abx² + (b² - ac)x - bc

=> p(-b/a) = ab(-b/a)² + (b² - ac)(-b/a) - bc

=> p(-b/a) = abb²/a² - b³/a + abc/a - bc

=> p(-b/a) = b³/a - b³/a + bc - bc

=> p(-b/a) = 0

Hence,

x = -b/a is a zero of the given polynomial p(x) = abx² + (b² - ac)x - bc.

Case2 : When x = c/b

=> p(x) = abx² + (b² - ac)x - bc

=> p(c/b) = ab(c/b)² + (b² - ac)(c/b) - bc

=> p(c/b) = abc²/b² + b²c/b - ac²/b - bc

=> p(c/b) = ac²/b + bc - ac²/b - bc

=> p(c/b) = 0

Hence,

x = c/b is a zero of the given polynomial p(x) = abx² + (b² - ac)x - bc.

Answer 2

Question :----

• Find the zeroes of the quadratic polynomial abx²+(b²-ac)x-bc ?

Concept used :----

• we will put the Equation equal to zero and use if any common Factor comes to Find both zeros of Equation .

**) The zero of the polynomial is defined as any real value of x, for which the value of the polynomial becomes zero. A real number k is a zero of a polynomial p(x), if p(k) = 0.

______________________________________

Solution:-

→ abx²+(b²-ac)x-bc = 0

→ abx² + b²x - acx - bc = 0

→ (abx² + b²x) - (acx + bc) = 0

taking common Now,

→ bx(ax+b) - c(ax+b) = 0

→ Again Taking (ax+b) common now,

→ (ax+b)(bx-c) = 0

Putting Both Equal to zero now we get,

→ (ax+b) = 0

→ ax = (-b)

Dividing both side by a , we get,

→ x = (-b)/a

Similarly ,

→ (bx-c) = 0

→ bx = c

Dividing both side by b we get,

→ x = (c/b)

Hence, zeros of Polynomial will be (-b/a) and (c/b) .

_____________________________

If A•x^2 + B•x + C = 0 ,is any quadratic equation,

then its discriminant is given by;

D = B^2 - 4•A•C

• If D = 0 , then the given quadratic equation has real and equal roots.

• If D > 0 , then the given quadratic equation has real and distinct roots.

• If D < 0 , then the given quadratic equation has unreal (imaginary) roots...