Question

If the roots of ax2

-bx-c =0 differ by 5, then prove that 25a2= b2

-4ac​

Answer 1

Answer:

Assume b+5a=0, then b=−5a and a=−15b. Also, we have that b2<4ac. Combine these together to obtain b2<−45bc and 25a2<4ac.

(1) If a>0,b>0, then we have: b<−45c< which implies c<0. False.

(2) If a<0,b<0, then we have: −25|a|>4c which implies c<0. False.

If a>0,b<0, you have the same thing as in (1) and if If a<0,b>0, you have the same thing as in (2). Thus it is not possible for b+5a=0. Proceed similarly.

Second part: let D>0, assume 5a+b=−D, so b=−(D+5a). Then for this equation to not have two distinct real roots: b2≤40a. replacing a in to obtain: (D+5a)2<40a Notice that this implies a>0. Simplify:

D+5a<±40a−−−√

Let u=a−−√, then we have 5u2∓40−−√u+D<0 is the condition, which is if we can find such u under some condition on D, then the same condition is also the condition for ax2+bx+10=0 to not have two distinct real roots. Solve this to get

u=110(40−−√−40−20D−−−−−−−−√)

this has a solution if D≤2, which means if D≤2, then the ax2+bx+10=0 does not have two distinct real roots. Thus the possible solutions are 5a+b=−1,−2, or (3) and (2). Please check my algebra carefully

Answer 2

Correct Question:-

If the roots of $ax^2-bx-c=0$ differ by 5, then prove that $b^2+4ac=25a^2.$

Solution:-

If $\alpha$ and $\beta$ are roots of a quadratic equation $ax^2+bx+c=0$ where $\alpha\geq\beta,$ then we have,

• $\alpha+\beta=-\dfrac{b}{a}$

• $\alpha\beta=\dfrac{c}{a}$

Then we see that,

$\longrightarrow \alpha-\beta=\sqrt{(\alpha+\beta)^2-4\alpha\beta}$

$\longrightarrow \alpha-\beta=\sqrt{\left(-\dfrac{b}{a}\right)^2-4\left(\dfrac{c}{a}\right)}$

$\longrightarrow \alpha-\beta=\sqrt{\dfrac{b^2}{a^2}-\dfrac{4c}{a}}$

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

$\longrightarrow \alpha-\beta=\dfrac{\sqrt{b^2-4ac}}{a}\quad\quad\dots(1)$

In the question, the roots of the quadratic equation $ax^2-bx-c=0$ differ by 5, i.e.,

$\longrightarrow \alpha-\beta=5$

By (1),

$\longrightarrow \dfrac{\sqrt{(-b)^2-4a(-c)}}{a}=5$

$\longrightarrow\sqrt{b^2+4ac}=5a$

$\longrightarrow\underline{\underline{b^2+4ac=25a^2}}$

Hence Proved!