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Attempt this one of many quizzes on factoring quadratics and sharpen your problem-solving skills. These free practice problems will help you understand the concept of factorization of quadratic equations better.

Factoring Quadratics Quiz 2

mathematics

 Find the values of x for the equation ${{\left( x+5 \right)}^{2}}-10=5x$ using the factorisation method.

 Convert the given equation in standard form $a{{x}^{2}}+bx+c=0$. Then, try to get the factors using the fact that the sum of factors must be b and product must be ac.

 $\text{Expanding and rearranging the terms in the equation, we get }\\ {{\left( x+5 \right)}^{2}}-10=5x \\\Rightarrow {{x}^{2}}+5x+15=0 \\ \text{To find two numbers such that their product is 15 and sum is 5.} \\ \text{15 has factors 1, 15 and 3, 5.} \\ \text{There is no possibility of getting the sum of two numbers as 5.} \\  \text{So, the method of factorisation is not possible here.} $

 A number is taken and seven times its square is found out. 20 times the number is deducted from it and 10 is added to it. To this, the number multiplied by 3 is deducted from it and another 4 is deducted. The final result obtained is 0. If the number taken was a whole number, then what was its value?

 Assume the number as any variable, follow the steps as mentioned in the question to form a quadratic equation. Solve it to get the number.

 $\text{Let us assume the number to be “n”.} \\ \text{Now, seven times the square of n is }7{{n}^{2}}. \\ \text{20 times of n }= 20n \\ \text{3 multiplied by n }= 3n \\ \text{Now, forming equation as per the operations performed on the number,} \\ \text{we have} \\ 7{{n}^{2}}-20n+10-3n-4=0 \\ \Rightarrow 7{{n}^{2}}-23n+6=0 \\ \text{Solving for “n”, we have} \\ \Rightarrow 7{{n}^{2}}-21n-2n+6=0 \\ \Rightarrow 7n\left( n-3 \right)+2\left( -n+3 \right)=0 \\ \Rightarrow 7n\left( n-3 \right)-2\left( n-3 \right)=0 \\ \Rightarrow \left( 7n-2 \right)\left( n-3 \right)=0 \\ \Rightarrow 7n-2=0\text{ and }n-3=0 \\ \therefore n=\dfrac{2}{7}\text{ and }n=3 \\ \text{Since n is a whole number, n = 3.} $

 Solve ${{x}^{2}}-11x+18=0$ . What is the speciality of the smallest root of this equation?

 Solve the given equation to obtain two values of $x$. Take the smallest root and then go through each option to check which one best suits it.

 $\text{Solving the equation, we have } {{x}^{2}}-11x+18=0 \\ \Rightarrow {{x}^{2}}-9x-2x+18=0 \\ \Rightarrow \left( x-2 \right)\left( x-9 \right)=0 \\ \therefore x=2,x=9 \\ \text{The smallest root is }x = 2.\\ \text{2 is an even number as well as a prime number.}$

 Find the roots of the quadratic equation $\dfrac{1}{y+1}-\dfrac{1}{4y-1}=2$.

 Solve by taking the LCM of the terms of the given equation. Simplify further to obtain a quadratic equation. Then get the values of $y$.

 $\text{Taking LCM and simplifying further, we have } \\ \dfrac{4y-1-y-1}{\left( y+1 \right)\left( 4y-1 \right)}=2 \\ \Rightarrow \dfrac{3y-2}{4{{y}^{2}}+3y-1}=2 \\ \Rightarrow 8{{y}^{2}}+3y=0 \\ \Rightarrow y\left( 8y+3 \right)=0 \\ \Rightarrow y=0,y=-\dfrac{3}{8} $

 A rangoli design is made in the shape of a circle. It has three concentric circles, the area of each circle is twice the area of the preceding one. If the difference between the sum of areas of circles and the sum of circumference of the circle is $21\pi $ cm, then find the radius of the first circle.

 Assume a variable for the radius of the first circle. Find the area of each circle using the given relation, then find the sum of areas and circumferences and equate their difference to $21\pi $.

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/1cfc3c09-399a-4ed2-9e90-30ab3f1062a63139281817385671492.png"/> $\\ \text{Taking the radius of the first circle as “r”cm.}\\ \text{Sum of areas of all circles }=\pi {{r}^{2}}+2\pi {{r}^{2}}+4\pi {{r}^{2}}\\ \text{Circumference of circles }=2\pi r+4\pi r+8\pi r \\ \text{Difference of areas and circumferences = }21\pi cm,\text{ so we have }\\ \Rightarrow \pi {{r}^{2}}+2\pi {{r}^{2}}+4\pi {{r}^{2}}-\left( 2\pi r+4\pi r+8\pi r \right)=21\pi  \\ \Rightarrow 7\pi {{r}^{2}}-14\pi r-21\pi =0 \\ \Rightarrow {{r}^{2}}-2r-3=0 \\ \Rightarrow {{r}^{2}}-3r+r-3=0 \\ \Rightarrow \left( r+1 \right)\left( r-3 \right)=0 \\ \therefore r=-1,3 $