**Question:**

Are the following statements True’ or ‘False’? Justify your answer.

(i) If the zeroes of a quadratic polynomial ax2 + bx + c are both positive, then a, b and c all have the same sign.

(ii) If the graph of a polynomial intersects the X-axis at only one point, it cannot be a quadratic polynomial.

(iii) If the graph of a polynomial intersects the X-axis at exactly two points, it need not be a quadratic polynomial.

(iv) If two of the zeroes of a cubic polynomial are zero, then it does not have linear and constant terms.

(v) If all the zeroes of a cubic polynomial are negative, then all the coefficients and the constant term of the polynomial have the same sign.

(vi) If all three zeroes of a cubic polynomial x3 +ax2 -bx + c are positive, then atleast one of a, b and c is non-negative.

(vii) The only value of k for which the quadratic polynomial kxz + x + k has equal zeroes is $\frac{1}{2}$

**Solution:**

(i) False, if the zeroes of a quadratic polynomial $a x^{2}+b x+c$ are both positive, then

$\alpha+\beta=-\frac{b}{a} \quad$ and $\quad \alpha \cdot \beta=\frac{c}{a}$

where $\alpha$ and $\beta$ are the zeroes of quadratic polynomial.

$\therefore \quad c<0, a<0$ and $b>0$

or $c>0, a>0$ and $b<0$

(ii) True, if the graph of a polynomial intersects the X-axis at only one point, then it cannot be a quadratic polynomial because a quadratic

polynomial may touch the X-axis at exactly one point or intersects X-axis at exactly two points or do not touch the X-axis.

(iii) True, if the graph of a polynomial intersects the X-axis at exactly two points, then it may or may not be a quadratic polynomial. As, a

polynomial of degree more than z is possible which intersects the X-axis at exactly two points when it has two real roots and other imaginary

roots.

(iv) True, let a, p and y be the zeroes of the cubic polynomial and given that two of the zeroes have value 0.

Let α =β = 0

and f(x) = (x- α)(x-β)(x-γ)

= (x-a)(x-0)(x-0)

= x3 – ax2

which does not have linear and constant terms.

(v) True, if f(x) = ax3 + bx2 + cx + d. Then, for all negative roots, a, b, c and d must have same sign.

(vi) False, let $\alpha, \beta$ and $\gamma$ be the three zeroes of cubic polynomial $x^{3}+a x^{2}-b x+c$.

Then, $\quad$ product of zeroes $=(-1)^{3} \frac{\text { Constant term }}{\text { Coefficient of } x^{3}}$

$\Rightarrow \quad \alpha \beta \gamma=-\frac{(+c)}{1}$

$\Rightarrow \quad \alpha \beta \gamma=-c$ $\ldots$ (i)

Given that, all three zeroes are positive. So, the product of all three zeroes is also positive

i.e., $\alpha \beta \gamma>0$

$\Rightarrow \quad-c>0 \quad$ [from Eq. (i)]

$\Rightarrow$ $c<0$

Now, sum of the zeroes $=\alpha+\beta+\gamma=(-1) \frac{\text { Coefficient of } x^{2}}{\text { Coefficient of } x^{3}}$

$\Rightarrow$ $\alpha+\beta+\gamma=-\frac{a}{1}=-a$

But $\alpha, \beta$ and $\gamma$ are all positive.

Thus, its sum is also positive.

So, $\quad \alpha+\beta+\gamma>0$

$\Rightarrow \quad-a>0$

$\Rightarrow \quad a<0$

and sum of the product of two zeroes at a time $=(-1)^{2} \cdot \frac{\text { Coefficient of } x}{\text { Coefficient of } x^{3}}=\frac{-b}{1}$

$\Rightarrow \quad \alpha \beta+\beta \gamma+\gamma \alpha=-b$

$\because$ $\alpha \beta+\beta \gamma+\alpha \gamma>0 \quad \Rightarrow-b>0$

$\Rightarrow$ $b<0$

So, the cubic polynomial $x^{3}+a x^{2}-b x+c$ has all three zeroes which are positive only when all constants $a, b$ and $c$ are negative.

(vii) False, let f(x) = kx2 + x + k

For equal roots. Its discriminant should be zero i.e., D = b2 – 4ac = 0

⇒ 1-4k.k = 0

=> k = ± 12

So, for two values of k, given quadratic polynomial has equal zeroes

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