Roots of a Quadratic Equation

Substituting \(x=2\) gives \(2^2-4(2)+4=4-8+4=0\). Therefore, \(x=2\) is a root of the quadratic equation. Option C is incorrect because substituting \(x=4\) gives \(16-16+4=4\), not zero. Exam tip: To test whether a number is a root, substitute it into the equation and check whether the result is zero.

Substituting \(x=-1\) into the equation gives \(2(-1)^2+(-1)-1=2-1-1=0\). Therefore, \(x=-1\) is a root of the quadratic equation. In an exam, check a proposed root by substituting it and verifying that the result is zero.

From the equation, \(x^2=36\). Therefore, \(x=\pm\sqrt{36}=\pm6\), so the two roots are 6 and -6. Option B incorrectly treats 36 and -36 as the roots, whereas 36 is the constant in the equation. Exam tip: when \(x^2=a^2\), the roots are \(x=\pm a\).

Factoring the quadratic gives \(x^2-8x+15=(x-3)(x-5)\). Thus, \((x-3)(x-5)=0\) implies \(x=3\) or \(x=5\), so the roots are 3 and 5. The negative pair is incorrect because it gives the wrong sign for the middle term. Exam tip: look for factors of 15 whose sum is 8.

The equation can be factorised as \(x^2+10x+21=(x+3)(x+7)\). Hence, \((x+3)(x+7)=0\) gives \(x=-3\) or \(x=-7\). Option B has the wrong signs. Exam tip: the sum of the roots must be \(-10\) and their product must be \(21\).

If the roots of a quadratic equation are 0 and 8, its factored form is \((x-0)(x-8)=0\). Thus, \(x(x-8)=0\), which expands to \(x^2-8x=0\). Option B has roots 0 and −8, while option C has roots 8 and −8. Exam tip: for roots \(\alpha\) and \(\beta\), use \((x-\alpha)(x-\beta)=0\).

Since \(x=3\) is a root, substitute it directly into the equation: \(3^2+3s-12=0\). Thus, \(9+3s-12=0\), so \(3s=3\) and \(s=1\). Therefore, option A is correct. Exam tip: whenever a root is given, substitute it for \(x\) in the quadratic equation.

For a quadratic equation \(ax^2+bx+c=0\), the sum of its roots is \(-\frac{b}{a}\). In the given equation, \(a=2\) and \(b=3\), so the sum is \(-\frac{3}{2}\). Option B misses the negative sign, while options C and D incorrectly use the constant term instead of the coefficient of \(x\). In an exam, apply the sum-of-roots formula \(-b/a\) directly.

For a quadratic equation \(ax^2+bx+c=0\), the product of its roots is \(\frac{c}{a}\). In this equation, \(a=6\) and \(c=-2\), so the product is \(\frac{-2}{6}=-\frac{1}{3}\). Option B misses the negative sign, while options C and D use an incorrect denominator. Exam tip: remember that the sum of the roots is \(-\frac{b}{a}\), whereas their product is \(\frac{c}{a}\).

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Using the quadratic formula \(x=\frac{-b\pm\sqrt{D}}{2a}\), when \(D=0\), both values of \(x\) become \(\frac{-b}{2a}\). Hence, the equation has two equal real roots. Option A applies when \(D>0\), whereas \(D<0\) gives no real roots. Exam tip: remember that \(D=0\) means equal and real roots.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=1\), \(b=4\), and \(c=4\). Therefore, \(D=4^2-4(1)(4)=16-16=0\). The value 16 results from forgetting to subtract \(4ac\), while -16 comes from using an incorrect sign. Thus, 0 is correct, and the equation has real and equal roots. Exam tip: identify \(a\), \(b\), and \(c\), including their signs, before substituting them into the formula.

For the quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). When \(D>0\), the equation has two distinct real roots. If \(D=0\), the roots are equal, while \(D<0\) gives non-real roots. In exams, first calculate \(D=b^2-4ac\) to determine the nature of the roots.

For every real \(x\), \(x^2\ge 0\). Hence \(x^2+25\ge 25\), so it can never equal zero. Therefore, \(x^2+25=0\) has no real roots. Option D is incorrect because both \(5^2+25\) and \((-5)^2+25\) equal 50, not zero. Exam tip: a quadratic equation has no real roots when its discriminant \(D=b^2-4ac\) is negative.

Factoring the equation gives \(x^2+6x=x(x+6)=0\). Hence, either \(x=0\) or \(x+6=0\), which gives \(x=-6\). Therefore, the roots are \(0\) and \(-6\). Option B has the incorrect sign for the second root. Exam tip: When a quadratic has no constant term, take the common factor first and set each factor equal to zero.

Rearranging the equation gives \(x^2-4x=0\). Factoring, \(x(x-4)=0\), so by the zero-product property, \(x=0\) or \(x=4\). Therefore, the roots are 0 and 4. Exam tip: first bring all terms to one side and then factorise the quadratic.

Dividing \(5x^2-45=0\) by 5 gives \(x^2-9=0\), so \(x^2=9\). Hence, \(x=\pm3\), and the roots are 3 and −3. In option B, 9 and −9 are incorrectly treated as the roots instead of the value of \(x^2\). Exam tip: whenever \(x^2=a\), write both roots as \(x=\pm\sqrt{a}\).

The equation can be written as \(x^2-12x+36=(x-6)^2\). Thus, \((x-6)^2=0\), giving \(x=6\), which is the repeated root. Exam tip: In a perfect-square quadratic equation, set the expression inside the square equal to zero to find the repeated root.

For a quadratic equation, the discriminant is \(D=b^2-4ac\). When \(D=0\), both roots are equal and real, with root \(-b/(2a)\). If \(D>0\), the roots are distinct real roots. Exam tip: identify the sign of the discriminant first.

For a quadratic equation \(ax^2+bx+c=0\), the sum of the roots is \(-\frac{b}{a}\). Here, the sum of the roots is \(11\), so the other root is \(11-5=6\). Indeed, \(x^2-11x+30=(x-5)(x-6)\), giving roots \(5\) and \(6\). Remember that \(11\) is the sum of the roots, not the other root.

Let the two real roots be α and β. Their product αβ is positive, so the roots must have the same sign: both positive or both negative. Their sum α + β is also positive. If both roots were negative, their sum would necessarily be negative, so that possibility is ruled out. Therefore both roots are positive. Option C would give a negative product, while option D would make the product zero, not positive. This conclusion uses the sign relationships between the sum and product of two real numbers, and it does not require calculating the roots individually.