Question

Hey All !

Let the roots of ax^2 + 2bx + c = 0 are p and q .
Also, the roots of Ax^2 +2Bx+C =0 are p+r and q+r.

Prove that A^2(b^2-ac) = a^2(B^2-AC).

Its a class 10th ques...


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Answer 1

Gɪᴠᴇɴ :-

• Roots of ax^2 + 2bx + c = 0 are p and q .
• Roots of Ax^2 +2Bx+C =0 are p+r and q+r.

To Prove :-

• A^2(b^2-ac) = a^2(B^2-AC)

Fᴏʀᴍᴜʟᴀ ᴜsᴇᴅ :-

• The sum of the roots of the Equation ax² + bx + c = 0 , is given by = (-b/a)
• Product of roots of the Equation is given by = c/a.
• (x - y)² = (x + y)² - 4xy

Sᴏʟᴜᴛɪᴏɴ :-

Given That, Roots of ax^2 + 2bx + c = 0 are p and q.

So ,

→ p + q = (-2b/a)

→ pq = c/a

And,

→ (p - q)² = (p + q)² - 4pq

Putting Both Values now,

→ (p - q)² = [(-2b/a)]² - 4(c/a)

→ (p - q)² = (4b²/a²) - (4c/a)

→ (p - q)² = (4b² - 4ac) / a²

→ (p - q)² = 4(b² - ac) / a² ------------ Equation (1)

____________________

Now,

We have also given That, Roots of Ax^2 +2Bx+C =0 are p+r and q+r.

So ,

→ (p + r) + (q + r) = (-2B/A)

→ (p + r) * (q + r) = (C/A)

Again using , (x - y)² = (x + y)² - 4xy we get,

→ [(p + r) - (q + r)]² = (-2B/A)² - 4(C/A)

→ [ p + r - q - r ]² = (4B²/A²) - (4C/A)

→ (p - q)² = (4B² - 4AC) / A²

→ (p - q)² = 4(B² - AC) / A² ------------ Equation (2) .

____________________

Now , Comparing Both Equation , we get ,

→ Equation (1) = Equation (2) = (p - q)²

Or,

→ 4(b² - ac) / a² = 4(B² - AC) / A²

→ A²(b² - ac) = a²(B² - AC) (Hence, Proved).

Answer 2

$\mathbb{\red{\huge{HEY\:MATE}}}$

$\boxed{HERE\:IS\:YOUR\:ANSWER}$

$\huge{Question:}$

⭐Let the roots of ax^2 + 2bx + c = 0 are p and q .

Also, the roots of Ax^2 +2Bx+C =0 are p+r and q+r.

Prove that A^2(b^2-ac) = a^2(B^2-AC).

$\huge{Answer:}$

✔$\small{Given:}$

✅Roots of ax^2 + 2bx + c = 0 are p and q .

✅Roots of Ax^2 +2Bx+C =0 are p+r and q+r.

✔$\small{To\:Prove:}$

⭐A^2(b^2-ac) = a^2(B^2-AC)

Now we know that:

✅sum of the roots = (-b/a)

✅Product of roots = c/a.

(x - y)² = (x + y)² - 4xy

✔$\small{Proof:}$

✅ Roots of ax^2 + 2bx + c = 0 are p and q:

=》p + q = (-2b/a)

=》 pq = c/a

Also,

=》 (p - q)² = (p + q)² - 4pq

✔Substituting the values:

=》 (p - q)² = [(-2b/a)]² - 4(c/a)

=》 (p - q)² = (4b²/a²) - (4c/a)

=》 (p - q)² = (4b² - 4ac) / a²

=》 (p - q)² = 4(b² - ac) / a² ----------(1)

✔Also, Roots of Ax^2 + 2Bx+C =0 are p+r and q+r :-

=》 (p + r) + (q + r) = (-2B/A)

=》 (p + r) * (q + r) = (C/A)

Now using , (x - y)² = (x + y)² - 4xy :-

=》 [(p + r) - (q + r)]² = (-2B/A)² - 4(C/A)

=》 [ p + r - q - r ]² = (4B²/A²) - (4C/A)

=》 (p - q)² = (4B² - 4AC) / A²

=》 (p - q)² = 4(B² - AC) / A² ----------(2)

✔Comparing (1) and (2) :

(1) = (2) = (p-q)²

⭐ 4(b² - ac) / a² = 4(B² - AC) / A²

⭐A²(b² - ac) = a²(B² - AC)

$\huge{Hence,Proved}$

✅Hope it helps you :)

#Regards❤