Roots of a Quadratic Equation

Correct answer: A. 4

Explanation: For a quadratic equation \(ax^2+bx+c=0\), the sum of the roots is \(\alpha+\beta=-\frac{b}{a}\). Here, \(a=3\) and \(b=-12\), so \(\alpha+\beta=-\frac{-12}{3}=4\). Option B has the wrong sign, while \(\frac{5}{3}\) and \(-\frac{5}{3}\) are related to the product of the roots, not their sum. Exam tip: Always check the sign of \(b\) before applying the formula.

Correct answer: A. \(b^2-4ac=0\)

Explanation: The discriminant is \(D=b^2-4ac\). If \(D=0\), the formula gives the same value \(x=\frac{-b}{2a}\) twice, so the roots are equal. In contrast, \(D>0\) gives distinct real roots. Exam tip: check the discriminant sign first.

Correct answer: A. \(k\le 4\)

Explanation: A quadratic equation has real roots when its discriminant satisfies \(D=b^2-4ac\ge0\). Here, \(a=1\), \(b=4\), and \(c=k\), so \(D=16-4k\). Therefore, \(16-4k\ge0\) gives \(k\le4\). Option D describes only a sufficient subset, not the complete condition, since roots are also real for \(0\le k\le4\). Exam tip: For questions about real roots, first apply \(D\ge0\).

Correct answer: A. Both negative

Explanation: The roots are (-3) and (-6) because ((x+3)(x+6)=0). When the sum is negative and product is positive, both roots are negative.

Correct answer: A. -8

Explanation: For the quadratic equation \(x^2+ax+16=0\), the sum of the roots is \(-\frac{a}{1}=-a\). Since both roots are 4, their sum is \(4+4=8\). Thus, \(-a=8\), giving \(a=-8\). Exam tip: In \(x^2+bx+c=0\), the sum of the roots is always \(-b\), not \(b\).

Correct answer: A. \\(\\frac{1}{2}\\) and \\(\\frac{3}{4}\\)

Explanation: Factor the quadratic as \\(8x^2-10x+3=(2x-1)(4x-3)\\)。 Thus, \\(2x-1=0\\) gives \\(x=\\frac{1}{2}\\), and \\(4x-3=0\\) gives \\(x=\\frac{3}{4}\\)。 Hence option A is correct. Exam tip: After factorising, set each factor equal to zero and substitute the roots back into the original equation to verify them.

Correct answer: A. \(-b^2\)

Explanation: For a quadratic equation \(ax^2+px+c=0\), the product of its roots is \(\frac{c}{a}\). Since the roots are \(b\) and \(-b\), \(\frac{c}{a}=b\times(-b)=-b^2\). Option B misses the negative sign. Exam tip: remember that the sum of roots is \(-\frac{p}{a}\), while their product is \(\frac{c}{a}\).

Correct answer: A. \u00028x^2+5x-14=0\u00029

Explanation: For roots \(\alpha\) and \(\beta\), the monic quadratic equation is \(x^2-(\alpha+\beta)x+\alpha\beta=0\). Substituting the given values gives \(x^2-(-5)x+(-14)=0\), which simplifies to \(x^2+5x-14=0\). Option B would give the roots a sum of \(5\), so it is incorrect. Exam tip: use the negative of the sum of the roots as the coefficient of \(x\), and use the product directly as the constant term.

Correct answer: A. 29

Explanation: Let the roots be \(\alpha\) and \(\beta\). By the relationships between roots and coefficients, \(\alpha+\beta=7\) and \(\alpha\beta=10\). Therefore, \(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=7^2-2(10)=49-20=29\). Hence, 29 is correct. Exam tip: Use the identity \(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta\) directly instead of solving for the roots.

Correct answer: A. (2)

Explanation: Putting (x=3) gives (9a-24+6=0), so (9a=18) and (a=2). Substitute in questions with an unknown coefficient.

Correct answer: A. 18 and 81

Explanation: In the standard form \(ax^2+bx+c=0\), we have \(a=1, b=-18, c=81\). Hence, the sum of the roots is \(-b/a=18\), and their product is \(c/a=81\). Equivalently, the equation is \((x-9)^2=0\), so both roots are 9; their sum is 18 and product is 81. Exam tip: identify the sum as \(-b/a\) and the product as \(c/a\) directly.

Correct answer: A. (2)

Explanation: Here (\alpha+\beta=\frac{7}{2}) and (\alpha\beta=\frac{3}{2}). Therefore (\alpha+\beta-\alpha\beta=\frac{7}{2}-\frac{3}{2}=2).

Correct answer: A. Root

Explanation: The condition \(q(t)=0\) means that substituting \(x=t\) into \(q(x)\) makes the equation true. Therefore, \(t\) is called a root of the quadratic equation. The degree is the highest power of the polynomial, while the constant term and coefficient are parts of its terms. Exam tip: To verify whether a number is a root of \(q(x)=0\), substitute it for \(x\) and check whether the result is zero.

