Are the following statements 'True' or 'False'? Justify your answer.
(i) If the zeroes of a quadratic polynomial `ax^(2) +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 ve 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 `x^(3) +ax^(2) - 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 `kx^(2) +x +k` has equal zeroes is `(1)/(2)`.

Explanation

(i) False, if the zeroes of a quadratic polynomial `ax^(2)+bx +c` are both positive, then,
`alpha + beta =- (b)/(a)` and `alpha. beta = (c)/(a)`
where `alpha` and `beta` are the zeroes of quadratic polynomial.
`{:( :.,c lt0,a lt 0, and, b gt 0),(or, c gt 0, a gt 0, and , b lt 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 quadrtic polynomial may touch the X-axis at exactly one point or intersects X-axis at ecactly two points or do not touch then X-axis.
(iii) True, if the graph of a polynomial intersects the X-axis at ecactly 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 `alpha, beta` and `gamma` be the zeroes of the cubic polynomial and given that two of the zeroes have valye 0.
Let `beta = gamma = 0`
and `f(x) = (x-alpha) (x-beta) (x-gamma)`
`= (x-alpha) (x-0) (x-0)`
`=x^(3) - ax^(2)`
which does not have linear and constant terms.
(v) True, if `f(x) = ax^(3) +bx^(2) +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) +ax^(2)-bx +c`.
Then, product of zeroes `= (-1)^(3) ("Constant term")/("Coefficient of" x^(3))`
`rArr alpha beta gamma =- ((+c))/(1)`
`rArr alpha beta gamma =- c` ....(i)
Given that, all three zeroes are positive. So, the product of all three zeroes is also positive
i.e., `alpha beta gamma gt 0`
`rArr -c gt 0` [from Eq.(i)]
`rArr c lt 0`
Now, sum of the zeroes `= alpha +beta +gamma = (-1) ("Coefficient of" x^(2))/("Coefficient of" x^(3))`
`rArr alpha + beta +gamma =- (a)/(1) =- a`
But `alpha, beta` and `gamma` are all positive.
Thus, its sum is also positive.
So, `alpha + beta +gamma gt 0`
`rArr -a gt 0`
`rArr a lt 0`
and sum of the product of two zeroes at a time `=(-1)^(2).("Coefficient of" x)/("Coefficient of" x^(3)) = (-b)/(1)`
`rArr alpha beta +beta gamma +gamma alpha =- b`
`:' alpha beta +beta gamma +alpha gamma gt 0 rArr -b gt 0`
`rArr b lt 0`
So, the cubic polynomial `x^(3) +ax^(2) - bx +c` has all three zeroes which are positive only when all constants a,b and c are negative.
(vii) False, let `f(x) = kx^(2) +x +k`
For equal roots. Its discriminant should be zero i.e., `D = b^(2) - 4ac = 0`
`rArr k = +- (1)/(2)`
So, for two values of k, given quadratic polynomial has equal zeroes.