Roots of a Quadratic Equation

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 \(\frac{c}{a}\).

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.

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.

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

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.

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

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.

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.

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.

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 \(=45\).

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 \(1\).

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 \(\frac{3}{5}\).

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 \(x^2-(\alpha+\beta)x+\alpha\beta=0\).

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.

For a quadratic equation ax² + bx + c = 0 with roots α and β, Vieta’s relation states that α + β = −b/a. In the given equation, a = 6 and b = −13. Therefore, α + β = −(−13)/6 = 13/6. The constant term c = 5 is used for the product of the roots, not their sum, so 5/6 and −5/6 are not appropriate. The negative option −13/6 would result from forgetting the minus sign in the formula. Hence option A is the unique correct answer.

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 \(\frac{c}{a}\).

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 \(b^2\).

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.”

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.

Two distinct real roots require \(b^2-4ac>0\). For option C, \(25-24=1>0\), so it has distinct real roots. Option B has discriminant 0 and hence equal roots. Exam tip: check the sign of the discriminant first.