Question

Solve This Problem ⛛ Also....... ​

Answer 1

Given :-

A quadratic equation

$\sf\bullet px^2+qx+r=0$

Their roots in ratio 3:4

To find

The condition on the coefficient of equation !

As we know that

$\bigstar{\boxed{\sf{ sum \ of \ zeroes (\alpha+\beta)= \dfrac{-coefficient \ of x}{coefficient \ of \ x^2}}}}$

$\bigstar{\boxed{\sf{ product \ of \ zeroes ( \alpha \beta)=\dfrac{constant \ term }{coefficient \ of x^2}}}}$

So , Now

$\sf \bullet px^2+qx+r=0 \\ \\ \sf \ p = coefficient \ of \ x^2 \\ \\ \sf q = coefficient \ of \ x \\ \\ \sf r= constant \ term$

• Zeroes (Roots ) = 3:4
• let the root be y
• then 3y and 4y

$\sf sum \ of \ zeroes (\alpha +\beta)=\dfrac{-q}{p}\\ \\ \dashrightarrow\sf (3y+4y)=\dfrac{-q}{p}\\ \\ \dashrightarrow\sf 7y=\dfrac{-q}{p}\\ \\ \dashrightarrow\sf y=\dfrac{-q}{7p} ------(i)$

$\sf Product \ of \ zeroes (\alpha \beta)=\dfrac{r}{p}\\ \\ \dashrightarrow\sf (3y\times 4y)=\dfrac{r}{p}\\ \\ \dashrightarrow\sf 12y^2=\dfrac{r}{p}\\ \\ \dashrightarrow\sf y^2=\dfrac{r}{12p}\\ \\ \dashrightarrow\sf y=\sqrt{\dfrac{r}{12p}}-----(ii)$

From equation (i) and (ii)

$\dashrightarrow\sf \dfrac{-q}{7p}=\sqrt{\dfrac{r}{12p}}\\ \\ \sf \ \ squaring \ both \ side \\ \\ \dashrightarrow\sf \bigg[\dfrac{-q}{7p}\bigg]^2= \bigg[\sqrt{\dfrac{r}{12p}}\bigg]^2 \\ \\ \dashrightarrow\sf \dfrac{q^2}{49p^2}=\dfrac{r}{12p} \\ \\ \sf \ Cross \ multiplication \\ \\\dashrightarrow\sf 12p\times q^2= r\times 49p^2\\ \\ \dashrightarrow\sf 12\cancel{p}q^2= 49\cancel{p^2}r\\ \\ \sf \ \ \ (p ) \ will \ be \ cancel \\ \\ \dashrightarrow\sf 12q^2= 49pr$

$\boxed{\sf{\dag \ \ 12q^2=49pr}}$

$\bigstar{\boxed{\sf{ Option \ A \ is \ correct \ option }}}$

Answer 2

GIVEN:-

• Quadratic equation : px² + qx + r = 0
• Ratio of roots should be = 3 : 4

TO FIND:-

The condition on the coefficients of the given equation

We know that

Sum of roots(α + β)=Coefficient of x/coefficient of x²

Product of roots (αβ) = Constant/coefficient of x²

Now, here

• Coefficient of x² = p
• Coefficient of x = q
• Constant = r

Let the roots of the given quadratic equation be x

Then the roots will be = 3x , 4x

Now,

Sum of roots (α + β) = 3x + 4x

→ - q/p = 7x

→ x = - q/7p

Product of roots (αβ) = 3x × 4x

→ r/p = 12x²

→ r/12p = x²

→ √r/12p = x

Hence, x = -q/7p , x = √r/12p

Then , - q/7p & √r/12p Both are equal

$\small{ \tt{ \to \green{ \dfrac{ - q}{7p} } \pink{ = } \red{ \sqrt{ \dfrac{r}{12p} }}}} \\ \\ \small{ \to \tt{ \green{ {\Big( \frac{ - q}{7p} \Big)}^{2} \pink{ = } \red{ \frac{r}{12p} }}}} \\ \\ \small \to \tt{ \green{ \frac{ {q}^{2} }{49 {p}^{2} } \pink{ = } \red{ \frac{r}{12p} } }} \\ \\ \sf{Cross \: multiply} \\ \\ \to \tt{ \green{12 {q}^{2} \cancel{p}\pink = \red{49 {p}^{ \cancel2}r } }}$

Hence,$\large\fbox{ \mathsf{ \orange{12 {q}^{2} = 49pr}}}$

Hence, option (a) is your answer