Roots of a Quadratic Equation

(p(2)=0) means the equation is satisfied when (x=2). Substitution is the key method in such questions.

Putting (x=2) gives (4-10+6=0) so it is a root. In exams always check the final sum after substitution.

(x^2-9=(x-3)(x+3)) so the roots are (3) and (-3). In exams quickly identify the difference of squares.

(x^2-4x+3=(x-1)(x-3)) so the roots are (1) and (3). Factorisation is the fastest method for easy questions.

Factor the quadratic: \(x^2+5x+6=(x+2)(x+3)\). Setting each factor to zero gives \(x+2=0\Rightarrow x=-2\) and \(x+3=0\Rightarrow x=-3\). The closest distractor (B) simply has the signs reversed (+2 and +3), which is the common sign error. Exam tip: verify quickly using sum and product of roots — sum = \(-b/a\) = \(-5\) and product = \(c/a\) = \(6\).

With roots (0) and (5) the equation is (x(x-5)=0) that is (x^2-5x=0). Remember the form (x-r) while forming an equation from roots.

Core idea: substitute the given root directly into the equation. With \(x=1\) we get \(1 - k + 2 = 0\) or \(3 - k = 0\), hence \(k = 3\). Option B (2) often arises from a careless mistake such as equating the expression to the wrong value or an arithmetic slip when adding terms. Exam tip: always substitute the root and simplify step by step, watching signs carefully.

For a quadratic equation the sum of roots is (-\frac{b}{a}). Keep the sign of (b) correct while using the formula.

Divide the equation by \(a\) (assuming \(a\neq 0\)) to get \(x^2+\frac{b}{a}x+\frac{c}{a}=0\). Comparing with \(x^2-(\alpha+\beta)x+\alpha\beta=0\) gives \(\alpha\beta=\frac{c}{a}\). Option B has the wrong sign; options C and D confuse the coefficient \(b\) with the constant term. Exam tip: product of roots = constant term ÷ leading coefficient, and sum of roots = -(middle coefficient) ÷ leading coefficient.

The discriminant is defined by \(D = b^2 - 4ac\). Two equal (repeated) real roots occur exactly when \(D = 0\); the repeated root is \(-b/(2a)\). The closest distractor A (D > 0) is wrong because that case gives two distinct real roots; C (D < 0) gives complex conjugate roots; D (D = 1) is just a specific value and does not generally imply equal roots. Exam tip: always compute \(D=b^2-4ac\) and compare with zero to determine the nature of roots quickly.

Discriminant is defined by \(D=b^2-4ac\). Here \(a=1,\;b=-4,\;c=4\). Thus \(D=(-4)^2-4\times1\times4=16-16=0\). So the discriminant is 0 and the quadratic has equal (repeated) roots. Option A (4) is incorrect because correct substitution and arithmetic give 0, not 4. Exam tip: always list values of \(a,b,c\) explicitly before computing \(b^2-4ac\) to avoid sign or multiplication mistakes.

The discriminant \(D=b^2-4ac\) determines the nature of roots. If \(D>0\) there are two distinct real roots, if \(D=0\) there is one repeated real root; hence real roots occur exactly when \(D\ge 0\). Option A is incorrect because \(D<0\) gives complex (non-real) roots. Option C is just a specific value and not a general condition. Option D is incorrect because \(a=0\) makes the equation linear, not quadratic. Exam tip: compute \(D=b^2-4ac\) first to decide root nature quickly.

(x^2) is never negative so (x^2+1=0) is not satisfied by any real number. Sign checking helps solve such questions quickly.

Substituting \(x=0\) gives \(3(0)^2+2(0)=0\). Hence the left-hand side is zero and \(x=0\) is a root. Alternatively factor: \(3x^2+2x=x(3x+2)=0\), giving roots \(x=0\) and \(x=-\tfrac{2}{3}\). Option (C) is incorrect because substituting \(x=2\) yields \(3(2)^2+2(2)=16\neq0\). Option (D) is wrong since the equation is explicit and substitution determines whether a value is a root. Exam tip: To check a root substitute the value into the polynomial or factor the polynomial to find all roots quickly.

Write the equation as \(x^2-x=0\) and factorize: \(x(x-1)=0\). Hence the roots are \(x=0\) and \(x=1\). Option B (1 and 2) is wrong because substituting \(x=2\) gives \(4\neq2\). Option C (-1 and 1) is wrong since \(x=-1\) does not satisfy the equation (\((-1)^2=1\neq-1\)). Exam tip: always bring the equation to \(=0\) and factorize to find roots quickly and reliably.

Divide both sides by 2: from \(2x^2-8=0\) we get \(x^2=4\). Taking square roots gives \(x=\pm 2\), i.e. 2 and −2. Option B would come from incorrectly assuming \(x^2=16\); option C is wrong because 0 does not satisfy the equation; option D is incomplete because it omits the negative root. Exam tip: when taking square roots always include both \(+\) and \(-\) solutions after squaring.

Factor the quadratic: \(x^2+2x+1=(x+1)^2\). Thus \((x+1)^2=0\) gives the double root \(x=-1\). A common mistake is choosing \(1\), which would correspond to \((x-1)^2\) and the polynomial \(x^2-2x+1\). Exam tip: check either factorisation or compute the discriminant \(b^2-4ac\); if it equals 0, the quadratic has a repeated root.

The equation ((x-2)(x-5)=0) gives (x^2-7x+10=0). You can also check the sum and product of roots.

Use factorisation or sum-and-product of roots. Factorising gives \(x^2-6x+8=(x-4)(x-2)\), so with one root 4 the other root is 2. The close distractor 1 is wrong because \(4\times1=4\), not equal to the constant term 8. Exam tip: For \(ax^2+bx+c=0\), use \(\alpha+\beta=-\dfrac{b}{a}\) and \(\alpha\beta=\dfrac{c}{a}\) to find the other root quickly.

A negative product occurs when one root is positive and the other is negative. (\alpha\beta<0) is a quick sign check.