Comparing the given equation (p+4)x² - (p+1)x + 1 = 0 with the standard form ax² + bx + c = 0, we identify a = p + 4, b = -(p + 1), and c = 1. Since it is a quadratic equation, the coefficient of x² cannot be 0, so p ≠ -4.

Comparing the given equation (p+4)x² - (p+1)x + 1 = 0 with the standard form ax² + bx + c = 0, we identify a = p + 4, b = -(p + 1), and c = 1. Since it is a quadratic equation, the coefficient of x² cannot be 0, so p ≠ -4.

A quadratic equation has real and equal roots if and only if its discriminant b² - 4ac = 0. Substituting the values of a, b, and c, we set up the equation for p.

Expanding the discriminant expression gives (p² + 2p + 1) - (4p + 16) = 0, which simplifies to p² - 2p - 15 = 0. Factoring this quadratic equation gives (p - 5)(p + 3) = 0, so p = 5 or p = -3. Both values satisfy p ≠ -4.

Expanding the discriminant expression gives (p² + 2p + 1) - (4p + 16) = 0, which simplifies to p² - 2p - 15 = 0. Factoring this quadratic equation gives (p - 5)(p + 3) = 0, so p = 5 or p = -3. Both values satisfy p ≠ -4.

For p = -3, the quadratic equation becomes x² + 2x + 1 = 0. Using the formula for equal roots -(b)/(2a), we get x = -(2)/(2(1)) = -1, so the roots are -1, -1.

Find the value(s) of p for which the quadratic equation given as: $(p+4)x² - (p+1)x + 1=0$ has real and equal roots. Also, find the roots of the equation(s) so obtained

Answer: The values of $p$ are $5$ and $-3$ . When $p = 5$ , the roots are $\frac{1}{3}, \frac{1}{3}$ . When $p = -3$ , the roots are $-1, -1$ .

Step-by-step solution

Step 1: Identify coefficients of the quadratic equation

Comparing the given equation $(p+4)x^2 - (p+1)x + 1 = 0$ with the standard form $ax^2 + bx + c = 0$ , we identify $a = p + 4$ , $b = -(p + 1)$ , and $c = 1$ . Since it is a quadratic equation, the coefficient of $x^2$ cannot be 0, so $p \neq -4$ .

Step 2: Set the discriminant to zero for equal roots

A quadratic equation has real and equal roots if and only if its discriminant $b^2 - 4ac = 0$ . Substituting the values of $a$ , $b$ , and $c$ , we set up the equation for $p$ .

Step 3: Solve the equation for p

Expanding the discriminant expression gives $(p^2 + 2p + 1) - (4p + 16) = 0$ , which simplifies to $p^2 - 2p - 15 = 0$ . Factoring this quadratic equation gives $(p - 5)(p + 3) = 0$ , so $p = 5$ or $p = -3$ . Both values satisfy $p \neq -4$ .

Step 4: Find the roots when $p = 5$

For $p = 5$ , the quadratic equation becomes $9x^2 - 6x + 1 = 0$ . When the discriminant is 0, the equal roots are given by $-\frac{b}{2a}$ . Here, $x = -\frac{-6}{2(9)} = \frac{1}{3}$ , so the roots are $\frac{1}{3}, \frac{1}{3}$ .

Step 5: Find the roots when $p = -3$

For $p = -3$ , the quadratic equation becomes $x^2 + 2x + 1 = 0$ . Using the formula for equal roots $-\frac{b}{2a}$ , we get $x = -\frac{2}{2(1)} = -1$ , so the roots are $-1, -1$ .

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Find the value(s) of p for which the quadratic equation given as: (p+4)x2−(p+1)x+1=0(p+4)x² - (p+1)x + 1=0(p+4)x2−(p+1)x+1=0 has real and equal roots. Also, find the roots of the equation(s) so obtained