Question

please solve this it's a humble request.
please Don't scam for 5 points.
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Answer 1

Answer:

The correct roots of the quadratic equation are 6 and - 1 .

Step-by-step explanation:

Let the given quadratic equation be :

$\sf \: a {x}^{2} + bx + c = 0$

Here,

x represents the roots of the equation.

a = coefficient of x²

b = coefficient of x

c = constant term

Now,

Given,

Incorrect roots of the equation = 3, 2

$\: \: \sf \: x = 3 \: \: \: \: \: \: \: \: or \: \: \: \: \: \: \: x = 2 \\ \therefore \: x - 3 = 0 \: \: \: - - - (1) \\ \therefore \sf \: x - 2 = 0 \: \: \: \: - - - (2)$

A quadratic equation is formed by multiplying the linear equations (1) and (2)

$\: \: \: \sf \: (x - 3)(x - 2) = 0 \\ \implies \sf \: {x}^{2} - 2x - 3x + 6 = 0 \\ \implies \sf \: {x}^{2} - 5x + 6 = 0$

Given that the correct value of the coefficient of

$\sf \: {x}^{2} = 1 \: \: \: \therefore \: a \: = 1$

Correct value of the constant term = - 6

$\sf \: \: c \: = - 6$

It is given that the first student copies only the value of the constant term incorrectly.

★ The coefficient of x is copied correctly by the first student.

$\sf \: \: b = - 5$

Now, putting the values of a, b, and c in the quadratic equation correctly, we will get,

$\sf \: \: \: \: \: 1 \times {x}^{2} + ( - 5)x + ( - 6) = 0 \\ \implies \sf \: {x}^{2} - 5x - 6 = 0 \\ \implies \sf \: {x}^{2} - 6x + x - 6 = 0 \\ \implies \sf \: x(x - 6) + 1(x - 6) = 0 \\ \implies \sf \: (x - 6)(x + 1) = 0 \\ \\ \therefore \: \sf \: x = 6 \\ \sf \: x = - 1$

Therefore, the correct roots of the given quadratic equation are 6 and - 1 .