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#### Which of the following is not the graph of a quadratic polynomial?

Solution.  Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 2.

If p(x) is quadratic polynomial then there are almost 2 values of x exists called roots.

Hence, the graph of p(x) cuts the x-axis almost 2 times.

But here we see that in option (D) the curve cuts the x-axis at 3 times

Hence, graph (D) is not a graph of a quadratic polynomial.

#### If one of the zeroes of a quadratic polynomial of the form $x^{}2+ax + b$ is the negative of the other, then it(A) has no linear term and the constant term is negative.(B) has no linear term and the constant term is positive.(C) can have a linear term but the constant term is negative.(D) can have a linear term but the constant term is positive.

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 2.

Here the given quadratic polynomial is $x^{}2+ax + b$               …..(1)

a = 1, b = a, c = b

Let x1, x2 are the zeroes of the equation (1)

According to question:

$x_2 = - x_1$

sum of zeroes = $x_2 + x_1=\frac{-b}{a}$

$x_1 - x_1=\frac{-b}{a}$    ( because x2 = - x1 )

$0=\frac{-a}{1}\Rightarrow a=0$             ( because b = a , a = 1)

Product of zeroes $= (x_1)(x_2) = \frac{c}{a}$

$= (x_1)(-x_1) = \frac{c}{a}$

$= -x_{1}^{2} =b$          ( because c= b, a = 1)

Put value of a and b in (1)

$x^{2}+ \left (-x_{1}^{2} \right )$

Hence, it has no linear term and the constant term is negative.

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#### If the zeroes of the quadratic polynomial $ax^2 + bx + c, c \neq 0$ are equal, then(A) c and a have opposite signs(B) c and b have opposite signs(C) c and a have the same sign(D) c and b have the same sign

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 2.

Here the given polynomial is $ax^2 + bx + c, c \neq 0$

$a = a, b = b, c = c$

We know that if both the zeroes are equal there

$b^2 - 4ac = 0$

$b^2 =4ac$     ….(1)

(A) c and a have opposite sign

If c and a have opposite sign then R.H.S. of equation (1) is negative but L.H.S. is always positive. So (A) is not a correct one.

(B) c and b have opposite sign

If c is negative and b is positive L.H.S. is positive but R.H.S. of eqn. (1) is negative. Hence (B) is not correct one.

(C) c and a have same sign

If c and a have same sign R.H.S. of eqn. (1) is positive and L.H.S. is always positive hence it is a correct one.

(D) c and b have same sign

If c and b both have negative sign then R.H.S. of eqn. (1) is negative and L.H.S. is positive. So this is not correct.

Only one option i.e. (c) is correct one.

#### The zeroes of the quadratic polynomial $x^2 + kx + k, k \neq 0$,(A) cannot both be positive(B) cannot both be negative(C) are always unequal(D) are always equal

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 2.

Here the given quadratic polynomial is $x^2 + kx + k$

$a = 1, b = k, c = k$

$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

$x=\frac{-k\pm \sqrt{k^{2}-4k}}{2}$

$x=\frac{-k\pm \sqrt{k(k-4)}}{2}$

$k(k - 4)$ must be grater then 0

Hence the value of k is either less than 0 or greater than 4.

If value of k is less than 0 only one zero is positive.

If value of k is greater than 4 only one zero is positive.

Hence both the zeroes can not be positive.

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#### The zeroes of the quadratic polynomial $x^2 + 99x + 127$ are(A) both positive(B) both negative(C) one positive and one negative(D) both equal

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 2.

Here the given quadratic polynomial is $x^2 + 99x + 127$

$x^2 + 99x + 127$

$a = 1, b = 99, c = 127$

$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

$x=\frac{-99\pm \sqrt{(99)^{2}-4(1)(127)}}{2(1)}$

$x=\frac{-99\pm \sqrt{9801-508}}{2(1)}$

$x=\frac{-99\pm \sqrt{9293}}{2}$

$x=\frac{-99\pm 96.4}{2}$

$x=\frac{-99+ 96.4}{2}$                                                            $x=\frac{-99-96.4}{2}$

$x=\frac{-2.6}{2}=-1.3$                                                          $x = -97.7$

The value of both the zeroes are negative.

#### If one of the zeroes of the cubic polynomial $x^3 + ax^2 + bx + c$ is –1, then the product of the other two zeroes is(A) b – a + 1(B) b – a – 1(C) a – b + 1(D) a – b –1

Solution.  Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 3.

Here the given cubic polynomial is $x^3 + ax^2 + bx + c$

Let $\alpha ,\beta ,\gamma$ are the zeroes of polynomial

$\alpha =-1$  (given)               …..(1)

put x = –1 in $p(x)=x^3 + ax^2 + bx + c$

$p(-1) = (-1)^3 + a(-1)^2 + b(-1) + c$

$0 = -1 + a - b + c ( Q=-1 is zero)$

$c = 1 - a + b$

We know that

$\alpha \beta \gamma =\frac{-d}{a}$

Here, a = 1, b = a, c = b, d = c

So, $\alpha \beta \gamma =\frac{-c}{1}$

$\beta \gamma =\frac{-(1-a+b)}{\alpha }$

$\beta \gamma =\frac{-(1-a+b)}{-1 } = 1 -a + b$            ( using equation (1))

Hence the product of other two is b – a + 1.

