Methods of Solving Quadratic Equations

(b^2-4ac) is called the discriminant and it tells the nature of roots. In exams, it is also written as (D).

For a quadratic equation \(ax^2+bx+c=0\), the nature of the roots is determined by the discriminant \(D=b^2-4ac\). When \(D=0\), \(\sqrt{D}=0\), so the quadratic formula gives both roots as \(-\frac{b}{2a}\); hence they are equal and real. Two distinct real roots occur when \(D>0\), so option B is incorrect. Exam tip: remember \(D=0\) as ‘equal real roots’.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is given by \(D=b^2-4ac\). Here, \(a=1\), \(b=-4\), and \(c=4\), so \(D=(-4)^2-4(1)(4)=16-16=0\). Therefore, the correct answer is 0. Exam tip: Substitute the value of \(b\) with its sign included.

It easily factors as ((x+1)(x+2)=0). In exams, factorisation is fast for questions with small coefficients.

Factorising the equation gives \(x^2+3x+2=(x+1)(x+2)\). Therefore, \((x+1)(x+2)=0\) gives \(x=-1\) or \(x=-2\). Option B has the signs wrong, while options C and D do not satisfy the equation. Exam tip: if \((x+a)=0\), the corresponding root is \(x=-a\).

In \(x^2+8x+16\), the coefficient of \(x\) is 8. Taking half of 8 gives 4, and adding its square, 16, gives \(x^2+8x+16=(x+4)^2\). Therefore, A is correct. In \((x+8)^2\), the middle term would be \(16x\), so B is incorrect. Exam tip: To complete the square in \(x^2+bx\), add \(\left(\frac{b}{2}\right)^2\).

Using the square root method, \(x^2=49\) gives \(x=\pm\sqrt{49}\). Hence, \(x=7\) or \(x=-7\), so \(x=\pm7\). Writing only the positive root (option B) is a common mistake because a positive number has two square roots. Exam tip: For \(x^2=a\), always write \(x=\pm\sqrt{a}\).

Here, \(4x^2-1=(2x)^2-1^2\). Using the difference-of-squares identity \(a^2-b^2=(a-b)(a+b)\), it becomes \((2x-1)(2x+1)\), so option A is correct. Option C expands to \(4x^2-4x+1\), not \(4x^2-1\). In an exam, first check whether the expression matches the difference-of-squares pattern.

((2x-1)(2x+1)=0), so (x=\frac{1}{2}) or (x=-\frac{1}{2}). In exams, solve each linear factor carefully.

The expression x² − 1 is a difference of two perfect squares because 1 = 1². The identity for this pattern is a² − b² = (a − b)(a + b). Taking a = x and b = 1 gives x² − 1² = (x − 1)(x + 1). Hence the equation can be rewritten as (x − 1)(x + 1) = 0, leading to x = 1 or x = −1 by the zero-product property. Therefore option A is correct. Option B is the identity for the square of a sum and would require a middle term 2ab. Option C is not a valid identity in general, and option D is not the zero-product rule; from ab = 0, the correct conclusion is a = 0 or b = 0.

(x^2+6x+8=(x+2)(x+4)), so (x=-2) and (x=-4). In exams, a positive middle term can give negative roots.

(x^2-3x=x(x-3)), so zero product rule gives (x=0). In exams, do not lose (x=0) by dividing by (x).

First factor the equation without dividing by a variable: x² − 3x = x(x − 3) = 0. By the zero-product property, either x = 0 or x − 3 = 0, which gives x = 3. If we divide the original equation by x, we implicitly assume x is nonzero, so the solution x = 0 is excluded before the solving process is complete. The reduced equation x − 3 = 0 therefore retains only x = 3 and misses x = 0. Thus option A is correct. The safe procedure is to factor first and then set each factor equal to zero. Options B is a valid remaining root, while −3 and 1 do not satisfy the original equation.

(3+5=8) and (3\times5=15), so with signs we get ((x-3)(x-5)). In exams, for a negative middle term, look for both negative factors.

((-2)+(-3)=-5) and ((-2)(-3)=6=ac), so the correct split is (-2x-3x). In exams, the product of the two numbers must be (ac).

Multiplying \((2x-3)(x-1)\) gives \(2x^2-2x-3x+3=2x^2-5x+3\), so the correct factorised form is \((2x-3)(x-1)=0\). In option B, the coefficient of the middle term becomes \(1\) instead of \(-5\), while options C and D do not produce the required leading and constant terms. In an exam, multiply the factors back to verify the middle and constant terms.

In ((x-2)^2=9), square root can be taken directly. In exams, recognize the form ((\text{expression})^2=k).

Taking the square root gives \\(x-2=\\pm3\\). Thus, \\(x-2=3\\) gives \\(x=5\\), while \\(x-2=-3\\) gives \\(x=-1\\). Option D is incorrect because the second value should be \\(x=-1\\), not \\(x=1\\). In an exam, remember to consider both cases represented by \\(\\pm\\).

The coefficient of \(x\) is \(-10\), whose half is \(-5\). Squaring it gives \((-5)^2=25\), so 25 must be added: \(x^2-10x+25=(x-5)^2\). Exam tip: in the completing-square method, add the square of half the coefficient of \(x\).

Using the perfect-square identity \((x-a)^2=x^2-2ax+a^2\) with \(a=5\), we get \((x-5)^2=x^2-10x+25\). Hence, option A is correct. Option B would produce a middle term of \(+10x\), while option D represents a difference of squares. In an exam, identify \(25=5^2\) and verify the middle term as \(-2\times5x=-10x\).