Roots of a Quadratic Equation

For a repeated root r the monic quadratic is \((x-r)^2=0\). With r=−4 we get \((x+4)^2=0\), which expands to \((x^2+8x+16=0)\). Option B would correspond to repeated root +4 (sign of the linear term is opposite), so it is incorrect. Options C and D have different coefficients and do not give root −4. Exam tip: to form a monic quadratic from roots \(\alpha,\beta\) use \(x^2-(\alpha+\beta)x+\alpha\beta=0\).

Putting (x=\frac{1}{2}) gives (1+\frac{p}{2}-3=0), so (p=4). When substituting a fractional root, solve each term carefully.

When \(p(a)=0\), substituting \(x=a\) satisfies the equation \(p(x)=0\). Therefore, \(a\) is called a root of the equation. Coefficients are numerical factors of terms, a constant is a fixed term, and the degree is the highest exponent. Exam tip: to check whether a value is a root, substitute it into the polynomial and verify that the result is \(0\).

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.

By the factor theorem, if \(x-9\) is a factor, then setting the factor equal to zero gives \(x-9=0\), so \(x=9\). Hence, 9 is certainly a root. Exam tip: a factor of the form \(x-a\) gives the root \(a\), whereas \(x+a\) gives the root \(-a\).

Since \\(5\\) is a root, substitute \\(x=5\\) into the equation: \\(5^2+5t-30=0\\). Thus, \\(25+5t-30=0\\), so \\(5t=5\\) and \\(t=1\\). In such questions, substitute the given root directly and check the signs carefully.

Since \(11\neq 0\), dividing \(11x^2=0\) by 11 gives \(x^2=0\). Thus \(x=0\) occurs twice, so 0 is a repeated root. Exam tip: In an equation of the form \(ax^2=0\), where \(a\neq0\), the root is always \(x=0\) with multiplicity 2.