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#### Given that one of the zeroes of the cubic polynomial $ax^3 + bx^2 + cx + d$ is zero, the product of the other two zeroes is(A) $-\frac{c}{a}$(B) $\frac{c}{a}$(C) 0(D) $-\frac{b}{a}$

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 3.

Here the given cubic polynomial is $ax^3 + bx^2 + cx + d$

Let three zeroes are $\alpha ,\beta ,\gamma$

$\alpha =0$ (given)    …..(1)

we know that

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

Put $\alpha =0$

$\beta \gamma =\frac{c}{a}$           ( using equation (1))

Hence the product of other two zeroes is $\frac{c}{a}$.

#### The number of polynomials having zeroes as –2 and 5 is(A) 1(B) 2(C) 3(D) more than 3

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s).

Let the polynomial is $ax^2 + bx + c = 0$                  ….(*)

We know that sum of zeroes $-\frac{b}{a}$

$- 2 + 5= -\frac{b}{a}$

$\frac{3}{1}=-\frac{b}{a}$

…..(1)

Multiplication of zeroes $=\frac{c}{a}$

$-2 \times 5=\frac{c}{a}$

$-\frac{10}{1}=\frac{c}{a}$                 …..(2)

Form equation (1) and (2) it is clear that

a = 1, b = –3, c = –10

put value of a, b and c in equation (*)

$x^3 - 3x - 10 = 0$                         ….(3)

But we can multiply of divide eqn. (3) by any real number except 0 and the zeroes remain same.

Hence, there are infinite number of polynomial exist with zeroes –2 and 5.

Hence the answer is more than 3.

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#### If the zeroes of the quadratic polynomial $x^2 + (a + 1) x + b$ are 2 and –3, then(A) a = –7, b = –1(B) a = 5, b = –1(C) a = 2, b = – 6(D) a = 0, b = – 6

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 2.

If 2 and –3 are the zero of $p(x) = x^2 + (a + 1)x + b$  then p(2) = p(–3) = 0.

$p(2) = (2)^2 + (a + 1) (2) + b$

$0 = 4 + 2a + 2 + b$

$0 = 6 + 2a + b$

$2a + b = -6$                        ….(1)

$p(-3) = (-3)^2 + (a + 1) (-3) + b$

$0 = 9 - 3a - 3 + b$

$3a - b = 6$                              ….(2)

$2a + b + 3a - b = -6 + 6$

2a + 3a = 0

5a = 0

a = 0

put a = 0 in (1)

2(0) + b = –6

b = –6

Hence a = 0, b = –6.

#### A quadratic polynomial, whose zeroes are –3 and 4, is(A) $x^2 - x + 12$(B)$x^2 + x + 12$(C) $\frac{x^2}{2} - \frac{x}{2} -6$(D) $2x^2 + 2x -24$

Solution.  Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 2.

(A)  $p(x) = x^2 - x + 12$

put x = –3                                                                            put x = 4

$p(-3) = (-3)^2 - (-3) + 12$                                 $p(4) = (4)^2 - 4 + 12$

$= 9 + 3 + 12 = 24 \neq 0$                                           $= 16 - 4 + 12 = 24 \neq 0$

(B) $p(x) = x^2 + x + 12$

put x = –3                                                                                    put x = 4

$p(-3) = (-3)^2 - 3 + 12$                                               $p(4) = (4)^2 + 4 + 12$

$= 9 + 9 = 18 \neq 0$                                                        = 16 + 16 = 32 $\neq$ 0

(C) $p(x)=\frac{x^2}{2} - \frac{x}{2} -6$

put x = –3                                                                    put x = 4

$p(-3)=\frac{(-3)^2}{2} - \frac{-3}{2} -6$                                 $p(4)=\frac{(4)^2}{2} - \frac{4}{2} -6$

$=\frac{9}{2}+\frac{3}{2}-6$                                                                  $=\frac{16}{2}-2-6$

$=\frac{9+3-12}{2}=0$                                               $= 8 - 8 = 0$

(D) $p(x) = 2x^2 + 2x - 24$

put x = –3                                                                    put x = 4

$p(-3) = 2(-3)^2 + 2(-3) - 24$              $p(4) = 2(4)^2 + 2(4) - 24$

$= 18 -6 - 24$                                                  $= 32 + 8 - 24$

$= -12 \neq 0$                                                       $= 16 \neq 0$

If –3, 4 is zeros of a polynomial p(x) then p(–3) = p(4) = 0

Here only (C) option satisfy p(–3) = p(4) = 0