We are given that the roots of the quadratic equation x² + px + q = 0 are in the ratio 2:3. Therefore, we can represent the roots as 2α and 3α for some non-zero constant α.

For a quadratic equation ax² + bx + c = 0, the sum of the roots is given by -b/a. In our equation x² + px + q = 0, a=1, b=p, and c=q. So, the sum of the roots 2α + 3α is equal to -p/1 = -p. This simplifies to 5α = -p.

For a quadratic equation ax² + bx + c = 0, the sum of the roots is given by -b/a. In our equation x² + px + q = 0, a=1, b=p, and c=q. So, the sum of the roots 2α + 3α is equal to -p/1 = -p. This simplifies to 5α = -p.

For a quadratic equation ax² + bx + c = 0, the product of the roots is given by c/a. In our equation x² + px + q = 0, the product of the roots (2α)(3α) is equal to q/1 = q. This simplifies to 6α² = q.

For a quadratic equation ax² + bx + c = 0, the product of the roots is given by c/a. In our equation x² + px + q = 0, the product of the roots (2α)(3α) is equal to q/1 = q. This simplifies to 6α² = q.

From equation (1), we have 5α = -p. We can express α in terms of p by dividing both sides by 5, which gives α = -p/5.

Now we substitute the expression for α from the previous step into equation (2). This means replacing α with -p/5 in the equation 6α² = q. Squaring -p/5 gives p²/25. Multiplying by 6, we get 6p²/25 = q.

Now we substitute the expression for α from the previous step into equation (2). This means replacing α with -p/5 in the equation 6α² = q. Squaring -p/5 gives p²/25. Multiplying by 6, we get 6p²/25 = q.

To obtain the desired identity, we multiply both sides of the equation 6p²/25 = q by 25. This results in 6p² = 25q which is what we needed to prove.

If the roots of $x^2 + px + q = 0$ are in the ratio 2:3, prove that $6p^2 = 25q$ .

Answer: The proof shows that if the roots of $x^2 + px + q = 0$ are in the ratio 2:3 $, then 6p^2 = 25q$ .

Step-by-step solution

Step 1: Define the roots

We are given that the roots of the quadratic equation $x^2 + px + q = 0$ are in the ratio 2:3. Therefore, we can represent the roots as $2\alpha$ and $3\alpha$ for some non-zero constant $\alpha$ .

Step 2: Apply Vieta's formulas for sum of roots

For a quadratic equation $ax^2 + bx + c = 0$ , the sum of the roots is given by -b/a. In our equation $x^2 + px + q = 0$ $, a=1$ , $b=p$ $, and c=q$ . So, the sum of the roots $2\alpha + 3\alpha$ is equal to -p/1 = -p. This simplifies to $5\alpha = -p$ .

Step 3: Apply Vieta's formulas for product of roots

For a quadratic equation $ax^2 + bx + c = 0$ , the product of the roots is given by c/a. In our equation $x^2 + px + q = 0$ , the product of the roots ( $2\alpha$ )( $3\alpha$ ) is equal to q/1 = q. This simplifies to $6\alpha^2 = q$ .

Step 4: Express $\alpha$ in terms of $p$

From equation (1), we have $5\alpha = -p$ . We can express $\alpha$ in terms of p by dividing both sides by 5, which gives $\alpha = -p/5$ .

Step 5: Substitute $\alpha$ into the product of roots equation

Now we substitute the expression for $\alpha$ from the previous step into equation (2). This means replacing $\alpha$ with -p/5 in the equation $6\alpha^2 = q$ $. Squaring -p/5 gives p^2/25$ . Multiplying by 6, we get $6p^2/25 = q$ .

Step 6: Rearrange to prove the identity

To obtain the desired identity, we multiply both sides of the equation $6p^2/25 = q$ by 25. This results in $6p^2 = 25q$ which is what we needed to prove.

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If the roots of x2+px+q=0x^2 + px + q = 0x2+px+q=0 are in the ratio 2:3, prove that 6p2=25q6p^2 = 25q6p2=25q.