Roots of the equation?

Question

If $p,q,r$ are real numbers satisfying the condition $p + q + r =0$, then the roots of the quadratic equation $3px^2 +5qx +7r=0$ are

(A)Positive

(B)Negative

(C)Real and distinct

(d)Imaginary

Actually im a 10 class student i don't know any of it, but my elder brother (IIT Coaching) cannot solve them, he told me post these questions on this site someone might know the answers and for now he is not in the town. So can you please help me. Thank you.

Answer 1

HINT:

Putting $r=-p-q,$

$$3px^2+5qx-7(p+q)=0$$

So, the discriminant is $$(5q)^2-4(3p)\{-7(p+q)\}=(5q)^2+84pq+84p^2=\left(5q+\frac{42}5p\right)^2+\{84-\left(\frac{42}5\right)^2\}p^2$$

$$=\left(5q+\frac{42}5p\right)^2+\frac{p^2(84\cdot25-42^2)}{25}$$

$$=\left(5q+\frac{42}5p\right)^2+\frac{42p^2(2\cdot25-42)}{25}>0$$

What can we make of it?

Answer 2

Hint : $$ p+q+r = 0 $$ This, implies that at least one of three must be of different sign than rest two. There is 3 possible cases for this $$ (p,q) \space |\space r $$ $$ (q,r) \space | \space p $$ $$ (p,r) \space | \space q $$

Now, we know that Nature of root can be known by value of D

$$ D = \sqrt{b^2-4ac} $$

Check for all possible combination. My answer is option (c)

Answer 3

I will take a different tack, which might be called the "Vi$\grave{e}$te approach". I will note that for resolving the choices in this problem, the numbers 3, 5, and 7 appearing in the coefficients have no special significance, other than that they do not change the signs of the coefficients from what is given by $ \ p \ , \ q \ , \ \text{and} \ r \ . $ (They do have the effect of making the discriminant of the quadratic polynomial look more intimidating to analyze.)

Since $ \ p \ , \ q \ , \ \text{and} \ r \ $ are real, the coefficients of our polynomial are real; as the door has been "left open" to consider imaginary zeroes, this means that such roots must form a "conjugate pair". If we call the zeroes $ \ \alpha \ \text{and} \ \beta \ , $ then the factored form of this polynomial is $ \ 3p \ (x - \alpha) \ (x- \beta) \ , $ which "multiplies out" as

$$ 3p x^2 \ - \ 3p \ [\alpha + \beta] x \ + \ 3p \alpha \beta \ = \ 3px^2 + 5qx + 7r \ \ . $$

We can eliminate choice (D) immediately. If the roots were pure imaginary, the real coefficients require that $ \ \alpha = b i \ \text{and} \ \beta = - b i \ . $ This would force their sum to be zero and, consequently $ \ q = 0 \ ; $ further, their product would be positive, requiring that $ \ p \ \text{and} \ r \ $ have the same sign. But under the condition $ \ p + q + r = 0 \ , \ q \ $ being zero forces $ \ p = -r \ . $ So we have an outright contradiction in this case.

If both $ \ \alpha \ \text{and} \ \beta \ $ are positive, then their sum and product are positive, requiring that $ \ p \ \text{and} \ r \ $ have the same sign and $ \ p \ \text{and} \ q \ $ have opposite signs. Under the specified condition, we can write $ \ p + r = -q \ , $ but this by itself can be satisfied by values of $ \ p \ \text{and} \ r \ $ of opposite sign as well. This is to say that the specified condition, $ \ p + q + r = 0 \ , $ does not impose a strong enough constraint to require that the two roots both be positive. (That is to say, they could be, but don't have to be.)

A stronger argument follows for having both $ \ \alpha \ \text{and} \ \beta \ $ negative: here the sum of the roots is negative and their product is positive, causing $ \ p \ , \ q \ , \ \text{and} \ r \ $ all to have the same sign, which is impossible under the imposed condition.

Finally, we show that choice (C) is correct, also by contradiction. Were the two roots equal $ ( \ \alpha = \beta \ ) , $ our polynomial would be

$$ \ 3p \ (x - \alpha)^2 \ = 3p x^2 \ - \ 6p \alpha x \ + \ 3p \alpha^2 \ . $$

We would thus have $ \ 5q = -6p \alpha \ \Rightarrow \ q = -\frac{6 \alpha}{5} p \ \ $ and $ \ 7r = 3p \alpha^2 \ \Rightarrow \ r = \frac{3 \alpha^2}{7} p \ . $ The specified condition implies that $ \ p \ (1 - \frac{6}{5} \alpha + \frac{3}{7} \alpha^2 ) \ = \ 0 . \ $ It is impermissible to have $ \ p = 0 \ . $ But the discriminant of the quadratic equation for $ \ \alpha \ $ is $ \ (-\frac{6}{5})^2 - 4 \cdot \frac{3}{7} \cdot 1 \ = \ \frac{36}{25} - \frac{12}{7} \ < \ 0 . $ So the roots cannot be equal and real. [This is the only place in the entire discussion where the numbers 3 , 5 , and 7 have much importance.]