In AP SSC Class 10 Maths Chapter 5 Quadratic Equations, we study about quadratic equations and methods to find their roots. Read on this article to know about quadratic equations and problem sums related to them. Students can refer to the AP SSC 10th Class Maths Chapter 5 Quadratic Equations notes and solutions to prepare for the board exams and ace the Maths paper.
What is a Quadratic Equation?
An equation in the variable x of the form [latex]ax^{2}+bx+c=0[/latex], where a, b, c are real numbers and a ≠0 is known as a quadratic equation.  A few examples of quadratic equations are [latex]3x^{2}−5x+7=0[/latex], [latex]2-x^{2}+400=0[/latex].
The roots of the quadratic equation [latex]ax^{2}+bx+c=0[/latex] is given by [latex]\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}[/latex]. This formula is known as the quadratic formula.
- If [latex]b^{2}-4ac> 0[/latex], we get two distinct real roots as follows [latex]\frac{-b+\sqrt{b^{2}-4ac}}{2a},\frac{-b-\sqrt{b^{2}-4ac}}{2a}[/latex]
- If [latex]b^{2}-4ac=0[/latex], we get two equal real roots.
- If [latex]b^{2}-4ac< 0[/latex], the roots that we get are imaginary.
Since [latex]b^{2}-4ac[/latex] determines whether the quadratic equation [latex]ax^{2}+bx+c=0[/latex] has real roots or not, [latex]b^{2}-4ac[/latex] is called the discriminant of the quadratic equation.
Class 10 Maths Chapter 5 Quadratic Equations Questions
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- Find the nature of the roots of the following quadratic equations. Find the real roots if it exists.
Solution:
The equation is of the form [latex]ax^{2}+bx+c=0[/latex],where a = 2, b = 3 and c = -5
The discriminant is given by [latex]\Delta= b^{2}-4ac[/latex],
Substituting the value, we get
[latex]\Delta =(3)^{2}-(4\times 2\times (-5))=9+40=49[/latex]Now the roots of the equation are found as follows:
[latex]x=\frac{-b\pm \sqrt{\Delta }}{2a}[/latex]Substituting the values in the equation, we get
[latex]x=\frac{-3-7}{4}=-\frac{5}{2}, x=-\frac{5}{2}[/latex] [latex]x=\frac{-3+7}{4}=\frac{4}{4}, x=1[/latex]-
- The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48. Is the situation possible? If so, determine their present ages.
Solution:
Let us consider two friends A and B and let us consider A’s age to be x.
We know that the sum of ages of two friends is 20, i.e. x + age of B = 20 ⇒ age of B = 20 – x.
Four years ago, the product of their age was 48 which means that
[latex](x-4)\times (20-x)=48[/latex]Solving, we get
[latex]-x^{2}+20x-112 = 0[/latex]Now, this is in the form of [latex]ax^{2}+bx+c=0[/latex].
Let us find the value of the discriminant, by substituting the value of a,b and c in the
[latex]\Delta =b^{2}-4ac[/latex] [latex]\Delta =(-20)^{2}-(4\times 1\times 112) = -48[/latex]As [latex]D< 0[/latex], the equation has no real roots. Hence the situation is not possible.
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