Roots of a Quadratic Equation

For every real \(x\), \(x^2\ge 0\), so \(x^2+36\ge 36\) and can never equal zero. Therefore, the equation has no real roots, making 0 the correct answer. As an exam tip, the discriminant can also be checked: \(D=b^2-4ac=-144<0\), which confirms that there are no real roots.

Factoring the equation gives \(x^2+x-12=(x+4)(x-3)\). Thus, \(x+4=0\) gives \(x=-4\), and \(x-3=0\) gives \(x=3\). Therefore, the roots are 3 and −4. In option B, the signs of both roots are reversed. Exam tip: Look for two factors of −12 whose product is −12 and whose sum is 1.

The equation \(x^2=121\) means that the square of \(x\) is 121. Since \(11^2=121\) and \((-11)^2=121\), both \(x=11\) and \(x=-11\) are roots. Option B is incomplete because it includes only the positive root. Exam tip: When solving \(x^2=a\) for positive \(a\), consider both signs, \(x=\pm\sqrt{a}\).

Let the roots be 6 and \(x\). Since their product is 48, \(6x=48\), so \(x=\frac{48}{6}=8\). Therefore, the other root is 8. The value 42 results from subtracting 6 from 48, but the question gives the product, so division is required. Exam tip: Divide the product of the roots by the known root to find the other root.

The sum of the roots equals the first root plus the second root. Therefore, the other root = 12 − 5 = 7, so option B is correct. Exam tip: subtract the known root from the sum of the roots; do not add it.

The two roots are \(r\) and \(r\), so their product is \(r \times r = r^2\). Option A, \(2r\), represents the sum of the roots, not their product. Exam tip: for equal roots \(r, r\), the sum is \(2r\) and the product is \(r^2\).

The equation can be written as \(4x^2-12x+9=(2x-3)^2\). Hence, \((2x-3)^2=0\), so \(2x-3=0\) and \(x=\frac{3}{2}\). Both roots are equal and repeated. In the exam, identifying the perfect square is the quickest method; the negative value \(-\frac{3}{2}\) is not a root.

If both roots are 5, the equation formed from the roots is \((x-5)(x-5)=0\), or \((x-5)^2=0\). Expanding it gives \(x^2-10x+25=0\), so option A is correct. Option B has the wrong sign for the coefficient of \(x\), since the sum of the roots is 10. Exam tip: For roots \(\alpha\) and \(\beta\), a monic quadratic is \(x^2-(\alpha+\beta)x+\alpha\beta=0\).

A number is a root of an equation when substitution of that number makes the equation equal to zero. Substitute x = −4 into 2x² + px − 8 = 0: 2(−4)² + p(−4) − 8 = 0. Since (−4)² = 16, this becomes 32 − 4p − 8 = 0, or 24 − 4p = 0. Hence 4p = 24 and p = 6. Therefore option A is correct. The negative value of the root affects the px term, giving −4p; ignoring this sign could lead to option B or another incorrect answer.

(p(r)=0) means the equation is satisfied when (x=r). Substitution is the safest way to check a root.

To check whether a number is a root, substitute it into the equation and verify whether the result is zero. For (x=3) , (3^2-7\times3+12=9-21+12=0) , so 3 is a root of the equation. The other root is 4, so option C is incorrect. Exam tip: For root verification, substitute the given value directly and check whether the expression becomes zero.

To test whether a number is a root, substitute it into the equation and check whether the result is zero. For \(x=-1\), \(2(-1)^2+5(-1)+3=2-5+3=0\). Therefore, \(x=-1\) is a root. In exams, carefully check the sign of the linear term and remember that the square of a negative number is positive.

The discriminant \(D=b^2-4ac\) determines the nature of roots. When \(D=0\), \(x=\frac{-b}{2a}\) occurs twice, so the roots are real and equal. Exam tip: check the sign of the discriminant first.

The equation can be factorised as \(2x^2-7x+3=(2x-1)(x-3)\). Therefore, \((2x-1)(x-3)=0\) gives \(x=\frac{1}{2}\) or \(x=3\), so option A is correct. The roots in option D do not have the required sum and product for this equation. Exam tip: verify the roots using sum \(\frac{-b}{a}=\frac{7}{2}\) and product \(\frac{c}{a}=\frac{3}{2}\).

Splitting the middle term gives \(3x^2+10x+3=3x^2+9x+x+3=(3x+1)(x+3)\). Therefore, \((3x+1)(x+3)=0\) gives \(x=-\frac{1}{3}\) or \(x=-3\), so option A is correct. Exam tip: After factorising, set each factor equal to zero and verify that the sum of the roots is \(-\frac{10}{3}\).

Because x = 2 is a root, substituting x = 2 must make the quadratic expression equal to zero. Thus 2² + k(2) − 14 = 0. Simplifying gives 4 + 2k − 14 = 0, so 2k − 10 = 0. Therefore 2k = 10 and k = 5. Option A is correct. The constant term is already −14 and should not be changed; only the given root is substituted into the x terms. Option B results from an incorrect sign, while options C and D do not satisfy the equation when k is checked by substitution.

For a quadratic equation \(ax^2+bx+c=0\), the sum of the roots is \(-\frac{b}{a}\). Here, \(a=5\) and \(b=-4\), so the sum is \(-\frac{-4}{5}=\frac{4}{5}\). Exam tip: substitute the signed value of \(b\), not just its magnitude, in the formula.

For a quadratic equation \\(ax^2+bx+c=0\\), the product of its roots is \\(\\frac{c}{a}\\). Here, \(a=3\) and \(c=-11\), so the product is \\(\\frac{-11}{3}\\). Option C incorrectly uses the coefficient ratio \\(\\frac{b}{a}=\\frac{8}{3}\\), which is not the product of the roots. Exam tip: remember that the sum of roots is \\( -\\frac{b}{a}\\), while their product is \\(\\frac{c}{a}\\).

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=2\), \(b=-4\), and \(c=2\), so \(D=(-4)^2-4(2)(2)=16-16=0\). Therefore, the roots are real and equal. Exam tip: when \(D=0\), a quadratic equation has two equal real roots.

Since (D=49>0), two distinct real roots will be obtained. When (D>0), the roots are different.