$CATEGORY: Maths - TicTacLearn/4. Quadratic equations/4. Quadratic equations--Solution of a Quadratic Equation by Factorisation

//Multiple Choice

The first step in solving a quadratic equation by factorization is\: {
~ Finding the square root of the constant term
= Splitting the middle term
~ Multiplying the coefficients
~ None of these
}

What are the factors of \(x^2−5x+6\=0\) ? {
= (x - 2)(x - 3)
~ (x + 2)(x - 3)
~ (x - 2)(x + 3)
~ (x + 2)(x + 3)
}

The roots of the equation \(x^2−4x−5\=0\)  are\: {
~ -1 and -5
~ 1 and -5
= -1 and 5
~ 5 and -5
}

Which of the following is a quadratic equation? {
~ x+5\=0
= \(2x^2−3x+1\=0\)
~ \(3x^3−x+2\=0\)
~ \(x^4+x^2−7\=0\)
}

What should be done if the quadratic equation does not factorize easily? {
= Use the quadratic formula
~ Ignore the equation
~ Multiply the equation by a constant
~ Guess the factors
}

The sum of the roots of \(x^2−7x+10\=0\) is\: {
~ 10
~ -7
= 7
~ -10
}

The product of the roots of \(x^2−4x+3\=0\) is\: {
= 3
~ -3
~ 4
~ -4
}

//True or False

The equation \(x^2+5x+6\=0\) can be solved by factorization. {T}

If the sum of the roots is negative, the quadratic equation must have a negative coefficient of x. {F}

The product of the roots of \(x^2−3x−10\=0\) is - 10. {T}

The roots of \(x^2+4x+4\=0\) are distinct. {F}

\(x^2−9x+14\=0\)  factors as (x−7)(x−2). {T}

The quadratic equation \(x^2+x−6\=0\) has roots -2 and 3. {F}

The factorization method always works for every quadratic equation. {F}

//Numericals



//Fill in the blanks

The equation \(x^2−6x+9\=0\) can be factorized as ______. {
= (x−3)(x−3)
~ (x+3)(x−3)
~ (x+3)(x+3)
~ None
}

The roots of the quadratic equation \(x^2−5x+6\=0\) are ______. {
~ 1, 5
= 2, 3
~ -2, -3
~ 1, 6
}

The factored form of \(x^2+7x+12\=0\) is ______. {
~ (x−3)(x−4)
= (x+3)(x+4)
~ (x+6)(x+2)
~ None
}

The sum of the roots of the equation \(x^2−px+q\=0\) is ______. {
= p
~ -p
~ -q
~ q
}

The product of the roots of the equation \(x^2−px+q\=0\) is ______. {
~ p
~ -p
= q
~ -q
}

The factored form of \(x^2−9\=0\)  is ______. {
= (x+3)(x−3)
~ (x−3)(x−3)(x-3)
~ (x+3)(x+3)
~ None
}

The quadratic equation \(x^2−2x−8\=0\) is factored as ______. {
~ (x+4)(x−2)
= (x−4)(x+2)
~ (x−4)(x−2)
~ None
}

//Match the following

Match the following items from Column A with their correct corresponding options from Column B\:
{
=\(x^2−5x+6\=0\) -> (x−2)(x−3)
=\(x^2+3x−10\=0\) -> (x−5)(x+2)
=\(x^2−4x−12\=0\) -> (x−6)(x+2)
=\(x^2−9\=0\) -> (x+3)(x−3)
}



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