GATE ECE 2020 | GA | Question: 9

Question

$a, b, c$ are real numbers. The quadratic equation $ax^{2}-bx+c=0$ has equal roots, which is $\beta$, then

• A. $\beta =b/a$
• B. $\beta^{2} =ac$
• C. $\beta^{3} =bc/\left ( 2a^{2} \right )$
• D. $\beta^{2} \neq 4ac$

Answer 1

In general, for a quadratic equation  $ax^{2}+bx+c=0,$ sum of roots is given by:  $\frac{-b}{a}$  and product of roots by:  $\frac{c}{a}$

Here, both the roots are $\beta$.

So,  according to the given quadratic equation, we have  

• $2\beta=\frac{b}{a}\quad \to (1)$ and  
• $\beta^{2}=\frac{c}{a}\quad \to (2)$

Multiplying equations $(1)$ and $(2)$ and simplifying, we get  $\beta^{3}=\frac{bc}{2a^{2}}$  
Option C is correct

Comments

but where did the minus sign disappear from -b/a?

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What about option D. Even that is correct since we cannot prove ß² = 4ac by any means. Or maybe that wanted to ask b ≠ 4ac

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@chokostar because here $b = -b$

sum of roots $= \dfrac{-b}{a}$ for $ax^2+bx+c=0$ but here equation is $ax^2-bx+c=0$.

Answer 2

Given quadratic equation is: $ax^{2}-bx+c=0$

Method 1.

Given it has two roots and both are $\beta$.

Using Shreedhara Acharya's formula we know,

$\beta =\frac{b- \sqrt{b^{2}-4ac}}{2a} \text{or}$

$\beta =\frac{b+ \sqrt{b^{2}-4ac}}{2a}$

So, $\beta +\beta =\frac{2b}{2a}=\frac{b}{a}$

$ \implies \beta =\frac{b}{2a}$

Now, $\beta *\beta =\frac{b-\sqrt{b^{2}-4ac}}{2a}*\frac{b+\sqrt{b^{2}-4ac}}{2a}$

                      $= \beta *\beta =\frac{b^{2}-(b^{2}-4ac)}{4a^{2}}=\frac{c}{a}$

                      $\beta ^{2}=\frac{c}{a}$

So, $\beta ^{3}=\beta *\beta ^{2}=\frac{b}{2a}*\frac{c}{a} \implies \beta ^{3}=\frac{bc}{2a^{2}}$

By this we can eliminate option A and B. 

Option C is true.

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Now option D,

$\beta ^{2}=\frac{c}{a}$        ……...(1)

$\beta ^{2}=\frac{b^{2}}{4a^{2}}$  ………...(2)

$\frac{c}{a}=\frac{b^{2}}{4a^{2}}$

or, $b^{2}=4ac$

Now we can say the $\beta^{2} \text{ value is either }\frac{b^2}{4a^2} \text{ or } \frac{c}{a}$.

 

Example:-

$4x^{2}-4x+1=0$ $(\text{here }a=4,b=4, c=1)$

Here roots($\beta$) are $\frac{1}{2},\frac{1}{2}$.

option A: $\frac{b}{a}=\frac{4}{4}=1\neq \beta$

option B: $a*c=4*1=4\neq \beta ^{2}$

option C: $\frac{bc}{2a^{2}}=\frac{4*1}{2*16}=\frac{1}{8}= \beta ^{3}$  (True)

option D: $4*a*c=4*4*1=16\neq \beta ^{2}$  (True for this case)

 

If we take the example 

$\frac{1}{2}x^{2}-x+\frac{1}{2}=0$ $(\text{here }a=\frac{1}{2}, b=1,c=\frac{1}{2})$.

$\beta =1$

$\beta ^{2}=4*\frac{1}{2}*\frac{1}{2}=4ac$

so option D is wrong.

Correct answer is option c.

 

 

 

 

 

Comments

I am saying is there any formal way of disprove option 4 rather than taking example.

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Counter example is indeed a formal way to disprove something.

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equal root means:  $b^2 = 4ac$     .......................(1)

and we need to find counterexample such that:  $\beta^{2} = 4ac$    ........................(2)

from eqn (1) and (2),

$\beta^{2} = b^{2} $    ........................(3)

$\boxed{\beta = \pm b}$

now, we know, $\beta^{2} = \dfrac{c}{a}$    ...........................(4)

from eqn (2) and (4),

$4ac = \dfrac{c}{a} \quad \Rightarrow \quad \boxed{a = \pm \dfrac{1}{2}}$

from (3) and (4),

$b^{2} = \dfrac{c}{a} \quad \Rightarrow \quad c = ab^{2} \quad \Rightarrow \quad \boxed{c = \pm \dfrac{b^{2}}{2}}$

so, in any equation following these $3$ conditions and having this format:

$\boxed{\pm \dfrac{1}{2}x^{2} + bx + \pm \dfrac{b^{2}}{2} = 0}$

will satisfy: $\beta^{2} = 4ac$.

ex: 1) $\dfrac{1}{2}x^{2} \pm 4x + 8 = 0$
      2) $\dfrac{-1}{2}x^{2} \pm 5x - \dfrac{25}{2} = 0$

PS: 1) while forming equations, keep  sign of $a$ and $c$ same means when $a = +ve$ then $c = +ve$ and vice versa.
2) we can also say that while following conditions for $a$ and $c$ we will automatically get $\beta = \pm b$ because by following conditions for $a$ and $c$:

$ \beta = \dfrac{-b + D}{2a}$  (where, $D =$ discriminant $= \sqrt{b^{2} - 4ac}$)

$ \beta = \dfrac{-b + \sqrt{b^{2} - 4ac}}{2a}$

$ \beta = \dfrac{-b + \sqrt{b^{2} - 4 \times \pm \dfrac{1}{2} \times \pm \dfrac{b^{2}}{2}}}{2a}$

$ \beta = \dfrac{-b + \sqrt{b^{2} - b^{2}}}{2a}$

$ \beta = \dfrac{-b + 0}{2a}$

$ \beta = \dfrac{-b}{2 \times \pm \dfrac{1}{2}}$

$\boxed{ \beta = \mp b}$