Correct answer: A. Yes

Explanation: Substituting \(x=5\) into the equation gives \(5^2-10(5)+25=25-50+25=0\). Therefore, \(x=5\) is a root. In fact, \(x^2-10x+25=(x-5)^2\), so 5 is a repeated root; substituting \(x=-5\) gives 100, not zero. Exam tip: a number is a root of a polynomial only when substitution makes the polynomial equal to zero.

Correct answer: A. Yes

Explanation: Substituting \(x=-3\) gives \(2(-3)^2+7(-3)+3=18-21+3=0\). Therefore, \(x=-3\) is a root of the quadratic equation. Option C is incorrect because substituting \(x=3\) gives \(18+21+3=42\), not zero. Exam tip: a number is a root of a polynomial if substitution makes the polynomial equal to zero.

Correct answer: A. 5 and 9

Explanation: Factorise the quadratic: \(x^2-14x+45=(x-5)(x-9)\). Thus, \((x-5)(x-9)=0\) gives \(x=5\) or \(x=9\). Therefore, the roots are 5 and 9. In option B, the product is 45, but the sum is 18 rather than 14. Exam tip: verify the roots using sum \(=14\) and product \(=45\).

Correct answer: A. 4 और \(\frac{1}{4}\)

Explanation: Factoring the quadratic gives \(4x^2-17x+4=(4x-1)(x-4)\). Thus, \(4x-1=0\) gives \(x=\frac{1}{4}\), and \(x-4=0\) gives \(x=4\). Therefore, option A is correct. The values in option D do not have the required sum and product for the roots. Exam tip: verify the roots using their sum \(\frac{17}{4}\) and product \(1\).

Correct answer: A. \(-3,\ -\frac{1}{5}\)

Explanation: Factoring the quadratic gives \(5x^2+16x+3=(5x+1)(x+3)\). Hence, \((5x+1)(x+3)=0\) gives \(x=-\frac{1}{5}\) or \(x=-3\), so option A is correct. Exam tip: Set each factor equal to zero and verify that the sum of the roots is \(-\frac{16}{5}\) and their product is \(\frac{3}{5}\).

Correct answer: A. \(x^2+5x-24=0\)

Explanation: For roots 3 and −8, the quadratic equation is \((x-3)(x+8)=0\). Expanding gives \(x^2+5x-24=0\), so option A is correct. Option B has roots 8 and −3 because the sign of the coefficient of \(x\) is different. Exam tip: if the roots are \(\alpha\) and \(\beta\), the monic quadratic is \(x^2-(\alpha+\beta)x+\alpha\beta=0\).

Correct answer: A. 4

Explanation: Substitute the given root (x=4) into the equation: (4^2+4k-32=0). Thus, (16+4k-32=0), so (4k=16) and (k=4). The value (-4) results from a sign error. Exam tip: Substitute the given root directly into the quadratic equation to find the parameter.

Correct answer: A. 13/6

Explanation: For a quadratic equation ax² + bx + c = 0 with roots α and β, Vieta’s relation gives α + β = −b/a. Comparing 6x² − 13x + 5 = 0 with the standard form, we have a = 6 and b = −13. Therefore α + β = −(−13)/6 = 13/6. Hence option A is correct. The constant term c = 5 determines the product αβ = c/a = 5/6, not the sum; this explains why option C is inappropriate. Option D incorrectly combines the constant term with a negative sign. Option B results from forgetting the minus sign in Vieta’s sum formula. Thus the leading coefficient and the x-coefficient, rather than the constant term, determine the required answer.

Correct answer: A. \(-\frac{9}{7}\)

Explanation: For a quadratic equation \(ax^2+bx+c=0\), the product of its roots is \(\frac{c}{a}\). Here, \(a=7\) and \(c=-9\), so the product is \(\frac{-9}{7}=-\frac{9}{7}\). Option B has the wrong sign, while \(-\frac{4}{7}\) and \(\frac{4}{7}\) incorrectly use the coefficient of the middle term instead of the constant term. Exam tip: remember that the sum of roots is \(-\frac{b}{a}\), whereas their product is \(\frac{c}{a}\).

Correct answer: A. 0

Explanation: For a quadratic equation \(ax^2+bx+c=0\), \(a=4\), \(b=-20\), and \(c=25\). Thus, \(D=b^2-4ac=(-20)^2-4(4)(25)=400-400=0\). Therefore, 0 is correct, and the equation has two equal real roots. Option 100 represents only \(b^2\) and does not subtract \(4ac\). Exam tip: Always put the negative value of \(b\) in parentheses while calculating \(b^2\).

Correct answer: A. विविक्तकर \(0\) होने पर दोनों वास्तविक मूल समान होते हैं, भिन्न नहीं।

Explanation: Here \(a=1, b=-6, c=9\), so \(D=b^2-4ac=36-36=0\). When \(D=0\), the real roots are equal; distinct real roots require \(D>0\). Exam tip: do not confuse “non-negative” with “positive.”

Correct answer: C. It has no real roots

Explanation: The discriminant is \(D=b^2-4ac=(-6)^2-4(1)(20)=36-80=-44\). Since \(D<0\), the quadratic equation has no real roots. Option B would be correct only if \(D=0\). Exam tip: the sign of the discriminant quickly determines the nature of the roots.