Lesson Problems on quadratic equations to solve them using discriminant
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<H2>Problems on quadratic equations to solve them using discriminant</H2> <H3>Problem 1</H3>Prove that equation 2x^2 - x + 3 = 0 has no real solutions. <B>Solution</B> <pre> Quadratic equation ax^2 + bx + c = 0 does not have real solutions if and only if its discriminant d = b^2 - 4ac is negative. In our case, the discriminant is d = b^2 - 4ac = (-1)^2 - 4*2*3 = 1 - 24 = -23. The discriminant is negative- hence, this equation 2x^2 - x + 3 = 0 does not have real solutions. </pre> <H3>Problem 2</H3>Find the smallest positive real value of 'k' such that equation 13x^2 - kx + 9 = 0 has at least one real solution. <B>Solution</B> <pre> The starting equation is 13x^2 - kx - 9 = 0. (1) Quadratic equation (1) has at least one real solution if and only if its discriminant is non-negative. Write the discriminant d = b^2 - 4ac = (-k)^2 - 4*13*9 = k^2 - 468. (2) Discriminant (2) is non-negative if k^2 - 468 >= 0, or k^2 >= 468, |k| >= {{{sqrt(468)}}}, |k| >= {{{2*3*sqrt(13)}}}, |k| >= {{{6*sqrt(13)}}}. (3) Since we are looking for positive values of 'k', inequality (3) takes the form k >= {{{6*sqrt(13)}}}. <U>ANSWER</U>. The smallest positive value of 'k' is {{{6*sqrt(13)}}} = 21.6333 (approximately). </pre> <H3>Problem 3</H3>Let 'c' be a real number. Find the maximum value of 'c' such that the graph of the parabola y = 2x^2 has at most one point of intersection with the line y = x+c. <B>Solution</B> <pre> This problem can be equivalently re-phrased in this way Let 'c' be a real number. Find the maximum value of 'c' such that equation 2x^2 = x+c has at most one solution in real numbers. It is equivalently to the condition that the discriminant of equation 2x^2 - x - c = 0 is non-positive real number. The discriminant of the last equation is d = b^2 - 4ac = (-1)^2 - 4*2*(-c) = 1 + 8c. Inequality d <= 0 for the discriminant is 1 + 8c <= 0. It implies c <= -1/8. The maximum value of 'c', satisfying this inequality, is -1/8. <U>ANSWER</U>. The maximum value of 'c', satisfying the imposed condition is -1/8. </pre> <H3>Problem 4</H3>Find the range of values of 'k' for which the expression 3 - 4k - (k+3)x - x^2 will be negative for all real values of x. <B>Solution</B> <pre> The given quadratic function 3-4k - (k+3)x - x^2 has the leading coefficient -1 at x^2, so it represents a parabola opened downward. In order for this quadratic polynomial be negative at all real values of x, the necessary and sufficient condition is that the discriminant d = b^2 - 4ac be negative. Then the square function has no real roots and remains negative for all real values of x. So, we write the discriminant d = (-(k+3))^2 - 4*(-1)*(3-4k) = (k+3)^2 + 4*(3-4k) = k^2 + 6k + 9 + 12 - 16k = k^2 - 10k + 21. It can be factored d = (k-7)*(k-3). So, the discriminant is negative (k-7)*(k-3) < 0 if and only if the parameter "k" is between the roots 3 < k < 7, when the factor (k-7) is negative, while the factor (k-3) is positive. At this point, the problem is fully solved. <U>ANSWER</U>. The range of values of k for which the expression 3 - 4k - (k+3)x - x^2 will be negative for all real values of x is 3 < k < 7. </pre> <H3>Problem 5</H3>For an integer n, the inequality 4x^2 + nx + 121 < 0 has no real solutions in x. Find the number of different possible values of n. <B>Solution</B> <pre> The given inequality is 4x^2 + nx + 121 < 0. It has no real solutions if and only if the discriminant of the quadratic polynomial in the left side is negative d = b^2 - 4ac < 0 n^2 - 4*4*121 < 0 n^2 < 16*121 |n| < {{{sqrt(16*121)}}} |n| < 4*11 |n| < 44 -44 < n < 44. This inequality has 43 + 1 + 43 = 87 solutions in integer numbers (43 negative, one zero and 43 positive integer solutions). <U>ANSWER</U>. There are 87 different possible integer values of n such that the given inequality has no real solutions. </pre> <H3>Problem 6</H3>Prove that this system of equations x + y = 8 and x^3 + y^3 = 200 - 4x^2 - 4y^2 has no solutions in real numbers. <B>Solution</B> <pre> Left side x^3 + y^3 can be composed as (x+y)*(x^2 - xy + y^2). The factor (x+y) can be replaced by 8, based on the first equation. After that, second equation can be written in this form 8(x^2 - xy + y^2) = 200 - 4(x^2 + y^2), or 12(x^2 + y^2) - 8xy = 200, 12(x+y)^2 - 24xy - 8xy = 200, 12(x+y)^2 - 32xy = 200. Again, we can replace here (x+y) by 8, and we get then 12*8^2 - 32xy = 200, 32xy = 12*64 - 200 = 568, xy = 568/32 = 17.75. But under the condition x+y = 8, well known minimax property says that the product xy may have the maximum value of 4*4 = 16 at x = y = 8/2 = 4. In other words, under given conditions, the original system has no real solutions. So, the given system of equations does not have solutions in real numbers.. </pre> My other lessons on quadratic equations in this website are - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/Introduction-Into-Quadratics.lesson>Introduction into Quadratic Equations</A> - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/proof-of-quadratic-by-completing-the-square.lesson>PROOF of quadratic formula by completing the square</A> - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/Learning-by-examples-on-HOW-TO-complete-the-square.lesson>HOW TO complete the square - Learning by examples</A> - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/HOW-TO-solve-quadratic-equation-by-completing-the-square-Learning-by-examples.lesson>HOW TO solve quadratic equation by completing the square - Learning by examples</A> - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/Solving-quadratic-equations-without-quadratic-formula.lesson>Solving quadratic equations without quadratic formula</A> - 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