Advertisement Remove all ads Advertisement Remove all ads Sum Solve the following equation: (x - 5)(x + 3) = (x - 7)(x + 4) Advertisement Remove all ads Solution(x - 5) (x + 3) = (x - 7)(x + 4) ⇒ x2 + 3x - 5x - 15 = x2 + 4x - 7x - 28 ⇒ -2x - 15 = - 3x - 28 ⇒ 3x - 2x = 15 - 28 ⇒ x = -13 Concept: Simple Linear Equations in One Variable Is there an error in this question or solution? Advertisement Remove all ads Chapter 14: Linear Equations in one Variable - Exercise 14 (A) [Page 165] Q 17Q 16Q 18 APPEARS INSelina Concise Mathematics Class 8 ICSE Chapter 14 Linear Equations in one Variable Advertisement Remove all ads a+b=-4 ab=-5 To solve the equation, factor x^{2}-4x-5 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved. a=-5 b=1 Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution. \left(x-5\right)\left(x+1\right) Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values. x=5 x=-1 To find equation solutions, solve x-5=0 and x+1=0. a+b=-4 ab=1\left(-5\right)=-5 To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-5. To find a and b, set up a system to be solved. a=-5 b=1 Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution. \left(x^{2}-5x\right)+\left(x-5\right) Rewrite x^{2}-4x-5 as \left(x^{2}-5x\right)+\left(x-5\right). x\left(x-5\right)+x-5 Factor out x in x^{2}-5x. \left(x-5\right)\left(x+1\right) Factor out common term x-5 by using distributive property. x=5 x=-1 To find equation solutions, solve x-5=0 and x+1=0. x^{2}-4x-5=0 All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction. x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-5\right)}}{2} This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -4 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}. x=\frac{-\left(-4\right)±\sqrt{16-4\left(-5\right)}}{2} Square -4. x=\frac{-\left(-4\right)±\sqrt{16+20}}{2} Multiply -4 times -5. x=\frac{-\left(-4\right)±\sqrt{36}}{2} Add 16 to 20. x=\frac{-\left(-4\right)±6}{2} Take the square root of 36. x=\frac{4±6}{2} The opposite of -4 is 4. x=\frac{10}{2} Now solve the equation x=\frac{4±6}{2} when ± is plus. Add 4 to 6. x=\frac{-2}{2} Now solve the equation x=\frac{4±6}{2} when ± is minus. Subtract 6 from 4. x=5 x=-1 The equation is now solved. x^{2}-4x-5=0 Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c. x^{2}-4x-5-\left(-5\right)=-\left(-5\right) Add 5 to both sides of the equation. x^{2}-4x=-\left(-5\right) Subtracting -5 from itself leaves 0. x^{2}-4x=5 Subtract -5 from 0. x^{2}-4x+\left(-2\right)^{2}=5+\left(-2\right)^{2} Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square. x^{2}-4x+4=5+4 Square -2.
\left(x-2\right)^{2}=9 Factor x^{2}-4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}. \sqrt{\left(x-2\right)^{2}}=\sqrt{9} Take the square root of both sides of the equation. x=5 x=-1 Add 2 to both sides of the equation. x ^ 2 -4x -5 = 0 Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0. r + s = 4 rs = -5 Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C r = 2 - u s = 2 + u Two numbers r and s sum up to 4 exactly when the average of the two numbers is \frac{1}{2}*4 = 2. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding:8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width:100%;max-width:700px' /></div> (2 - u) (2 + u) = -5 To solve for unknown quantity u, substitute these in the product equation rs = -5 4 - u^2 = -5 Simplify by expanding (a -b) (a + b) = a^2 – b^2 -u^2 = -5-4 = -9 Simplify the expression by subtracting 4 on both sides u^2 = 9 u = \pm\sqrt{9} = \pm 3 Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u r =2 - 3 = -1 s = 2 + 3 = 5 The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s. |